[Analyse des infiniments petits. English]. The method of fluxions both direct and inverse. 1730 by [L’Hôpital, Guillaume François Antoine de, Marquis de Sainte-Mesme].

[Analyse des infiniments petits. English). The method of fluxions both direct and inverse. 1730
by [L’Hôpital, Guillaume François Antoine de, Marquis de Sainte-Mesme].

Publication date 1730
Topics Language & Literature, Literary And Political Reviews, General Interest Periodicals–United Kingdom, Law, Philosophy & Religion, Fine & Performing Arts, Social Sciences, History, History–History of North And South America, Books, microfilm
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[Analyse des infiniments petits. English). The method of fluxions both direct and inverse. 1730..
Digitized from IA40312802-87.

Tak

METHOD.

FLUXIONS

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N DCC Xx.

PREFACE.

E Analyſis explain’d in the
BA following Work ſuppoſes the
common Analyſis, but is of a
very different Nature from it;
the latter being confin’d to Fi-
nite Quantities, whereas the
former extends to Infinity it-
ſelf. By means of this Analyſis we compare
the infinitely ſmall (Differences or) Parts of fi-
nite Magnitudes, and find their Ratio’s to each
other; and hereby likewiſe learn the Ratio’s
of finite Magnitudes, thoſe being in reality ſo
many infinitely great Magnitudes, in reſpect
of the other infinitely ſmall ones. This Ana-
lyſis may ever be ſaid to go beyond the Bounds
of Infinity itſelf; as not being confined to in-
finitely ſmall (Differences or) Parts, but diſco-
vering the Ratio’s of Differences of Differen-
ces, or of infinitely ſmall Parts of infinitely
ſmall Parts, and even the Ratio’s of infinitely
ſmall Parts of theſe again, without End. So
that it not only contains the Doctrine of Infi-
nites, but that of an Infinity of Infinites. It
is an Analyſis of this kind that can alone lead
us to the — of the true Nature and

Principles of Curves: For Curves being no o-
A 2 ther

PREFACE.

ther than Polygons, having an infinite Num-
ber of Sides, and their Differences ariſing al-
rogether from the different Angles which their
infinitely ſmall Sides make with each other, it
is the Doctrine of Infinites alone that muſt en-
able us rodetermine the Poſition of theſe Sides,
in order to get the Curvature formed by them;
and thence the Tangents, Perpendiculars, Points
of Inflexion and Retrogreſſion, reflected and
refracted Rays, c. of the Curves.

Polygons circumſcribed about or inſcribed
in Curves, whoſe Number of Sides infinitely
augmented till at laſt they coincide with the
Curves, have always been taken for the Curves
themſelves. But the Notion reſted here, with-
out farther Improvement, for many Ages: And
it was the Diſcovery of the Analyſis of Infi-
nites that firſt pointed out the vaſt Extent and
Fecundity of this Principle.

What the Ancients, and particularly Ar-
chimedes, have done herein, may claim our
Wonder; but then they have only conſi-
der’d a few Curves, and thoſe too, flight-
ly enough: They have left us little other than
a Courſe of particular Propoſitions, which give
no Indication of any uniform and conſiſtent
Method. Yer ought they by no means to be
reproached on this ſcore, conſidering the great
Force of Genius * they ſhew’d in penetrating
ſuch Obſcurities, and ſetting foot in a Land
utterly unknown before: If they did not tra-

*. Theugh I have twice or thrice read over Archimedes’s
Treatiſe of Spirals with the utmoſt Attention, to compre-
hend the Art employ’d in his ſubtle Demonſtrations relating
4% the Tangents of Spirals, yet cou’d J never riſe from him

toi thut ſome ſaſpician that I had not here taken the whole
Force of” the Demonſt ration, &. Bu LLIAL DUS Praf.
de Lineis Spiralibus. | |

vel

PREFACE.

vel far, nor took the moſt direct Way, yet
whatever Vieta * may imagine, they did not
deviate far; and the more difficult and thorny
the Routs they took, the more ſurprizing it is
they were not loſt. In a word, the Ancients
appear to have gone as far as their Time could
admit of, and done the fame that our later Ge-
nius’s would, if under the fame Circumſtan-
ces: And had the Ancients lived in ours, it is
reaſonable to believe, they would have had our
Views; this being a Conſequence of the natu-
ral Equality we find in Genius’s, and the ne-
ceſſary Time required for the Succeſſion of
Diſcoveries. *Tis therefore no Surprize the
Antients have not gone farther in this Affair;
but very ſtrange indeed that great Men, un-
doubtedly as great as any of the Ancients, ſhou’d
ſo long have fat down here, and with a kind
of ſuperſtitious Veneration contented them-
ſelves to read and comment the Works of An-
tiquity, without allowing themſelves any other
Uſe of their Faculties, than barely ſcrv’d them
as Followers, that durſt not think for them-
ſelves, or carry their Views beyond the Diſco-
veries of their Predeceſſors.

In this manner many employ’d themſelves; they
wrote; the Number of Books increaicd, yet no
farther Progreſs made; the Labours of many Ages
ſerving only to fill the World with obſequious
Comments and multiplied Tranſlations of O-
riginals, that in themſelves were often worth-
leſs. Such was the State of Mathematicks,
and principally of Philoſophy, till the ‘Time
of Deſcartes; who, at the Inſtigation of his

FSi ver? Archimedes, fa/laciter concluſit Euclides, Sc.
Suppl. Geom.

A 3 great

PREFACE.

great and commanding Genius, left the Anci-
ents, to follow the ſame Guide, Reaſon, that
had conducted them: And this happy Bravery,
tho’ treated as a Revolt, was follow’d by an
infinite Number of new and uſeful Views, both
in Philoſophy and Geometry. People now
begun to open their Eyes and think for
themſelves.

To keep to Mathematicks : Deſcartes
here began where the Antients left off,
viz. with the Solution of a Problem, at
which, according to Pappas, | in Colle. Ma-
them. Lib. 7. at the Beginning] all of them
ſtuck. *Tis well known to what a Pitch he
carried Algebra and Geometry, and hoy ealy,
by the Introduction of the former into the lat-
rer, he has render’d Solutions of innumerable
Problems, which no one before him could
maſter. But as he principally applied himſelf
to the Reſolutions of Equations, Curves were
by him confider’d no farther than as they might
aſſiſt him in finding their Roots: So that com-
mon Algebra being ſufficient for this, he did
not endeavour to And any other, except in che
Buſineſs of drawing Tangents to Curves, where
he has happily applied it; and the Way he
diſcover’d for that purpoſe appear’d to him ſo
excellent, that he did not ſcruple to ſay, © this
Problem * was the moſt uſeful and general,
not only that he then knew, but even that
he he ever had a Deſire to know in Geome-

try.” |

The Geometry of Deſcartes having brought
the Conſtruction of Problems by the Reſolu-
tions of Equations into great vogue, and gi-
ven a conſiderable Inſight into the Affair, the
major part of Geometricians now apply them-
(elves to ſtudy and improve it with their own

| | Diſco-

— — — — —äöäů— — — — 0

PRE F A c E.

Diſcoveries, which thus daily advance it to-
wards Perfection.
Monſieur Paſchal indeed directed his Views
uite another way: He examined Curves in
themſelves; and under the Form of Polygons,
found out the Lengths of ſome, the Spaces
contained under them, the Solids deſcribed by
thoſe Spaces, their Centres of Gravity, &c.
And from a bare Conſideration of their Ele-
ments, that is, their infinitely ſmall Parts, he

diſcovered ſome general Methods relating there-

to: Which are the more ſurprizing, as he ſeems
to have come at them without Algebra, by
the ſole Force of Imagination.

Soon after the Publication of Deſcartes’s
Method of Tangents, Monſieur De Fermat
diſcovered another, which Deſcartes himſelf
at length allows, (Seck. 71. Tom. 3.) to be
more {imple than his own on ſeveral Occa-
ſions. Yet this itſelf is not ſo fimple as Dr.
Barrow afterwards made it, from a cloſe Con-
ſideration of the Nature of Polygons, which
naturally repreſent to the Mind a little Triangle
cConſiſting of a Particle of a Curve, (contained
between two infinitely near Ordinates,) the
Difference of the correſpondent Abſciſs’s; and
this Triangle is ſimilar to that formed by the
Ordinate, Tangent, and Subtangent: So that
by one ſimple Analogy, this Method of Dr.
Barrow’s performs the Buſineſs, without the
Calculus required in the Method of Deſcartes
and De Fermat. FOE

Dr. Barrow reſted not here: He alſo invented
a kind of Calculus ſuitable to the Method,
(Lect. Geom. page 80.) tho! deficient as well as

that of Deſcartes, in clearing Equations of

Fractions and Surd Quantities.
–. – ——— The

vij

“05 ©

vii

Acta E-
rudit. Lipſ.
Ann. 1684.
p. 467.
See Com-
mercium
Epiſtoli-
cum.

PR EF ACE.

The Defect of this Method was ſupplied

by that of Mr. Leibnitz’s, * [ or rather the great
Kir Iſaac Newton. T] He 1 where Dr.
Barrow and others left off: His Calculus has
carried him into Countries hitherto unknown
and he has made Diſcoveries by it aſtoniſhing
to the greateſt Mathematicians of Europe. The
Meſſieurs Bernouli were the firſt who perceived
the Beauty of the Method; and haye carried
it ſuch a length, as by its means to ſurmount
Difficulties that were before thought inſu-
perable. |

This Calculus is of vaſt Extent; as being
ſuited ro mechanical Curves, as well as geo-
metrical ones. Radical Signs are no Incum-
brance at all in it, but ſometimes a Conveni-
ency. It extends to any number of indeter-
minate Quantities; and the Compariſon of in-
finitely ſmall Quantities of all kinds is perform-
ed by it with equal Facility. Whence ariſe
an Infinity of ſurprizing Diſcoveries with re-
gard to Tangents, as well Curves as right

Lines, to Problems de maximis & minimis,

Points of Inflection and Regreſſion, Evolutes,
Cauſticks by Reflection and Refraction, c.

as will appear in the Work itſelf.

I have divided it into ten Sections; the firſt
whereof. contains the Principles of the Me-
thod; the ſecond ſhews the uſe thereof, in
finding the Tangents of all kinds of Curves,
Jet what will be the number of indeterminate
Quantities in the Equation expreſſing their
Natures. ‘ Tho’ Mr. Craigs, indeed, (in Lib.
de Quadr. Figurar. Curvilin. part 2.) thinks it
only applicable to geometrical Curves. The
third ſhews the uſe of the Method in ſolving

all Problems de maximis && minimis. The

fourth

“ox

— —

—

PR EF AGE.
fourth determines the Points of Inflexion and
Retrogreſſion of Curves. The fifth ſhews
how to find the Evolutes of Monſieur Hugens
in all kinds of Curves. The ſixth and ſeventh
ſhew the Merhod of finding Cauſticks by Re-
flection and Refraction, whereof M. T/chirn-
hauſen is the Inventor. The eighth contains
the farther uſe of the Method in finding Points
in Curves that touch an infinite number of

Right Lines, or Curves of a given Poſition.

The ninth contains a Solution of ſome Pro-
blems that depend upon the foregoing Diſco-
veries: And the tenth exhibits a new way of
uſing the Calculus Differentialis, (or Method
of Fluxions) in geometrical Curves: From
whence 1s deduced the Method of Deſcartes
and Hudde, which is only applicable to ſuch
kind of Curves.

In the 24, zd, 4th, th, 6th, 7th, and 8:6
Sections, there are indeed only a few Propo-

ſitions: But then they are all General. And,

as it were, ſo many Methods, eaſily applica-
ble to any number of particular Propoſitions.
But I have only applied them in ſome ſelect

Examples, being perſuaded, that in Mathema-

ticks general Methods are heſt; and that the
Books containing Details, or particular Pro-
poſitions, miſemploy the Time both of the
Reader and Author. Whence I ſhould not
have added the Problems of the ‘9th Section,
if they were not curious and exceedingly ge-
neral. Thus the 107h Section likewiſe con-
tains nothing but Methods which the Calculus
Differentialis gives to the manner of Deſcartes
and Iludde, for drawing of Tangents. And if
theſe are leſs general, all the preceeding Me-
thods ſhew, that the Fault is not in our Cal-

culus

x PREFACE |
culus, but in the Manner of Deſcartes, where+

to it is confined.
On the other hand, there can be no better
Proof of the vaſt Uſe of our Calculus, than |

this great Variety of Methods, as compre-
hending the Whole of what Deſcartes and
Hudde have done in the Affair of Tangents.
And the univerſal Proof it gives us of the uſe
of arithmetical Progreſſions therein, leaves no
room to doubt of the Certainty of this laſt
Method. | | [
T intended to have added another Section,
to ſhew the ſurprizing uſe of this Calculus in
Phyſicks, and to what degree of Exactneſs it
may bring the ſame; as likewiſe the uſe there-
of in Mechanicks; But Sickneſs has prevented
me herein. However, I hope to effect it hereaf-
ter, and preſent it the Publick with Intereſt. And
indeed the W hole of the preſent Treatiſe is on-
ly the Firſt Part of the Calculus of M. Leib-
nuitz, or the Direct Method, wherein we de-
ſcend from Whole Magnitudes to their infi-
nitely ſmall Parts, of what kind ſoever. com-
paring them with each other, which is called
the Calculus Differentialis: But the other Part,
called the Calculus Integralis, (or Inverſe Me-
thod of Fluxions,) conſiſts in aſcending from
theſe infinitely ſmall Parts to the Magnitudes,
or W holes, whereof they are the Parts. This
Inverſe Method alſo I deſigned to publiſh; –
but Mr. Leibnitz’s having wrote to me, that he
was at work upon this Subject, in order for a
Treatiſe de Scientia Infiniti, I was unwilling to |
_ deprive the Publick of ſo fine a Piece, which
muſt needs contain whatever is curious in the
Inverſe Method of Tangents, Rectifications
of Curves, Quadratures, Inveſtigation of Su-
) perficies

e 8
— —
1

— ws.

PR E F ACE.

perficies of Solids, and their Solidities, Cen-
tres of Gravity, Sc. Neither would I ever
have publiſhed the preſent Treatiſe, had he
not intreated me to it by Letter; as likewiſe
becauſe I believed it might prove a neceſſar
Introduction to whatever ſhall hereafter be l.
covered on the Subject.

I muſt own my ſelf very much obliged to
the Labours of Meſſieurs Bernoulli, but parti-
cularly to thoſe of the preſent Profeſſor at
Groenengen, as having made free with their
Diſcoveries as well as thoſe of Mr. Leibnitz.
So that whatever they pleaſe to claim as their
own, I frankly return them.

I muſt here in juſtice own, (as Mr. Leib-
nitz himſelf has done, in Journal des Sgavans

for Auguſt, 1694.) that the learned Sir //aac |
Newton likewiſe diſcover’d ſomething like the
_ Calculus Differentialis, as appears by his excel-
lent Principia, publiſhed firſt in the Year 1687.

which almoſt wholly depends upon the Uſe of
the ſaid Calculus. But the Method of Mr. Leib-

_ #itz’sis much more eaſy and expeditious, on ac-

count of the Notation he uſes, not to men-

tion the wonderful Aſſiſtance it affords on ma-
ny Occaſions.

When this Treatiſe was nearly printed off,

Mr. Nieuwentiit’s Performance happened to

come to hand; the Title whereof being Ana-
Iyfis Inſinitorum, gave me the Curioſity of run-
ning it over; upon which I found it very dif-
ferent from mine: For the Author not only u-
ſes a Notation different from Mr. Leibnitz’s,
but abſolutely rejects ſecond, third, &c. Dif-
ferences (or Fluxions;) and as the greater Part
of my Book is built upon that Foundation, I
ſhould have thought my ſelf obliged ro anſ ys

PREFACE
his Objections, and ſhew their Inſufficiency,
if Mr. Leibnitz’s had not already fully done it
to my hands, in the Acta Eruditorum An. 1697.

p. 310 and 369.

To conclude: The two Poſtulata or Sup-
ſitions laid down at the Entrance of this
ork as the ſole Foundation thereof, ap

to me ſo ſelf evident, as not to leave the leaſt

Scruple about their Truth and Certainty on
the Mind of an attentive Reader. They might

however have been demonſtrated after the Man-

ner of the Antients, if I had not intended to

be ſhort in things already known, and apply

my ſelf principally to ſuch as are new.

TH E

TH ANIL A TING

TO THE

READER.

WR 7 7s needleſs for me to ſay any
bing in Commendation of the
we /4thor’s Piece, the Character
—— whereof is ſo well eftabliſh’d :
Aid had he wrote likewiſe a ſe-
SEE cond Part, or the Inverſe Me-

thod of Fluxions, or Calculus
Invogralis „ (as Foreigners call it) the Whole
would, no doubt, have been an excellent Introdu-
ion to this admirable Doftrine.

My Deſign at firſt was to have publiſhed the
Tranſlation alone; but conſidering that it would
be imper fect without the Inverſe Method, I there-
fore have ſupplied it, and ſhewn its UL 72 in the
Quadratures of Curv’d-lin’d Spaces, the Nectiſi-
cations of Curves, Cubations of Solids, Quanti-

ties of their Superficies, and the Inveſtigations of

Centres of Gravity and Percuſſion, the Whole
concluding with a few miſcellaneous Problems.

The firſt Section, being a general Introduction,
handles the Doctrine of infinite Series, which is
ſo weceſſary to what follows, that without it, a

Perſon

XIV

The TRANSLATOR

Perſon can have but a very ſlender Skill in the
Inverſe Method of Fluxions.

In the ſecond Section, concerning the finding of
Fluents, you have not only the uſual Ways of
performing the Buſineſs, (as well when they can-

nut be had in finite Terms, as otherwiſe,) but

likewiſe the Manner of wing the Tables of the
late learned Mr. Cotes for that end: By which,

and the Tables of Logarithms and natural Sines,

elegant Expreſſions of Fluents may be had,
without the Labour of throwing Duantities into
Series, which in many Caſes is not a little emba-
raſſing. | I

In the ſucceeding Soctions you have a Number
of ſelect Examples, whoſe Solutions will more di-
vert than trouble, and at the ſame time ſufficient-
ly inftrutt the induſtrious Learner. . Among theſe
you have the analytical Proceſſes of all the Pro-
blems of Mr. Cotes, about Quadratures, Refti-
fications, Cubations, and the. Duantities of the
Superficies of round Solids, (the bare Conſtructi-
ons whereof he has given us in his admirable Trea-

iſe, entituled, Harmonia Menſurarum, ) from

whence we gain the neat and elegant Conſtructions
he has given in the aforeſaid Treatiſe.

In the Seftion concerning the Inveſtigations of
the Centres of Gravity, you will find a Mean ob-
ſerv’d in the Choice of Examples, both as to

Number and Facility of Solution: You have here

alſo the Determination of the Centres of Gravity

of Hyperbolick and Elliptick Spaces in the Mea-

after Mr. Cotes’s Way.

ſures of Ratio’s and Angles. In the next Sefti-
on you have nothing new : Thoſe that have a mind
to try, may perhaps get not inelegant Conſtructi-
ons of Fluents, determining the Centres of Per-
cuſſion of ‘ Hyperbolick and Elliptick Spaces, &c.
The

tothe READER.

The firſt four Problems at the End may ſerve
to give the Learner a Taſte of the Inverſe Me-
thod of Tangents, about which I might have been
more full, but theſe alone will teach the Manner
of ſolving others of the like nature. And it is
for this Reaſon 1 have given alſo ſome phyſical
Problems, which may direct him to maſter thoſe
of a more difficult Kind, whenever they occur,
As the illuſtrious Author has omitted the Ex-
ponential Calculus, or Manner of – finding the
Fluxions of Exponential Quantities, ſuch as x
Da, X = V, Sc. where the Index s of the
variable Quantities are alſo variable; thinking,
as 1 ſuppoſe, this Branch of Doctrine to be of ve-
ry little or no uſe, ſo I have been filent in this
Matter alſo, which it is much better to be, than
take up the Reader’s Time in learning what is on-
ly mere Speculation. Thus you have a ſummary
View of what is contained in the Appendix. Once
for all; that I may not be thought a Plagiary, 1

freely own, many things are here taken from Sir

Iſaac Newton, c.

In a word, I am of opinion that the Tranſla-
tian, together with the Appendix or ſecond Part,
may ſerve very well for an Introduction to the
Doctrine of Fluxions, and will ſo far qualify the

Learner, as to render him capable of underſtand-
ing with moderate Eaſe the more ſublime Diſco-
wveries in Mechanicks, Phyſicks, &c. that have
been found by this Art. Such a York becomes
tbe more neceſſary, becauſe there are but two Eng-
liſh Treatiſes on the Subject, (that I know of)
the one being Hays’s Introduction to Mathema-
tical Philoſophy, and the other, Ditton’s Inſti-
tution of Fluxions; the former of which is by
far too prolix for Learners, abounding in general
Rules and Examples, many of which will rather

confound

ij he TRANSLATOR

confound than inſtruct him; and is at the ſame
time deficient in things that would not a little for-
ward him; not to mention the Book’s being out of
Print, and very likely for ever ſo to continue.
On the other hand, Mr. Ditton’s Book is by
much too ſparing in Examples of the Uſes of the
Methods Thoſe few he has given us being by no
means fit for Beginners. He is alſo too redundant
at his firſt ſetting out, in the Explanation of the
Definition of Fluxions, as Sir Iſaac Newton has

: Which, tho” it be true and exatt, it is next to

impoſſible for one who has not been converſant a-
bout Infinites to apprehend it. That of our Au-
‘ thor is much eafier, tho leſs Geometrical, whe
calls a Differential (or Fluxion) the infinitely
ſmall Part of a Magnitude z not deterring his
Readers at firft from proceeding, by dwelling long
on the Explication of an intricate Definition, but
comes immediately to the Algorithm, or Arithme-
rick of the Art; and thence to plain Examples
f Solutions by it. Lo Nee 9
But I would not here be thought in any wiſe
to leſſen the Value of Sir Iſaac Newton’s Defi-
nition: When the Learner has made ſome Pro-
7255 I would have him then make himſelf Ma-
fer of it. F | f
‘* Thus much for the Wort; preparatory to the
due Peruſal whereof, it may not, be amiſs to give
Some Notion of the general Nature and Origin of
Fluxions, according to the Senſe o the great Au-
thor and Inventor thereof, Sir Iſaac Newton.
Inn order to this, we are to confider Quantities
not as made up of very ſmall Parts, but as de-
ſcribed by a continued Motion. For Example, a
Line is deſcribed, not by the Appofition of little
Lines or Parts, but by the continual Motion of a
Point, A Surface or plain Superficies is de rh

to the READER XVI}
ed by the Motion of a Line, (nt according to
its own Direction.) And a Solid, by the comi-
nual Motion of a Superficies, (not according to
its own Direction.) |

Now the Velocities of the Increaſes or Incro-
ments of Magnitudes thus moving in very ſmall
equal Particles of Time, at the firſt Inſtant of
the Generation of thoſe Increments are called
Fluxions, and thoſe Magnitudes Fluents, or
Flowing Quantities. | |

_ Theſe Fluxions are nearly proportional to the
Jncrements of the Fluents or Flowing Quanti-
ties, generated in very ſmall equal Parts of
Time ; but accurately as the Velocities wherewith
they ariſe and begin to be generated; that is, they
are thoſe very Velocities; as was ſaid before.
The reaſon why the Increments generated in ſmall
equal Parts of Time, are not exactly proportional
to the Velocities wherewith they are generated, is,
becauſe thoſe Velocities are not conſtantly equable,
which, if they were, the little Spaces deſcribed
in equal Particles of Time, would then be exatly
as the Velocities, (which is an allowed Principle
in Mechanicks;) but the Velocities are mutable,
or accelerated continually, and fo the ſaid little
Spaces deſcribed, cannot be conſtantly proportional
to the Yelocities wherewith they were firſt deſcri-
bed, or proportional to any one of thoſe mutable
Pelocities.

Net if the Particles of Time be taken very ſmall,
and the Acceleration be ſo likewiſe, Fluxions may
be taken as „ R to thoſe Increments, and
fo the [aid Increments, may repreſent Fluxions in

|
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1
|

|
:
1
1

all Operations; the Reſults of which will be as ex-
alt, as if they had been determined from the Ve-
locities of the Motions wherewith the ſaid Incre-
ments begin to be generated. But to ſcrupulous

See Com-

mercium

Xviij The TRANSLATOR

Perſons the Proceſſes will not be allowed ſo geo-
metrical. Neither indeed is it ſo ſatisfactory to
throw out Quantities from an Equation, on ac-
count of their being infinitely leſs than others, (as
all who uſe an Increment for a Fluxion do,) as to
reject them, becauſe they really vaniſh and be-
coms equal to nothing, as Sir Iſaac Newto
does. |
Upon this latter Foundation is built the Calcu-
lus Differentialis, fir/# publiſbed by My. Leib-

nitz, in the Near 1684.3 having been ſince fol-

lowed by almoſt all the Foreigners : Who repreſent
the firſt Increment, or Differential, (as they call
it,) by the Letter d, the ſecond by dd, the third
by ddd, c.; the Fluents, or Flowing Quanti-
ties, being called Integrals. But fince this Me-
thod in the Practice thereof, does not differ from
that of Fluxions, and an Increment or Differential
may be taken for a Fluxion; out of regard to
Sir Iſaac Newton, who invented the j –

fore the Year 1669, 1 have altered the Notati

Epiſtoli- of our Author, and inftead of d, dd, d, Cc.

cum.

put his Notation, viz. x, &, Xx, Cc. or ſome o-
ther of the final Letters of the Alphabet, point-

“ed thus, and called the infinitely ſmall Increment,

or Differential of a Magnitude, the Fluxion of

, | | |
So far of the general Nature of Fluxions : I

hall conclude with propoſing the Solutions of the
Fo following Problems to ſuch as are capable.

I. The Latitude of a Place and the Day of
the Year being given, to find the Time of
that Day the Heat of the Sun ſhall be the

_ greateſt, admitting it be as any Power of

the Sun’s Duration above the Horizon, and as

any Power of the Sine of his Altitude.
Lag | II. To

to the REA D E R. XxIix

II. To find the Nature of one of the
Superficies of a Wall of uniform Matter, of
a given Height, Length, Breadth at Bottom
(viz. the Ground) and Top, the other oppoſite
upright Superficies being a right-angled Pa-

rallelogram, as alſo the Section at the Ground
and the Top, that ſhall ſtand firmer, or be
leaſt diſpoſed to fall, when the Wind blows
directly againſt the upright plain Superficies
of it, than when the Superficies (to be found)
is of any other Nature; allowing the Particles
of the Air to be equally diſtant from each o-
ther, of equal Magnitudes, and all to move
with equal Velocities parallel to each other;
and the Wall to break off at the Surface of
the Ground, with a ſufficient Force of Wind.

Note, The three Sides of the infinitely ſmall 1
Triangle, mentioned in Page vij. aforegoing, |
are an infinitely ſmall Part of the Curve,
the Difference between two infinitely near
Ordinates, and that of two infinitely near

Abſciſs’s,

TABLE

22 — — — n
att hs 2 FW >!
m _— pan. —
— — „4 mw. wore TY —_—— As Es * e
3 ˙˙ ˖ EY : = , X —— X & a N
3 1 2 * 4 —— s . — * 5 „ me” Wy .
ws 2 5 1 * Prey 2 r “ung . 1 * N – » SE — ” —
En, EN RO IDELY — arm. — 2 — LEES EI:
n 2 oy 5 A a ou me MO RY 8 ; “— CA VEAL Bf; . d 88 . 4 —
— — Nw SI a. 5 „ 12 * — : F . p — —
1 Cong, de p
2

ONTE
. 0 – N * *
0 . * J X | 1.” “\ i Ge Y * i *

” t, £ o – i

— —
—

— — — —

SECTION I.

SG Ss WAY hk

8 K 0 I. 1.
The Uſe of Fluxions in drawing Ti . to all
) aaa Curves, n

|
SECT 117. 1

-Of the Uſe of * in ; finding the greateſ
and- leaf} Ordinates in a Curve, to which t
Solution of Problems gs, maximis and mini- =

mis may be reduced,” 444 |

Aer | |

of the Uſe of Fluxions in finding of the Points
of Inſlexion and Retrogreſſion of Cn, 73
T.

” 8 ” – a>
o A of

Tho U Uk of Fluxions in Geemetrical Corps after |

The r

2 bs

SECT. V. 7
The U, 7 of Fluxions in the Doftrine 7 Evolute
— Imoolute ”_ _Fuge 94
8 E c T. VI. 23
The Uſe of Fluxions in finding of Cauſticks by
by Refiexion, | 137
SECT. VI. Ji
T, 1 Uſe of Flixions in frnding 7 Canfick; by
ay raction, 15
1 SECT. VII.

The Uſe of Flaxions in finding the Points of

_ Curves touching an infinite Number of Curves,

or right Lines given in Poſition, 173
SEC M

The Solution of ſome Problems depending upon the
Methods TOs 191
8 E 0 7. x.

a new manner; from whence is deduced the

Method of Deſcartes and Hudde, 2216

The CoNTERNTS.

In the APPENDIX.

SECT. I.

JP the Reduction of fraftional Expreſſions and
ſurd Quantities to infinite Series, Page 1

SECT. Il
Of finding the Fluents or flowing Quantities of
given fluxionary Expreſſions, 21
SECT. III.
Uſe of the Inverſe Methol in dhe n of
Curve- lin d Nan, 35
SECT. IV.

Uſe of Huxions in the Reftification of Curves, 89

ETST–V:;

Of the Uſe of Fluxions in the Cubation of So-
lids, and in the Quadrature of their Surfaces,

| 116
| SECT. VI.
Of the Uſe of Fluxions in finding the Centres of
Gravity of Figures, 153
SECT. VII.
Of the Uſe of Fluxions in finding the Centres of

— of Figures, 169
| SECT.

:
1
ii
1
|

eren

Of the „ Roſalavie of ſome * Problems
a Page 181

A }

A TREA-

TRE AT IS E

O F
FLUXIONS.
“EW >, ine
SECT. L

Of finding the Fluxions 0 ff Quantities.
DEPINITION I.

— Ariable Ouantities are thoſe that
N continually increaſe or decreaſe;
20″ J) and conſtant or ſanding Quantities,
105 al — 5 are thoſe that continue the ſame
while others vary. As the Ordi-

nates and Abſeiſſes of a Parabola are variable

Quantities, but the Parameter is a conſtant « or
ſtanding Quantity. |

DE TIN. II.

1 HE infinitely ſmall Part whereby a varia-
ble Quantity is continually increaſed or
decreas’d, is called the * Fluxion of that Quan-

tity. B | F
| or
dee the Tranſlator’s Preface.

4 |

Frsc. 1.

For Example

A Treatiſe

Line AMB, whoſe Axis or Diameter is the
Line AC, and let the right Line P M be an
Ordinate, and the right Line pm another in-
finitely near to the former. EI

Now if you draw the right Line MR pa-
rallel to AC, and the Chords 4M, Am; and
about the Centre A with the Diſtance AM,
you deſcribe the ſmall circular Arch MS:
then ſhall Pp be the Fluxion of PA; Rm
the Fluxion of Pm; & m the Fluxion of AM;
and Mm the Fluxion of the Arch AM. In
like manner, the little Triangle M 4m, whoſe
Baſe is the Arch Mm, ſhall be the Fluxion
of the Segment 4M; and the ſmall Space
MPpm, will be the Fluxion of the Space
contained under the right Lines AP, PM,
and the Arch AM.

CoROLLARY.

1 T is manifeſt, that the Fluxion of a con-
ſtant Quantity, (which is always one of
the Initial Letters a, 6, c, Sc. of the Alpha-
bet) is o: or (which is all one) that conſtant
Quantities have no Fluxions.

5
SCHOLIUM.

PHE Fluxion of a variable Quantity expreſ-

‘ fed by a ſingle Letter, which is commonly
one of the later Letters of the Alphabet, is re-
preſented by the ſame Letter with a Dot or Full-
point over it : as the Fluxion of x is x, that of
V 75 y, that of 2.is x, and that of is u. And

, you call the variable Quantities AP, x;
PM, y; AM, z; the Arch AM, uz the mixt-

2 lin’d

und Space AP M,sz and the Segme

of F LUXIONS.

nt AM, t:

then will x expreſs the Value of Pp, y the 2
lue of Rm, 2 the Value of Sm, u the Value of
the ſinall Arch Mm, s the Value of the little
Space MP pm, and t the Value of rhe ſmall

mixt-lin d Triangle M Am.

PosrvLars I.

G RANT that two Quantities, whoſe
Difference 1s an infinitely ſmall Quan-
tity, may be taken (or uſed) indifferently for

each other: or (which is the ſame thing) that |
a Quantity, which is increaſed or > Ave
only by an infinitely ſmall Quantity, may be

conſider’d as remaining the fame.
For Example: Grant that Ap may be ta-

ken for AP; pm for PM; the Space Apm
for APM; the ſmall Space MP pm tor the

ſmall Rectangle MPpR; the ſmall Sector
AMS for the ſmall Triangle 4 Mm; the An-

gle p Am for the Angle PAM, &c.

PosTtuUuLATE II.

3-Q RANT that a Curve Line may be con-

ſider d, as the Aſſemblage * an infinite
Number of infinitely ſmall right Lines: or
(which is the ſame thing) as a Poly on of an infi-
nite Number of Sides, each of an infinitely
ſmall Length, which Dede the Curvature
of the Line by the Angles they make with

each other.

For Example: Grant that the Part Mm of
the Curve, and the Circular Arch MS, may
be conſider d as ſtraight Lines, on account of

their 5 2 ＋ . that the little

Trian-

0 Art. 1.

Triangle »SM may be looked upon as a
right-ln’d Tang Ne n

PR OP. 1.

connected together with the Signs + and

4 7 find the Fluxions of ſimple Quantities

Iris required to find the Fluxion of a+ x +y
. If you ſuppoſe x to increaſe by an infinite-
ly ſmall Part, viz. till it becomes x + x

then will y become y+ y; and 2, 2 2: and

the conſtant Quantity à will * ſtill be the
ſame a. So that the given Quantity a + x +

will become a + x + x +5 + 9 —2—2z3
and the Fluxion of it (which will be had in
raking it from this laſt Expreſſion) will be
x +3 — Z; and ſo of others. From whence
we have the following

Rl. E I.

For finding the Fluxions of ſimple Quantities con-
netted together with the Signs + and —.
Find the Fluxion of each Term of the

Quantity propoſed; which connected together

by the ſame reſpective Signs will give another

Quantity, which will be the Fluxion of tha

given. | |

P R O p. II.

ſ- find the Fluxions of the Produtt of ſe-
1 ® wveral Duantities multiplied, or drawn in-
to each other. SY

The Fluxion of #y is yx *: for y

becomes y + 5, when x becomes x + *; and

therefore æj then becomes xy +-y AFT.
| 0 ** T Which

y
d <7
p.
h

$2+-+x25) Plus the Prod

of FLUXIONS.
Which is the Product of x & into y+5,
and the Fluxion thereof will be yx + x +
55, that is, * yx+ xy: becauſe + is a Quan-
tity infinitely ſmall, in reſpect of the other

Art. 2.

Terms y and xy: For if, for Example, you

divide yx and xj by x, we ſhall have the
Quotients y and y, the latter of which is infi-
nitely leſs than the former.

Whence it follows, that the Fluxion of the
Product of two Duantities, is equal to the Pro-
duct of the Fluxion of the firſt of thoſe Quanti-
ties in the ſecond Plus the Product of the Fluxion
of the ſecond into the firſt. |

The Fluxion of xyz is yz4# + x25 |

E. For by conſidering the Product xy as
one Quantity, we muſt (as has been before
ſhewn) take the Product of the Fluxion y z +
x y into the ſecond Quantity 2, (which will be

o& of the Fluxion
z of the ſecond Quantity z into the firſt
Quantity xy (which is xz 😉 and therefore
the Fluxion of xyz will be yz++x25 + x52.

The Fluxion of xyz# is #39zx+uxzy
A EE TEA. Which is proved as in the
Caſe aforegoing, by conſidering the Product
y as one Quantity. Underſtand the fame

of others. Hence we have this

mes
For the Fluxions of Quantities of ſeveral Di-

menſz0ns,

The Fluxion of a Quantity of ſeveral Di-
menſions, or (which is the ſame) of the Pro-
duct of ſeveral Quantities multiply’d into one
another, is equal to – Sum of the Products

3 —

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Art. 5.

Square « of the

A Treatiſe
of the Fluxion of each of thoſe Quantities

into the Product of the others.
So the Fluxion of ax is xo + ax, that is,

ax; and that of a+ N Ku. is ee
„

P R o v. III.

1. find the Fluxion of a Frath ion.

The Fluxion of © — is E=2. For fup- |

poling © TK, Ro will x we 55 and

ſince theſe variable Quantities x and E ought
always to be equal to one another, whether
they increaſe or decreaſe, it follows that their
Fluxions, that is, their 4 ſmall Incre-

ments or Decrements ſhall 9 —.— be equal.

And therefore will * x =y2 +2, and 2 =
— r ſubſtituting z for its Va-

wes “Which was to be done, Ge. Hence
we have thi 22

1 III.

For the Fluxions of Quantities divided 50 one
another, or Fractions.

The Fluxion of a Fraction is equal to the
Product of the Fluxion of the Numerator in-
to the Nenominator Minus the Product of
che Fluxion of the Denominator into the Nu-
merator: the whole being * by the
minator.

of FLUXIONS.

As the Fluxion of ü and that of

| * 1 |

Lewd 7 1c: 5 2.15
Pao nr N.

7 O find the Fluxion of any whole or bro-
ken Powers of a variable Quantity.

Before we lay down a general Rule for
finding the Fluxions of all forts of Powers,

we mult explain the Analogy that there is be-

tween their Exponents or Indices. In order
to this, if there be a Geometrical Progreſſion,
Having 1 for its firſt Term, and the ſecond
Term be any Quantity x; and if the Indices
or Exponent be orderly ſet under them, it is
plain that the faid Exponents will form an
Arithmetical Progreſhon.

Prog. Geom. 1, x, xx, x, xt, x, x, x”, Kc.
Prog. Arithm. o, 1, 2, Fo 47 ſy 6, 7, &c.
And if the Geometrical Progreſſion be con-
tinued downwards from Unity, and the A-
rithmetical Progreſſion downwards from 0,
the Terms of this laſt ſhall be the Exponents
of thoſe to which they anſwer in the other;

as — I is the Exponent of 5 — 2 of —»
Se. *

| bin $2 ont. Þ. A+

Prog. Geom. #, 1, 2 —» 35 = Kc.
Prog. Arith. 1,0, 1, —2,—3,—4, &c.
But if ſome new Term be brought into the
Geometrical Progreſſion, we muſt find an an-
ſwerable Arithmetical one for the Exponent

of it. , L £38] |
B 4 As

A Treatiſe ©
6.5 if the Term be „x, its Exponent on
1: 3/x, will have for its — Vx, 4

3 — 7.
WE. . n = 2 = =<, 4
80 that theſe Beeten Vr and x, V and
F
„ V/! and x

„ and Ke. g
nify the fame thing.

Progr. Geom. 1 V, A 1, Vest Var; F., Ve * Tm hv.
rg == RY * 1. bo +, 1. o, J, 4, « x,

= A — * 1 1

Prog. Geom. — 2

Prog. Ar, — 1 n =p „
Where you may obſerve, that as / x is
a Geometrical Mean between 1 and x, ; is
an Arichmetical Mean between their Expo-
nents o and 1. In like manner, as / x
is the firſt of two mean Proportionals
tween 1 and .x 1 is the firſt of two A-
rithmetical ber wien their Expo-
nents o and 1: Underſtand the ſame of — 4
Now from the Nature of theſe two Progreſ-

ſions it follows,

That the Sum of the 3 of any
two Terms of the Geometrical Progreſſion,

is the Exponent of the Product of theſe wo
Terms drawn into each other, as * +3 or x

ia the Produt᷑t _ into a., and a +105
is the Product of x* into xt, and x e

„ js the Product of x “* into & , Ec.
In! like F Ig 7 or & : is the Product

WES 50

it;

of FLUXIONS.
at, and x++2* or e is the Product of **
into into x*, that is, the Cube of it, and

Square, Cube, c. of that Term; and con-
ſequently that the , 3, Cc. of the Exponent
of any Term of the Geometrical Progreſſion,
ſhall be the Exponent of the Square Root,
Cube Root, c. of that Term. ;

That the Difference of the Exponents
of any two Terms of the Geometrical Pro-
greſſion, ſhall be the Exponent of the Quoti-
ent ariſing by the Diviſion of one of thoſe

Terms by the ather, as fI= ** ſhall be
the Exponent of the Quotient of x* divided

by £7, and SEE =, hall be che Ex-
ponent of the Quotient of the Diviſion of

7. divided by 1 * whence you ſee that it
is the ſame thing to multiply x by «© A as

to divide f by ad: and ſo of others. This
being well underſtood, there may happen two
Caſes in finding the Fluxions of Powers.

I. When the Power is a whole one, that
is, when its Exponent is a whole Number.
Now the Fluxion of xx is 2 xx, of x* is 3x»x,
of xd is 4x*x, &c. for ſince the Square of x
is only the Product of x into x, the Fluxion
thereof ſhall be & x x + x 5, that is, 2 * 4. In
like manner, fince the Cube of + is only the
Product of # into x into x, the Fluxion of ar

PT — –

2 –

Ari. 5.

|
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|
|

: ® fre. f.

A T1 reatiſe.

will be x * xx& * &, that is, 3xxX;
and ſince it is the fame of any other of theſe
Powers whatſoever, it follows, that if m re-
preſents any whole Number, the Fluxion of

„will be ax2—”” .

If the Exponent be negative, the Fluxion
of x” of of >= ſhall e r

When the Power is broken, bd 7

when the Exponent i is a Fraction. It is requie’d

to find the Fluxion of vo or & (be-

ing any Fraction ) Suppoſe * z, and by rai-

ſing both Sides of the pages to the 2

u, we ſhall’ have K =-, and finding

Fluxions of both Sides, (as in the firſt 8400

then will 1 =-, and 2 —

e 26 e, by

ſubſtituting 1 F for its Value n2″ *
WR the A La be I + the Fluxion of

221

7 5 WP, „5
— 2 K .

5 dor of ſhall be — 1

. Mr vel Fromhencewehavothe

following general |

Fl Lei ie SEWS de
= the Fluxions of all Kinds of Powers. 1

‘The Fluxion of any Power (whole or bro-

; ken) of a variable Ys is _ to the

Product

of FLUX ION.
Product of the Exponent of that Power made
by that ſame Quantity, raiſed to a Power leſ-
ſen’d by 1, and multiply’d by its Fluxion.
As if m expreſſes any whole Number or
Fraction poſitive or negative, and x any varia-
ble Quantity, the F luxion of * I be al-

ways mx .
EXAMPLES.

The Fluxion of t the | Cube of ay—zxx,
that is, of x, is zxay xx xa —2XX
= 3a” yy) —644×83) enn
A GA, 110i.

The Fluxion of N uf 77457, N
is FY inn or
ei * bt
2
The Fluxion of ZE or of == *

is ix 7 Fax) re or
a 34%9) +

2/Faxyy –

The Fluxion of; 222 or of Pas rr

1 1 ITEN * FEES, 2s

and+2xx-
3 Ver er:
k —
The Flxionof Var NN + Vibe orof
ar- Tar- E, D is D

—Tx 88 2×5 2

T Ly at baxyy or

4 * ＋ 2 x
TT
F ayyX be 24xyy
ear x 2a I xx bo vi Þ any

Laſh,

U
—

A Treatiſe

Laſtly, the Fluxion of QEED — accord-
| ha .

ic, ng to this Rate Tha KL a”
ax+zxx —_ x20),
| 3 2 D i ax TH .

_— ——

—

SCHOLIUM.-

Hur we are to obſerve, that in find-
: ing the Fluxions of Quantities, we have
hitherto ſuppoſed one of the variable Quanti-

ties x as increaſing; while the others y, z, &c,
do ſo likewiſe: That is, when the be-
come x + x, the ) and 2 Cc. become y+y,
z ＋ 2, Cc. So that if it ſhould happen that
ſome of them do decreaſe, while others in-

creaſe, then the Fluxions of the former muſt

be look’d upon as negative Quantities, with
reſpect to the Fluxions of the latter increaſing
Quantities z and conſequently the Signs of the
Terms affected with the Fluxions oFtheſe de-

| creaſing Quantities, muſt be changed.

To make this plain, let „ increaſe while
y and 2 decreaſe; that is, when x becomes
x +, let y and z become y ) and 2 — 2:

Firſt, get the Fluxion-x y z TZ + 35 2 &z
Oy Art.” 5.) and change the Signs of the Terms

ed with y and z. And this Expreſſion
thus alter’d, vis. yz x—x y & 2 3, Hall be
the Fluxion ſought, _

SECT.

of FLUX10NS. I3

SECT. I.

Of the Uſe of Fluxions in drawing
Tangents to all kinds of Curve Lines.

DEFINITION,

F one of the ſmall Sides M m of a Poly- Hg. 2.
gon, whereof a Curve Line conſiſts *, be Art. 3.
continued out; the ſaid ſmall Side thus conti-
nued, is a Tangent to the Curve in the Point
M or m.

P R O p. I.

9.J7 is required to draw a Tangent MT from Egg. z.
a given Point M, in a Curve Line A M,
whoſe Nature is expreſſed by any Equation, re-
preſenting the relation of an Abſciſs AP to its
Correſpondent ordinate PM.

Draw the Ordinate M P, and conceive the
right Line MT meeting the Diameter in the
Point 7, to be the Tangent . more-
over, let m p be an Ordinate infinitely near
M P, and draw the ſmall right Line M R
parallel to 4 P. Now call the N Lines
AJP, «„; PM, y; and then will Pp or M R
be = x, and Rm =y, and the ſimilar Tri-
angles m RM, MP 7, will give this Propor-
tion m R (3): RM (A) :: N (9): 77

— Now by Means of the Fluxion of
the

F 1c. 4.

Art. 8.

A Treatiſe

the given Equation, we can get a Value of x,
expreſſed in Terms that will be all affected
with j; which being multiplied by , and di-
vided by y, will give us the Value of the Sub-
tangent P T in known Terms, freed from
Fluxions; by which the ſought Tangent MT

may be drawn.

SCHOLIUM.

HEN the Point T falls on the con-
trary Side of the Point A, where the

begin, with reſpect to P, then it is plain,
that while x increaſes, y does decreaſe; and

conſequently, the Signs of all the Terms of
the Fluxion of the a Equation affected
with 5, muſt be * changed: Otherwiſe, the

Value of & in j will be negative; and there-
fore that of P T GE . Notwithſtanding,

to avoid Trouble, * U be beſt to get the
Fluxion of the given Equation, according to
the Rules laid down in $2. I. without any
Alteration: For, if at the End of the Ope-
ration, it happens. that the Value of PT be
poſitive, then the Point 7 muſt be aſſumed on

the ſame Side the Vertex A of the Diameter,

FC. z.

as was ſuppoſed in the Operation; but if it be

negative, it muſt be taken on the other Side
the Point 4. All this will be plain by the
following Examples. | 18

Ex AMP IL E I.

II. 1. I I expreſſes the Relation of

AP to PM, the Curve 4M will he

a Parabola, the given Quantity a, being the

Parameter;

—

of FLUxXIONs. 15
Parameter; and if you throw both Sides of
the Equation into Fluxions, then will 2 & bg

OH, 255 n
19 3, and x = —— and Þ (2) =

— = 2 x, by ſubſtituting ax for yy the Va-

lue thereof. Hence, if you make P T equal
to the double of A P, and draw the right
Line MT, this ſhall be a Tangent to the
Curve in the Point M. |

If aa x n., be an Equation expreſſing F : c. 4
the Nature of an Hyperbola between the Aſym-
totes. By throwing both Sides of the Equa-
tion into Fluxions, we ſhall have x y + y x=0z

and therefore P 2) = x. Whence, if

you take PT =P A on the other Side of the
oint A, and draw the right Line MT, this
will be a Tangent to the Hyperbola in the
Point M.

Let y = x expreſs the Nature of all
Kinds of Parabolas, where the Exponent m,
repreſents a poſitive whole Number or Fra-
Eton. As allo of all Kinds of Hyperbola’s,
when that repreſents a negative Mader or
Fraction. Now the Equation being thrown
into Fluxions, will be h ; and

therefore P T (2) m y = x, by ſub-

ſtituting x for y”, which is equal to it. If n
be =, the Equation will be y* =#x, ex-
preſſing the Nature of one of the Cubick Pa-
rabola’s, and the Sub-tangent PT Ax. If
mn = —2, the Equation will be 4* =x y y,
expreiling the Nature of a Cubick Hyperbola,
and the Sub-tangent PT -x. *

A Treatiſe
If a Tangent be to be drawn to the Point
A the Vertex of the Diameter of a Parabola,

ou muſt find the Ratio of & to) in that Point.
For it is plain, when that Ratio is known,

chat the Angle made by the Tangent and the

Fic. 5.

Diameter, will alſo be determined. In this
Example, x: :: my”—* :1. Whence it
appears, that y being =o in A, the Ratio of
4 to & will be then infinitely great, when is
_ than 1; and infinitely ſmall when the

fame is leſs: That is, the Tangent to the Curve

in A, muſt, in the former Caſe, be parallel to
the Ordinates, and coincide with the Diame-
ter in the latter Caſe. |

ExaneLs II.

11.1 AM AB be a Curve Line of ſuch a Na-

err
5 (5 : A3 (a): AD (5). Then

—— Sax xx; and throwing both Sides of
the Equation into Fluxions, we have =
=ax—2 xx; whence P 12 D

y ab==2bx
EY 2AXwme) XX

= — by ſubſtituting ax — x x for

= and PT— AP, or AT= =

2—2 *
Now if 7P (s) H
| | ＋
(3) :: 4B (a): 4D (b); then will — =xx4—x*, and throwing both Sides of this

Equation

of FLUxX10NSs. 17

4 5
Equation into Fluxions, we have J —

= 3 A ; and ſo
222 F * * A —%*

y z ** i D
Fxxa—x ok D,

and 4 T =
ZA—JX & J&E—Fx
24x
3

AN v generally, if m be the Exponent of
the Power of AP, and z that of the Power of

A mern

PB, we ſhall have _ = x * =

which is a gen eneral Equation for all Kinds of
Ellipſes. An — both Sides of this Equa-

mu
tion into Fluxions, we have = 2 —5 — — —

— = 16 Ri —- and ſo
. 25 Va-
lue thereof, there will come out 5 5 7

N Xa— x

— 122 — ͤ—
MX X 4 -& — n — x* XX

by ſubſtituting x = for

mM EnxXa—=x m +-1Xax—xx
= or PT 7

ma – u ma -n –

EXAMPLE III.

| 13 HE ſame Things bein poſed as in p 1 Cc. 6.
T the foregoing Tha ing frp here the *
Point B falls on the other 4 Side A, with re-

C ſpect

A Treatiſe 1

ſpect to > the Point P, and we ſhall have this

7 * ＋ thi

Equation ——.— — Xx 4 Pr 9 which ex-

preſſes the Nature of all Hyperbolas 4 re-
2 ect to their Diameters. W hence, as above,

r
nA * |

ma n Xx
Now if A be ſu BP) to be infinitely
great, the Tangent 7 M will meet the Curve
at an infinite Diſtance ; that is, it will be-
come the Aſymtote CE; in which Caſe AT
ax | “+: of |
Err 7 will be = „ Ac,
becauſe @ being infinitely leſs than , the
Term ma will be o in regard to Dx.
By the ſame Reaſon, the Equation of the Curve
—_ will be 2 „bat.. And ma-
king mu = p, for Brevity ſake, and ex-
ing the Root p of both Sides, then will
5 yl the Fluxion of which will be
Y = b. So that if you draw A pa-
L to the Ordinates, and conceive a ſmall
Triangle to be at the Point wherein the Afym-
tote C E meets the Curve, the lowing — |
portion will ariſe x : 75 : or Va: Vb:

(=« ): AE 2 _— . Now the

Value of CA and ch being thus determined,
the Aſymtote CE may be drawn. |
If =I, and » = 1, the C e will bethe
common Hyperbola, and ＋ C will be = a 5
and AE ]; that is, one half of.the con-

jugate Diameter, which is a known Truth

eſtabliſned from other Principals.
Font | EXAMPLE

of Fi UxIONS.

; ‘E X’A NP LE IV.

1 T there be an Equation y— « San F. &

i (AP being = x, PM), and a is
a given right Line) expreſſing the Nature of
the Curve AM. The Hluxion of this ſhall be

3% —3xx*=ax5+ayxs. Whence = =

3 —axy „ F. 2 .
Jax Tay? and AT >. « ) —
Dr

3 3xxHay
ting 3 ax) for 33 — 34.

Now -: AP and P M be ſuppoſed infinite-
ly great, the Tangent TM will become the
Aſymptote C E, and the right Lines AT, AS,
will become 4 C, AE, and theſe will deter-
mine the Poſition of the Aſymptote. Now
AT, which call 7, is e from whence

by ſubſtitu-

we get y= I hen AT becomes
5 d& – ak a

4 — becauſe then a7 is o with reſpect to a x-
And putting down 2 for y in 1 — x? =

a, and there will come out 27 * — AN
zt xx, and ſtriking out the Term za xx,
ſince x being infinite, it is o with regard to
the xwo others 277* * and a; then will
AC (2) be =34. In like manner 4S
2 * . | kc . axy F
( 7 ) which call s, is = D
whence we get xi LL, ſince y
being infinite with reſpect to s, the Term as
will be o compar’d to 4); and putting that
es C2 Value

FI 0. 7.

A 7. reatiſe

Value in the Equation of the Curve, we ſhall
get AE (s) a. Whence if you take 4C
=AE=7a, and draw the right Line CE, it
will be the Aſymptote to the Curve AM.

Theſe two latter Examples will ſerve as
Guides in finding the Aſymptotes of other
Curve Lines. nk

P R O p. II.

1 7. J. in the Propoſition aforegeing, the Abſeiſ-

ſes AP be conceived to be Parts of a
Curve Line that we know how to draw the Tan-

_ gents (PT) of. It is requir’d to draw the Tan-

gent MT from the given Point M in the Curve
0

Draw the Ordinate MP, and the Tangent
PT, and ſuppoſe the right Line MT, which
cuts it in Z, to be the Tangent ſought: like-
wiſe ſuppoſe another Ordinate p infinitely

near to the former, and let the little right

Line MR be parallel to P: then call the

en Quantities AP, x, and PM,), and as
Ee we {hall have Pp or MR =x, Rm=
5, and becauſe of the ſimilar Triangles mRM,

MP, we have thisProportion,viz. mR(5): RM
(#)::MP (): PTV Now proceed
with the Equation expreſſing the Relation of
the Abſciſſes 4P(x) to the Ordinates PM (9),
as in the aforegoing Examples, and moreover
the following ones. | 23

We EKAMPLI I.

Lr = be . thisthrown
EE. 3 9 into

of FLUXIONS.
into Fluxions will be 555

— ==; and reducing the ſame i into an
a
Analogy 5 * (MP : PT): ——
a

122 22.
3 Foy > 24 Ane tete the

Relation of the given Ordinate MP to the
Subtangent PT ſought, is expreſſed in known
Terms freed from Fluxions. Which is what
was nr to be done. 7

ExAMPLE Il.

Line x be =. This thrown into
Fluxions, and we have x =D, and PT
EE ==; If the Curve & P B bea se-

mide * the Ordinates M P, continued out
to ©, de perpendicular to the Diameter AB ;
then the Curve AMC ſhall be a Semi-Cycloid.
And when þ=4, the Cycloid will be a com-
mon one: when b is greater than a, the Cy-
cloid will be a Prolate one, and when 1 It islels,
a Curtate one. |

San:

ä 18. HEN the gen Point of the
W cloid is in 75 P het of the Q.
cle, if you draw the Chor ; 1 ir
will be parallel to the Tangent 7 MT. For the
Triangle MPT bring then an Iſoſceles *
3

Fe. 7:

AT, rbatiſe

the . Angle TP © ſhall be the Double

of the internal potts Angle 2 Y, But

the Angle 4P © is equal to the Angle AP T,

becauſe half of the Arch AP. 1s the Meaſure

of each of them; and thereforc it is the one

Half of the An le T PD. Whence the An-

gles TMO, 4P2, ſhal be equal to each o-

ther; and conſequently. the. Lines M. 25 4 P
Thall be | (a6

Prop, II.

12 AP be any b Live, yy the

right Line KNAQ a Diameter of it;
and ſuppof ing the Method of drawing Tangents
(PK) 70 it known; likewiſe let AM be ants

tber Curve ſuch, that drawing any how the Or-

dinate M Q cutting the former Curve in the Point

N, the relation of She Arch AP to the Ordinate

MQ Je expreſſed by any Equation. It is requir d

to draw the Tangent MN from a gh ven Poigt ag

Call the known Quantities PK, ts K 256

the Arch 4P,x; M Oy: then ſuppoling
another Ordinate ꝗ infinitely near MD, and
FIG PO, MS. parallel to 4 , we : ſhall

4 2 PP DS, and ſince the Triangles

and P po, SM and M £ Mare ſimi-
re PK(): K ©(5)::Pp($);PO

RY pied THT Jun Mg

“0: ;QN== > =. Now by means of the Fluxi-

on of the given Equation find a Value of x in
Terms that arg all. 7920 by J, and if you

0 e this Value in oy for , the j will

yanh,

3 %
of FLUXIoONs 23
vaniſh, and the Quantity of the Subtangent

8 IV.

2 T there be two Curves A QC, BC N, Flo. 8.
the right Line TE ABF being a Dia- |

meter; and e the Method of drawing the
Tangents QE, NF, to be known z moreover, let

how be another Curve Line MC uch, that the
Relation of the Ordizates M P, Pp, NP, be
expreſſed by any given Equation. 1t is required

to draw the Tangent MT from a given Point M

in this latter Curve.

| IMAG1N the final Triangles 7 2.0 4: 75
MR m, Nn, at the Points Q, M

call the known Quantities P E, 955 2E, 6;

P. Q, x3 PM, ); PN, x; then will Oꝗ be

=X* Rm=y, La becauſe * when Ain a.

x and y increaſe, z decreaſes. And ſince the

Triangles © E and 20 , NP Fand SN,

M PP. and m RM are ſimilar ; therefore 9P EY
(9: 2 740 (x):09 or MP or SN =

“And NP * PF): 50 (. SN

“= === Whaees ariſes 2 = =)
4 k 7x

and R (5): RM i): ; MP (yh: PT

. Now if — = be ubſtirared for
in the Equation of — “Que thrown intro

*Juxions, y we ſhall have a Value of x my;

\ 0 C4 which

4
|
N
:

tk -x Z

A er Jo
which being put in —=, and the ys. Std

ſtroy one another; a fo the Value of the
Subtangent P T will be had 3 in known Terms,

LS EXAMPLE. x
21, L* xz; this thrown into ibn

FZX—SZ K
_ is W e

by putting — — == for z; whence We er

ay
SAL, 1 therefore PT (425) 2

E
2sStyy

= eee

Again, let there be given this general Equa-
tion, viz. 1 g this thrown into by

_ Xions’ willbe n PT: 0 = h ⁰

4 mn 2 x A „A WS ub
n * = — = x

by putting — 7-7 23 whence we get P 7

2 r mts:

x3 mx r mr—ns
if x” 2″ be put for y”+”.

_ you may obſerve, that if che Curves
ADC, BCN, become ri cht Lines, the Curve

MC will be one of the Conick Section kind,

viz..an Ellipſis, when the Ordinate CD, drawn
from the Point of Concurrence C, falls be-

tween the Extremities Aand B; an Hyperbola,

when it falls on either Side; and a Parabola,
when one of the Extremities A or is infi-

nitely diſtant from the other, or when one of

of FLUXIONS. as *
the right Lines CA or C is parallel to the
Danke AB.

|
PROP? Yo |

ET APB be a Curve be inning at the Fi c. 9. |
Point A, and ſuppoſe the Method of
drawing Tangents P H) 20 it known, and let the |
Point F be aſſumed without this Curve, always |
having the ſame Situation, and if there be another il
Curve CMD ſuch, that any right Line FM |
being drawn, the Relation of the Part F M | |
thereof, to the Part AP of the Curve, is ex-
preſſed by any given Equation. It is required to
draw the Tangent MT from the given Point M.
|

Draw FH perpendicular to FP, meeting
the given Tangent P H in the Point HF, and
the 85 ought Tangent Min the Point 7 ſup-
poſe a right Line F Rm Op making an Angle
infinitely ſmall with FP, and from the Centre
F deſcribe the ſmall Arches PO, MR, of a
Circle; the little Triangle pO.P ſhall be ſimi-
lar to the right-angled Triangle PV; for
the At les Ap, Hp F, are equal, becauſe ® 47. 2.
they differ only by = Angle P Fp, which is
ſuppoſed to be infinitely tmall; and moreover
i Angle p Op is a right Angle, ſince the Tan-
nt in O (which is the Continuation of the
little Arch PO conſider’d as a right Line) is
pendicular to the Radius FO. By theſame

Re: ſon, the Triangles m R M., MF 7 will be
ſimilar. Now it is evident, that the little
Triangles or Sectors FP O, FMR are ſimilar.

And – the known Quantities PH, t; HF, 33;

then

t
3
©
8
_
®
=_ |
Si
)

— _ –
ed. nas ao ad
—— —

A Treatiſe

then ſhall PH (H): HO P54) 0

2 And FP (z): FM(y):: PO EZ):

MR”. And n R 5): E Di

F . 10.

F M. OE FT = =_ And by throwing the

= Equation 15 ee, what i is ſtill to
done wy” be effected.

EXAMPLE.

2 A 1˙5 F. the Curve 4 PB be a Cir cle, the Point

F being the Centre; it is plain that the
Tangent P V does become parallel and equal
to the Subtangent FH, becauſe II thall be
_ 1 — PF; and fo, in this Caſe,

FT= 5 — 27. by calling FP (2), 43

ſince it is now a conſtant Quantity. This be-
ing ſuppoſed, if the whole Periphery , or any
determinate Part thereof, be called P, and you
make h: x::4: y, the Curve CM D, which in

ttnhis Caſe is F M D, 2 be thẽ Spiral of Ar-

chimedes, and y = =p which thrown in F luxi-

wing 9 WT: whence ariſes Jun AL 25

x), by putting == HIP ”3 ; pl therefore F T

Jy 7599 ==. And fo we get the following

Conftrattion.

From

of FLUXIONS.
From the Centre F with the Radius FM,
deſcribe the circular Arch M 2, bounded in
Aby the Radius FA joining the Points A, V;
and take F. = Arch M9, I fay, the right
Line MT will be a Tangent in M. For be-
cauſe of the ſimilar Sectors FP A, F 9, the
following Proportion will ariſe F P (a) : FM

(A (as): MO AE.
4
If you ſuppoſe generally, that b: x:: a: jm
(the — — m expreſſing any whole Num-

ber or Fraction) the Curve T will be a
Spiral of all Kinds ad infinitum. And then y

A
==, and this thrown into Fluxions, and

mby® 3

Amn

4 „ e e.
m = ; whence ariſes 17

| An x 5, by putting for ** * and there-

| br ee

Px 62. 1 7

|
|
|
|
|

ET there be a W Line „ab the F 1 C. 11.

Method of drawing Tangente (P H) to it
being known, and let F be a given Point taken
wvithout the ſame, always keeping its Situation.
Let there be likewiſe another Curve CMD ſuch,
that auy how drawing the right Line FP! be
Relation of FP to FM be 2 by any gives
Equation. It is required to draw a 25 gent
M T Lr the go $7 M.

Draw the right Line FH e

F * and ſuppoſe (as in the laſt Propoſi-
tion)

A Treatiſe

Both tha Erie Triangles Y Gp, M Rm ſimilar

to the Triangles HFP, TFM; then calling

the known Quantities F Hs; FP,x; FM, T

and rr N have P F (x): ‘FH (5): 50 (5):

And FP 005 FE(S): FA O): 2 ,

OPS. And FP (0): TAG) AFG. *)
x
R. And m R (Y): I ..

FM (9): Ess What is farther to be

done, may be effected by nig 4 the gin

Equation into Fluxions.

EXAMPLE.

1 r inſtead of the Curve 4 PB, you would

ha ve a ſtraight Line PH, and the Equation
expreſſing the Relation of FP to PM be) x
Daz; that is, if PM be always equal to the
ſame given right Line a; 30 will; be= =X3

py therefore FT (2) 2 3 — U Whence

XXY) 4}

we get the following Conſtruction.
Draw ME parallel to P H, and MT paral-

lel to PE- F fay, the ſame will touch the
Curve in „

For FP 00. FH(::FM 005 15222 = Y.

XN
It iS therefore manifeſt, that. the Curve GMD

is the Conchoid of. Nicomedes, the right Line
PH ns the OM IT rho and the Point F

ait t want

pRO r.

of FLUXIONS.

PRO P. VII.

JET ARM be a Curve, the Method of F 1 d. 12.

drawing Tangents (M) to it being known,
and let the right Line EPAH T be a Diameter
thereof, without which let the Point F have a
conſtant Poſition; and from the ſame let the inde-
finite right Line FPS M iſſue, cutting the Dia-
meter in P, and the Curve in M. Now if you
conceive the right Line FPM to revolve about
the Point F, and at the ſame time to move the
Plane PA M conſtantly parallel to itſelf along the
indefinite immoveable right Line ET, ſo that the
Diſtance PA be every where the ſame; it is evi-
dent that (M) the continual Inter ſection of the
Lines FM, AM, by this Motion, will deſcribe
the Curve CMD. From the given Point M of

which, it is required to draw MT to touch the
Curve.

The Plane PAM being ſuppoſed to be come
to a Situation pam infinitely near, and the
right Line R & drawn parallel to A P, it is
evident (from the manner of Generation) that
Pp=Aa—Rmy; and therefore that R n
—Pp. Now if you call the known Quanti-
ties FP or Fp, x; FM or Fm,y; PH,s;
MH, t; and the Fluxion Pp,z: Then from
the ſimilar Triangles F Pp and FSm, MP H
and MSR, MHT and M Rm, we ſhall have

Fp (x): Fm ():: Pp OS whence

S And PH: HM ():: SR

]: NM D And MR
( * ):RM TIS 8

Iv2,—tx7, a | r
( — . 92 MH (1): HT ==
Whence if FE be drawn parallel to A, and
you take HTP E; the Line MT ſhall be
the Tangent ſought. |

If A were a right Line, the Curve CHD
would be an Hyperbola, and the Line ET
would be one of the Aſymptotes. And if it
vere a Circle, the Point P would be the Cen-

tre, and the Curve CM would be the Con-

cChoid of Nicomedes, the Line ET being its

Aſymptote, and the Point F the Pole of it. But
if it were a Parabola, the Curve CMD would
be one of the parabolick Kind, mentioned by
Deſcartes in Lib. 3. Geom. and at the ſame time

would be deſcribed below the right Line E,
by the Interſection of FM with the other half

Fis. 13.

of the Parabola. |

PRroe. VIII.

L TAN be a Curve, whoſe Diameter is

AP, and let F be a Point without them,
having a conſtant Situation: Moreover let CMD
be another Curve ſuch, that any how drawing
the right Line FMPN, the Relation of the Parts
FN, FP, FM of it, is expreſſed by any given
Equation. It is required to draw a Tangent M
from any given Point M in it.

Throꝰ the Point F draw the Line HK per-
pendicular to FN, meeting the Diameter A P
in K, and the given Tangent NH in H. From

the Centre F, with the Diſtances FN, FP, FM,

deſcribe the ſmall Arches N 2, PO, MR,
terminated by the right Line Fx ſuppoſed to
make an infinitely {mall Angle with FN. This
being ſuppoſed. I

: Call

of FLUX10NS.

Call the known Quantities FA, s; FA, t;
Fp, x; FM,); FN,z then becauſe of the
ſimilar Triangles PFK and OP, FM R and
FO, FO and FN9, HFN and Nn,
m R M and M FT, we ſhall have the following
Proportions: P F(x): FK ():: 0 (0: 0 P=

i* And FP (x): FM(y)::PO (= MR=
x * –/ | x

And FPO. Fx H. POE
XX 7 c
1 | SX
And HE (): FN (Y:: Ng, 2
eee AndmR(5):RM(=

Tx XX

. F (Y 2. . 2 “=
FM (y):F ns ow by throwing

the given Equation into Fluxions, we ſhall
find a Value of) in x and Z; in which ſubſti-
KX
ſes, z decreaſes; and then all the Terms will
be affected with x : So that at length this Va-
lue being put in- Z, the & will go out; and
h * bh

therefore the Value of F will be expreſſed
in known Terms freed from Fluxions.

If the right Line AP were ſuppoſed a Curve,
and the ‘Tangent’P X had been drawn, we

ſhould: find that F would always have the

for ⁊æ; becauſe when x increa-

. ameValue, and the Reaſoning would have been

the fame.

EXAMPLE

— — — ———— — — —
_— —ͤ— ——— ——

ExanyLs.

51 0. 14. 28. L* che Curve AN be a Circle ator

thro? the Point F (ſo ſituated — re-

788 to the Diameter AP, that the Line FB

dicular to the ſame Diameter, paſſes
chr G the Centre of the Circle) and let PM
be always equal to PN; it is manifeſt that the
Curve CMD, which in this Cafe becomes
FM 4, will be the Ciſſoid of Diocles, and the

Equation thereof ZH ix; which thrown

into Fluxions, will be y—=2x—z—

21XXX Terz * by ſubſtituting— SZEX 4
1XX 7xX

above (Art. 27.) for z. And therefore F

(2=)= — 4.
XXxy NN 4-SEg

If the Point M coincides with the Point A,

the Lines F M, FN, FP, will be each equal

to FA; as alſo the right Lines F K, FH; and
therefore we ſhall bave in this Caſe KF Ga _

x+
„ * that is, take FT= AF, and

draw the right Line 4 7, the ſame will touch |
the Curve in the Point. A.

Tangents may be drawn likewiſe to the

Ciſſoid by means of the firſt Propoſition, by
drawing NE, MI, perpendicular to the Di-

ameter FB, and ſeeking an Equation expreſ-

ſing the Relation between the Ordinate LM,

and correſpondent Abſciſs FL. And this may
be thus done; firſt call the known Quantities
FB, za; FIL or BE, x, LM, y; then from
the Similarity of the Triangles ‘FE N, FLM,
and the Nature of the Circle, we have F L
(#):LM (9): : FE: FN:: *
EB ().

of FLUxX1oNs.
EB (x). Whence we get yy=

—; which
thrown into Fluxions, will = 2773 =

GaxxX—2%*x – and |
— — therefore LO. 05

24—x
. 24x —.—
zar ol EVE] —

, by putting — — — for yy.

PR O P. IX.

„Linie, be two Curves ANB, CPD, F10. 15.

and a ftraight Line FKT, in which
are taken the three Points A, C, F; and let

there be ſome other Curve E. M G ſuch, that a

right Line F MN being drawn from any Point
M of it, and the right Line MP parallel to
F K the Relation 0 the Arch A N. to the
Arch CP, is expreſſed by any given Equation.

It is required to draw the Tangent MT from 4

given Point M in the Curve E. G.

Thro’ the Point 7 ſought, draw the right

Line T H parallel ro FM; and thro” the wen
Point M, the right Lines MR K, MO

rallel to che Tangents in P and N, and 17.100
FmOn infinitely near to FMV, and mm Rp

parallel to MP.

Now call the known Quantities FM, 5;
PN, tz MA, u; CP, x; AN, y; (then will
Pp or MR A, Nu=5). And becauſe of
the ſimilar Triangles FENn, and FMO; Mon,
and MHT; MRm, — MXT; therefore

FN (4): FM (s):: Vu (5): : MO =.

And MR (5): MO(= 7: MK .

SUY

—

r | D Nov

A. Treatiſe

Now by help of the Fluxion of the given
Equation, we (hall have a Value of in Terms,
every of which will be affected with &; which

being ſubſtituted in =, and the Z*dodeſtroy
C | * i

one another. And therefore the Quantity of
M H will be expreſſed in known Terms. From
whence we have the following Conſtruction.

Draw M H parallel to the Tangent in NM,
and equal to the Expreſſion before found;
draw HT parallel to FM, meeting the right
Line FX in T; through which, and the given
Point M, you muſt draw the Tangent M1;
which will be that ſought.

E X AMP I. E.

8 zo. Ir the Curve AN B be a Quadrant of a

Circle, the Point F being the Centre;
and the Curve CPD becomes the Radius APF,
perpendicular to the right Line FX GIB,
and the Arch AN (, be always to the right
Line 4 P (x), as the Quadrant I MB (b), to
the Radius A F (a); then the Curve EMG,
ſhall beceme AM the Quadratrix of Dino-
rutus; and () will be =.
„ OR FX ax
ſince FP or MK (u) =a—x, and FN (t)=

But from the ſuppoſed Analogy ay b,

and 85 e. Now putting and “= for
their Equal x and j, in the Value of MH,
and there ariſes ME . From whence

we get-the following Conftruftion.

_ Draw

of FLuxions. 35
Draw M 7 perpendicular to FM, and equal
to the Arch N deſcribed from the Centre
J, and draw HT parallel to FM. I ſay, the
Line MT will touch the Curye in M: For
becauſe of the ſimilar Sectors FNB, FM,
we have this Proportion, viz. FN (a) : F

O N B by): N.

a
Cora A.

| 31 r you deſire the Determination of the F 1 C. 17:
1 Point &, where the Quadratrix 4 MG
meets the Radius FB, you muſt conceive a- =
nother Radius Fg 6 infinitgy near FGB; and
then drawing g parallel to FB, from the Na-
ture of tlie Quadratrix, and the ſimilar Tri-
angles FB, gf F, right- angled at B and f, we
ſhall have this Proportion AB : AF: : Bb :
T:: FB or AF: g for FG. |
Whence if you take a third Proportional
to the Quadrant A B, and the Radius A E, it

Hall be equal to FG; that is, FS.

which Means the Conſtruction of Tangents
may be ſhortened thus:

Draw 7 E parallel to MZ7; then from the

ſimilar Triangles FMA, FTE, we have MX F 16. 16.

Gu ET er MH(ESL)
.. __bss jo GY pn

F 6 — . by putting : for its
Equal x, and afterwards dividing the whole by
b—y. From whence’it is manifeſt, that the
RR FT is a third- Proportional to FG and

D 2 PR O P.

36 A Treatiſe:.

, P R O p. X.
F C. 18. 32. LE T AMB be a Curve ſuch, that the
| Relation of the right Lines MF, MG,
MH, Sc. drawn from any Point M taken in
it, to the Foci F, G, H, &c. is expreſſed by any
given Equation. It is required to draw the Per-
pendicular MP from the given Point M, to the
Tangent in that Point.

Take the infinitely ſmall Arch Mm in the
Curve AB, and draw the right Lines F Rm,
m I, HmO; and from the Centres J, &, H,
deſcribe the ſmall Arches MR, MS, MO;
likewiſe from the Centre M, with any Diſtance,
deſcribe the Circle CDE cutting the Lines
MF, MG, MH, in the Points C, D, E; from
which let fall the Perpendiculars CL, DA,
2 to MP. This being done, you may ob-

erve, | 0 8 K
10, That the right – angled Triangles MRm,

NL. C, are ſimilar; for if from the right An-

gles L Mn, R MC, the common Angle LMR

e taken away, the Angles remaining R Mm,

LMC are equal; and they are right- angled at

Rand L. After the ſame way we prove that

the right- angled Triangles M&S m and MX D,

Mon and MILE, . Therefore ſince

Mm is a common Hypothenuſe to the little

Triangles MR mn, Mm, MO m, and the Hy-

pothenuſes MC, MD, ME, of the Triangles

MLC, MX D, MIE, are equal; it is evi-

| dent that the Perpendiculars CL, DA, EI,

= have the ſame Relation to each other, as the
Fluxions Rm, Sm, On. a

2e That

oo Eh… Ss

of FLUXIONS.

20, That the Lines iſſuing from the Foci
ſituate on the ſame Side the Perpencicular
M, do increaſe at the ſame time the others
decreaſe, or contrariwiſe. As (in Fig. 18.) FM
increaſes by its Fluxion Rm, while GM, 7M,
decreaſe by theirs, viz. S m, Om.

Now it the Equation ax+xy—2z2z=, be
ſuppoſed to expreſs, for Examples ſake, the

Relation of the right Lines FM (x), @ M (9),

HM (z). This Equation thrown into Flux-
ions, will be ax + yx+xy—222z==0. And
then it will follow that the Tangent in M
(which indeed is only the Continuation of the
little Side Mm of the Polygon, that the Curve

AZ is conceiv’d * to be made of) mult be * Arr. 3.

ſo ſituate, that if the Parallels R, mn , mO,
to the right Lines FM, & M, HM, be drawn
from any Point m in it, terminated in R, S,
and O, by Perpendiculars MR, MS, MO, to
the ſaid right Lines, we ſhall always have this
Equation ã x Rm-+xx$m—22xO0m=0;
or (which comes to the ſame, in putting CL,
DK, EI, for their Proportionals Rm, Sm, Om)
the Perpendicular (AP) to the Curve, mult
be ſituate ſo, that a+y x C L+xx D K— 22x
EI. From whence wc have the follow-

ing Conſtruction.

If the Point C be conceived to be laden with g 10. 18,
a Weight a+z, which multiples the Fluxion 19.

x of the right Line FM on which it is ſituate,

and the Point D with the Weight x, and. che

Point E, taken on the contrary Side M with
regard to the Focus 7 (fince the Term 22
is negative) with the Weight 22. I ſay, the
right Line M paſſing thro’ the common
Centre of Gravity of = Weights ſuppoſed to

FIC. 19.

4A Freatiſe \9

be in the Points C, P, E, will be the Perpen-
dicular required.

For it is plain (by the Priaciple of Mecha-

nicks) that every right Line pu aſſing thro? the

Centre of Gravity of ſeveral Weights, ſo ſe-
parates them, that the Weights on one Side,
each multiplied by its Diſtance from that Line,
are equal to the Weights on the other Side,
each multiplied alſo by its Diſtance from the
faid Line. Whence, ſuppoſing x, y, E to in-
creaſe together; that is, conceiving the Poci
F, G, H, to be all on one fide MP, as we
have always done, in throwing the given E-
quation into Fluxions, according to $e Rules
before delivered: Therefore the Line M P
will leave on one Side the Weights in C and
D, and on the other the Weight in E, and

ſo we ſhall have Ty xCLExxD K—22%
EI o, which is the Equation to be con-

ſtructed.

Now ſince the Conſtruction is jult + in this

Caſe, I fay it is likewiſe ſo in any other Caſe.

For Example, if the Situation of the Point M
in the Curve be ſo alter’d, that + increaſes
while y and 2 decreaſe; that is, that the Foci
G and Z fall on the other Side MP. Then i it
follows, 1. That the Signs of the Terms of

the given Equation affected with y and ⁊ 25 Or

their Proportionals D A, E Z mult be chan ed;

Whence the Equation to de conſtru

this latter Caſe, will be 2) “x C LDR
ZEIT. 2. That the Weights, in

and E will change their Situation with re

to MP; and fo from the N NUrC of the Cen-

tre of Gravity, we have akSyxC LDR
hx EI, which is the zz to be
con-

19 —

of FLUXIONS.
conſtructed. And ſince this is fo in all the
Caſes. poſſible, therefore, &c.

Hence it appears, that the Reaſoning is the
fame, ler he will be the Number of the
Foci, and the given Equation, ſo that we thus
denounce the general Conſtruction. |

Find the Fluxion of one Side of the given
Equation (the other Side being o) and from
the Centre M, with any convenient Diſtance,
deſcribe the Circle CDE interſecting the

right Lines M, MG, MH, in the Points

C, D, E, in which are conceived the Weights
that have the ſame Relation as the Quantities,

multiplying the Fluxions of the Lines on which

they are ſituate. I fay, the Line M paſſing
through their common Centre of Gravity,
ſhall be the Perpendicular required. But here
we mult obſerve, that if one of the Weights
be negative in the Fluxion of the given Equa-
tion, the ſame muſt be ſuppoſed to be on the
contrary Side of the Point M with regard to
the Focus. |

If inſtead of the Foci F, G, I, you ſuppoſe
right Lines or Curves, to whieh the right Lines
MF, MG, MH, fall at right Angles, the

ſame Conſtruction fill takes place. Friis. 20.

For drawing from the Point taken infi-

nitely near M the Perpendiculars mf, mg, mh,

to the Foci, viz. the Lines in the Foci; and
from the Point M, the little Perpendiculars
MR, MS, MO, to thoſe Lines. It is evident
that Rm will be the Fluxion of MF, becauſe
the right Lines MF, Rf; being Perpendiculars
between the Parallels F , MR, are equal to
one another; and in like manner Sm is the
Fluxion of M, and Om that of M; and
what remains may be proved as above.

os D4 Inſtead

ck — —

Fi. 21.

A:T reatiſe V

Inſtead of all or ſome of the Foci F, G, II,
we may ſuppoſe Curves beginning in F, G, II.
And the Curve AIM ſuch, that the Tan-
gents MV, MA, and the right Line M be-
ing drawn from any Point M in it; the Rela-
tion of the mixed Lines FV M, HA M, and
the right Line GM be expreſſed by a given
Equation. For by drawing the Tangent m «

from the Point #2 taken infinitely near M, it

is manifeſt that the ſame will meet _ other
Tangent in the Point Y, (it being only the
Continuation of the little Arch Yu, confider’d
as a little right Line.) And therefore if from
the Centre / be deſcribed the ſmall Circular
Arch MR; Rm ſhall be the Fluxion of the
mixed Line FV M, which becomes FY/uRm.

And all the reſt may be demonſtrated as afore-

oin
Ti 2 Problem was fir ſt farted by Mr. Tſchirn-
hauſen, in his Book de la Medecine de l’eſprit;
and Mr. Fatio in the Journal des SGavans, has
given a very ingenious Solution thereof. But
their Solutious are only particular Caſes of *
11 Conſtructiun bere laid down, T3;

Ex Ar LX I.

33 L* Ane +, 8 (the
right Lines a, h, c, F being given) the

Fluxion of which is -@x x +byj5 e o.
Now – Auppoling the Weight ax in C, the
Wag. t % in D, and the Weight cz in E,
eights that are to one another, as thoſe
Reckangles The Line M going through
the common Centre of Gravity of —_— ſhall

be ee bo che Curr in the an N.

5 0 But

of FLUX ILONs.
But if FO be drawn parallel to CL, and

the Radius MC be , then (becauſe of the

ſimilar Triangles, MCL, MFO) FO will be
=xxCL; and in like manner drawing GR
parallel to DA, and H to EI, we thall have
G R= N DA, and HS Xx EI: ſo that by
imagining the Weights a, b, c in the Foci
E, &, H, the Line MP paſſing through the
Centre of Gravity of the Weights a x, by,cz
conceiv’d in C, D, E, ſhall alſo paſs through
the Centre of Gravity of theſe latter Weights.
Now this Centre is an invariable Point as to
Situation, becauſe the Weights in F, G, ,
viz. a, h e are conſtant ſtraight Lines, being
the ſame in all Poſitions of the Point M.
Therefore the Curve AM muſt be ſuch,
that all the Perpendiculars to it do cut each
other in one Point; that is, it is a Circle ha-
ving that Point for its Centre. From whence
we have the following remarkable Property of
a Circle.

If there be ever ſo many Weights a, , c,
on the ſame Plane ſituate in F, G, V, and
a Circle AMB be deſcribed about their com-
mon Centre of Gravity. And if the right
Lines MF, MG, MH, &c. be drawn from
any Point M in it: I fay, the Sum of their
Squares, cach multiplied by the correſpondent

Weight, will always be equal to the fame

Quantity,

ExXAMPL E II.

T ET the Curve AMZ be ſuch, that the F : 6. 23

– right Line M being drawn from any
Point M in it to the Focus F of a ſtanding
Poſition, and the Perpendicular MG 5 the
3214 | ocus

FI c. 24.

be the Pe

A Treatiſe

Focus G, taken as a ſtraight Line; the Rela-

tion of MF to MG will be always the fame,

as of the given Quantity a to the given one b.
Call FM, x; and & M, y; then will ::: a: b.
And therefore ay bx. And this thrown in-
to Fluxions will be aj — . = 0. Now

conceiving the Weight b in C taken on the

other Side M with reſpect to F, and the Weight
4 in O (at a like Diſtance from and draw-

ing the Line MP through the common Cen-
tre of Gravity of thoſe Weights; and it will
rpendicular requr’d.

For from the Nature of the Balance it is

plain, if the Cord CD be fo divided in
P that CP:DP::a:b, and the Point will
be the common Centre of Gravity of the
Weights in C and D.
Ihe Curve A MB is a Conick Section;
viz. a Parabola, when a =; an Hyperbola,
when 4 exceeds &; and an Ellipſis, when the
lame is leſs.

mo “EXAMPLE III.

1 r you faſten the Ends of a Thread

FEZLY MGMXYTH in the Points F and
H, fixing a little Peg in the Point E, and then
extend the Thread by means of the Pin M ſo,
that the Parts FZ V, HY X wrap about the
Curves beginning in Fand IV; and if the Part
Me be doubled or folded together, the Mo-
tion of the Pin M will deſcribe a Curve AM.
Now it is required to draw a Perpendicular

M to this Curve from à given Point M in

it. The Poſition of the Thread . be-
ng AT in that n a xo 1

Here

of FLUXIONS. 43

Here 1 obſerve, that the ftraight Parts
MY, MX of the Thread are always Tangents
in J and X, and if the mixed Lines FE AM,
and HTA M be called x and z, the right Line
MG, y; and a right Line equal to the Length
of the Thread, a. Then we ſhall always have
x +29 +z a: whence we know that the
Curve AMB is contained under the general
Conſtruction. Therefore finding the Fluxion
of the Equation, viz. x +25 +2 =o, and
conceiving the Weights 1 in C, 2 in D, and
1 in E. I fay, the Line MP paſting through
the common Centre of Gravity of thoſe
Weights, will be the Perpendicular requir’d.

PR OPp. XI.

T ET APB, EQF be any two Lines, Fic. 25.
the Method of drawing Ta PG,

QH, to the ſame being known; and let PQ be

a right Line, having a Point M given in it.

Now if the Ends P and Q of this right Line

move along the Lines AB, EF; it is plain that

the Point M by that Motion will deſoribe a

Curve CD. I is requir’d to draw a Tangent

MT to the ſame from any given Point M.

Imagine the moveable right Line PM © to
come to a Situation infinitely near pm, and
draw the ſmall right Lines PO, MR, QS,
perpendicular to PO, and we ſhall have the
ſmall right Angles pO P, RM, 489; likewiſe
take P&R M2, and draw the right Line
HK perpendicular to P ©, and continue
out OP to 7, whercat it is ſuppoſed to inter-
ſe& the Tangent ſought MT. This being
done, it is evident that the little Tight

mes

FI. 26.

A Treatiſe
Lines Op, Rm, $4, will be equal to each o-
ther, becauſe by donſtruction P M and N 2,
are every where the ſame.

Call the known Quantities P Mor K 9,9; ;
Mor PA, I; KC, f; XH, g; and the lit-
tle right Line O p, Rm, or $q,y, Then the
ſimilar Triangles PKG and pO P, QX A and
28 willgive PK(b):KG(f)::pO(z):OP

2 And Q (0: XH: 7860. .=

N. Now from common Geometry we know

that MR LEES SI = 21 8
So the ſimilar Triangles » RM, MPT will

give m R()): E Y) M (a): PT

A Which was to be found.
a ＋

bags. > XII.

1. T BN,FQ ge any two Lines, the

right Lines BC, ED, cutting each o-
ther at right Angles in the Point A, being the
Axes of them; and let LM be a Curve ſach,
that the right Lines MGQ, MPN drawn
from any Point M taken in it, parallel to AB,
AE, the Relation of the Spaces E GQF, (the
Point E being a Stable, one in the right Line AE,
and the Line EF parallel to AC) APN D,
and the right Lines AP, PM, PN, GQ, be
expreſſed by a given Equation. * 18. required ta
draw the Tangent M LE to the given Point M in
90 Curve LM.

of FLUXIONS.
Card or MM „*; PM or AC, y; PNʒu;
e Pete: EG Fs; the Space
7 N. he given Subtangents Pa
G A, U. Then will Pp or NS or MR Dx,

g or Rm or OA =; S-

boa of Fe fone: Tropa as N 92 13

F 15 we are to take
notice that the Values of Rm and $7 are ne-
gative, becauſe when AJP ( increaſes, PM(y)
and PN (u) do decreaſe. This bein . ſuppo-
ſed, throw the given Equation into |

wherein fubſtitute 1 &,. — 29. 2 [= for
a

their Equals 7, 5, and ⁊, and we ſhall have a

new Equation expreſſing theRelation (ought)

of 9 to &, 0 N= |

EN Ar J.

1252 s+22 be =!+ux; this thrown
into Fluxions is zz =t+ur+x1,

and putting for 3, i A, d their Equals, we ſhall

find—zy— ESEF _ = 264 — EE; from

1
whence we, get PT 05 2 55 e

ban Zabu :

= E XAMPLE II.
39 L* 5 be t: then i Af; that is, — 2)
1; MAS) and therefore PT (=) *
5 1 Now ſince this Quantity is negative,
there-

UX10Ns,

PPP INS oO i EI ve np eb

Art. 10.

Fo. 27.

any Paar
Arch MG from the Centre A, and drawing the
Line G (parallel io E F) * perpendicular 70 AB,

be Relation of the Spaces EG QFss; ANB, t;

A Treatiſe:

therefore the Point T muſt & be taken on the

other Side the Point 4, where the #” begin.
It we ſuppoſe the Line F to be an Hyperbola,

that G ©,

whoſe. — are AC, AE,

6 andche Line E N Da ſtraight Line

parallel = A B;- whence PN (4) will be al-

ways equal to che given right Line c. Then

it is plain that the right Line 4B will be an

Aſymptete to the _ LM, and the Sub-
| e ÞT ()* en iber! is en
to a ſtanding Quantity.

And ſo in this Caſe LM will be the Is
garitbmetick Curve, © |

| ‘P 1 0 7. XIII.

„Lr B N, F Q bt any 160 Lie, the
g. Line B A being their Axis, inwhich
take the = Points A and E; and let LM be a
Curve ſuch, that drawing a right Line AN. from
M talen in it, 4% bing the circular

and the right Lines AM or AG, y; AN, z;

u; is expreſſed by any given Equatian, It

W 10 draw the Tangent M T from a

given Point M in the Curve L M. |
Draw the ri ght Line 4TH perpendicular

to AM N. LP coneeive another rig du Lane

Amn infinitely near 4 N N, another Arch

gz another Perpendicular gg, and the little

Arch N S deſcribed from the Centre A. Now 1
2 Ca

of FLUXIONS.
call the given Subtangents AH, a; and GK, 553

then will Rm or Gg, Su=X: And be-
cauſe of the ſimilar Triangles AN, NSA;

K 62 and 20 ; 3 $ 1 ; Og

=— #2] an =#), ANnor

4A NX1 1 All theſe Values

muſt be put in the Fluxion of the given Equa-

tion, with which a new one will be formed;

from whence we get a Value of Z in 5. Now
becauſe of the Sectors and ſimilar Triangles
ANS and AMR, 9 and MAT, AN(z)

: AM(g):: NS
22

1K G0 U 0 4

_ 2

Equal! in), the Floxions will vaniſh, and the

Oamtity’ of the Subtangent 47 ſought, will
be expreſſed in known Terms. Which was
to be found.

EXAMPLE I.

o
7 =

; MR ==2L! = ind!

Now 1 inſtead of 2 you Choe its

| a.

7

L.. 140 t be Et; this throw in in-

to Fluxions, is »+y – 223—t;
t ſubſtituting as neceſſary we have 2

22 And laſtly, putting this Va-

we in 228, there ariſes 4 7 Lee

en ö

Ex Ari

— — — — — —— — — — —

—— oo i to — —

11 1 *
1 >

I ge WE >
——_—

rhe Curve L “ak is the Logarithmical 2205

F 1 C. 28.

4 T1 reatiſ, e

EAN II.

wo t; 3 _ „—¹5 VIZ. —

8, or , therefore 47 =
WI. —— *

If 85 Line BN be a n the Point 4
being the Centre, and the Radius be 4 B =

AN=c, and if F be an Hyperbola ſuch,

that C2 (4 =D, then it is manifeſt, that the

Curve LM . an infinite Number ofRevo- |
lutions about the Centre A before it comes to

the ſame; becauſe the Space FE G © becomes
infinite when the Point & falls in A, and 4Tis=

2 J =. Whence we may obſerve, that the Ratio

of FAM: to AT is conſtant ; and therefore-the
Angle 4 MT’is a ſtanding Quantity E; and ſo

s |

PRO p. XIV.

£7 AMD, BMC be any two Curves

touching one ancther in the Point M, and
tet L be a 255 ſtable Point in the Plane of the
Curve BMC. Now if you conceive the Curve
BMC to revolve on (or roll along) the Curve
AM /o, that the rzvolving Parts AM, BM,
be always equal to each other. Now it is mani-

| feſt that the Point L, moving along in the

Plane BMC, will deſcribe a kind of Cycloid
ILK. This being premiſed, I ſay, if the right
Lins L M be drawn from the deſcribing Point
P L,

of Fi.Ux10Ns.

L, to the Point of Contatt M in every diffes
rent Poſition of the Curve BMC, the ſame
will be perpendicular to the Curve ILK.

| For the infinitely ſmall equal Parts Mm,
Mm of the Curves AMD, BMC, may be.

taken * for two little right Lines making an

infinitely ſmall Angle at the Point M. Now
that the “lrrle Side Mm of the Curve or Poly-
gon BMC, may fall upon, or coincide with
the little Side Mam of the Curve or Polygon
AMD; it is neceſſary for the Point L to de-

ſcribe a little Arch LI, ut the Point of

Contact M, as a Centre; conſequently
the ſaid ſmall Arch will be a part of the Curve
TLXK; and the right Line ML, which is per-
pendicular to it, will alſo be perpendicular to
the Curve (IL in the Point I. Whien
was to be demonſtrated. –

PRO p. XV.

ET MLN be any right-lined Angle, Fi c. 97

whoſe Sides LM, LN, touch any two
Curves AM, BN. I “theſe Sides move “along
the Curves, ſo as to touch them continually, it
is evident, that L the Vertex of the Angle —
deſcribe the Curve IL K. Now it is re
to draw L C perpendicular to this Curve, t b.
ſition of the Angle MN being given.”

Deſcribe a Circle thro’ the Vertex L, and
the Points of Contact M, N, and draw the
right Line C L thro” the Gentre of the ſame 3
: LE it will be e to the Curve

E For

— — ——— — — —

F 4c. 30.

A Treatiſe

For conceiving the Curves AM, BN, as
Polygons of infinite Numbers of Sides, of
which Mm and Nu are each one; it is plain
that if the Sides LM, LN, of the right-
lined Angle L MN, of a given Quantity,
move about the ſtable Points M, N, (the
Tangents LA, L N, being ſuppoſed the Con-
tinuations of the little Sides Mf, Ng) until
LM, the Side of the Angle, falls on the lit-
tle Side Mm of the Polygon 4 M, and the
other Side LN upon the little Side Vu of
the Polygon B N; the Vextex L will deſcribe
2 little Part LI of the circular Arch ML.
Therefore the ſaid fmall Part LI will be com-
mon to the Curve ILA; and conſequently
the right Line CL, which is perpendicular
to it, will be perpendicular to the Curve in
125 nt T. Which was to be demon-

ated, a} © 269

PRO P. XVI.

LZE7 ABCD be a flexible! Cord having
different Weights A, B, C, Sc. hung
to it at any given Diſtances A B, BC, Sc. Now
F this Cord be drawn along an Horizontal
Plane by one End D thereof, which End moves
along a. given Curve DP in the ſaid Horizon-
tal Plane, it is maniſeſt, that the ſuid Weights
will ſtretch. the Cord while it is drawing a-
long, and will deſcribe the Curves AM, BN,
CO, Fc. tt 1s required to draw Tangents 40
them, the Magnitude of the Weights, and the
Poſition of the Cord ABCD being given.
In the firſt Particle of Time, the End D
moves forwards towards P, the Weights A,
| ö | 5 B, C,

FLuxieNs. FI
B, | deferibe, or endeavour to deſcribe
the le Sides Aa, Bb, Ct, of Polygons, which
the Curyes M, BN, CO, are th ppos’d to
be; ‘and therefore to draw 4 Tangents 4 B,
B 2 CA, to chem, 1 18 only the Determination
of the Weights A, By C, in the ſaid firſt In-
ſtant or Partiele of Time, or the Poſition of
rhe fra Lines they endeayottt to deſcribe :
To which; we muſt obſerve, As.

1%; That the Weight A, in the firſt Parti-
cle of Time, is drawn in the Direction A B
and ſinee there is nothing to divert it from
that Direction, becauſe it draws no other
Weight irſelf, it muſt keep to that Directi-
on; and ſo the right Line An wil rouch
the Curve AM in A. .

2 That the Weig t B is drawn accord-
ing tothe Direftion BC; but becauſe it draws
the Weight A after it, “which does nor move
in that Direction, and ſo muſt induce ſome
Alterariori to that Direction; the Direction of
B will not be in the Line BC, but in another
Line BG, whoſe Poſition may be thus found.

Deſeribe 2 Rectangle E F, with BC for the
Diagonal, and having one Side BY in the Con-
tinuation of AB; then if the Force wherewith
the Weight B is drawn according to the Di-
rection BC, be expreſſed by BC; it follows by
theLawsof Mechamicks,that the Force BE may
be divided into two others B E and B E, vis.
when the Weight B is drawn by the F orce
BC according to the Direction BC, it is the
fame, as if it was drawn at the ſame er by
the Force BE in the Direction B E
the Force BF in the Direction FN Now
the Weight A gives no Diſturbance to the
Direction BE, it * drawn 5

FTreatiſe

to itz and conſequently the Force I E in that
Direction will receive no Alteration; but it
oppoſes the Force B F in the Direction BF by
the whole Weight thereof. Therefore in or-
der for the Weight B with the Force BF to
overcome the Reſiſtance of the Weight A,
the Force F muſt be divided into two Parts,
having the ſame Proportion to each other, as
the Weights 4 and B. Whence divide E G
in the Point & ſo, that C & be to GE, as the
Weight A to the Weight B; then it is plain
that E & will expreſs the remaining Force
wherewith the Weight B endeavours to move
in the Direction B V, when it has overcome
the Reſiſtance of the Weight 4. Therefore
the Weight B is drawn in the ſame Time by
the Force BE in the Direction BE; and by
the Force EG in the Direction BF or EC;
and ſo it will endeavour to move along B G
with the Force BG: That is, B G will be the
Direction, and — Boren will touch the
Curve BN in B.

3e. To find the Tangent CX. With CD
3s 2 Di onal, make the Rectangle H, the
Side C being in BC continued. Now the
Weight B does not at all diſturb the Force
CH, wherewith the Weight C is drawn in
the Direction CH; but the Force C1 in the
Direction C, is diſturbed the greateſt poſſi –
ble * Weig ht B, and in ſome meaſure by
the Weight A allo. To find out the Quan-

tity of . draw AL perpendicular to B C
continued out. (Here we may obſerve, that if
A expreſſes the Force where with the Weight
A is drawn according to the Direction A B;
B will expreſs the Force where with the ſaid
Weght A 1s drawn in the Direltion BC. *

of FLUXI1ONS.

that the Weight C, together with the Force
C 1, muſt overcome the whole Weight B, to-
gether with a Part of the Weight A, which
is to the Weight A as BL to B 4, or BF to
BC. Therefore if B 7 C:: DR:
KA, it is plain that C & will be the Direction

of the Weight C, and conſequently will be a

Tangent to the third Curve CO in the Point
2 1

If there were a greater Number of Curves,
the Tangents to the fourth, fifth, &c. might
have been found after the ſame way; and the
Tangents of the Curves: deſcribed by the in-
termediate Points between the Weights, may
be found by Art. 30. |

80 * © %\

. 13
Treutiſe
., : N EY , 2 *

S#&./ #4 ” * C n *
1 — N – * Y
1 ju

of the 9 of Paci, in e finding the
greateſt and leaſt Ordinates in a
urve, to. which the Solution of

. 4 Fg de MAXIM * e
ane be reduces. . }

WEEN ag I.

F16. 31, ET MDM be a Curve, whoſe Ordi-
2 nates PM, E D, PM are parallel to each
4, other; and let this Curve be ſuch, that while

the Abſcif AP continually increaſes, the Or-
dinate PM increaſes likewiſe, until it comes
to a certain Point E, and afterwards decreaſes ;
or, on the contrary, if rhe fare decreaſes un-
til it comes to a certain Point E, and after-
wards increaſe.

Then the Line E is called the greateſt or
leaſe Ordinate, 2 a maximum of minimum.

D &+*1%. IL.
| bo ra 8 as PM, be propoſed, conſiſt»

ing of one or more indeterminate Quanti-

ries, A AP; and while AP continually in-

creaſes, the faid Quantity PM increaſes like-

Wiſe until it comes to à certain Point E, after

which it conſtantly decreaſes, or contrariwiſe 4
an

of FLUx10Ns.
and if it be required to find ſuch a Value or
Expreſſion of A, that the Quantity E D,
of which it conſiſts, may be . or leſs
than any other Quantity PM fo
manner from AP. Thus is called a Problem
de maximis and mMinims. t

GENERAL PROPOSITION.

* E Nature of the Curve MD M being

given: to find AE ſuch a Value of AP,
that tbe Ordinate E D be greater er leſſer than
any other PM of the ſame nature.

When AP, PM increaſe together, it is

evident * that Rm the Fluxion of Pm will be * Art. 8,

rmed in like

poſitive with regard to the Fluxion of 4 P. 10.

And on the contrary, if M decreaſes while the
Ordinate AP increaſes, the Fluxion of PM
will be negative. Now every Quantity that
continually increaſes or decreaſes cannot from
being poſitive become negative, without firſt
paſſing thro’ Infinity or nothing; viz. thro’ o
when the Quantity in the Beginning conſtant-
ly goes on decreaſing, and thro’ Infinity when
it continually increaſes in the Beginning.
Therefore the Fluxion of a Quantity Are
ſing a maximum or minimum, muſt be equal to
o, or Infinity. Now becauſe the Nature of
the Curve MD M is given, we can find (by
$e8. 1 or 2.) a Value of Rn, which bein
firſt made equal to o, and afterwards to Inf:
nity; from thence in both the Suppoſitions
the requir’d Value of ZE will be had.

E 4 Scho-

F C. 31,
32

A Treatiſe
SHOLA1U M.

4 HE Tangent in D is parallel to the

Axis AB, when the Fluxion Rm
becomes o in that Point: But when it becomes

34+ ‘* Infinite, the Tangent coincides with the Or-

dinate E D. Whence we may obſerve, that
the Ratio of m R to RM, wiz. that of the
Ordinate to the Subtangent in the Point D,
is o or Infinite. 55
It eaſily appears that a Quantity continu-
ally decreaſing, from poſitive cannot become
negative without firſt paſſing thro’ꝰ o; but that
a — continually increaſing muſt paſs

_ thro? Infinity to become negative, does not fo

F 16. 31,

Art. 10.

FI d. 35.

eaſily appear. And therefore to aſſiſt the Ima-
3 let Tangents be conceived to iſſue

om the Points M, D, M; now in Curves,
where the Tangent in D is parallel to the Axis
AB, it is manifeſt that the Subtangent PT
increaſes ſo much the more, as the Points M
and P accede to D, E; and ſo when M coin-
cides with D, the ſame becomes infinite; and

when at 4 75 AB is greater than AE, the

Subtangent PT from poſitive becomes * ne-
gative, or contrariwiſe.

EXAMPLE I.

LEA * + =axy (AP being x,
PM=y, AB a) expreſs the Na-
ture of the Curve MD M. The Fluxion of
the ſame will be 3 ** + 3935 =axj Ta,

and j dent Lodod o, when the Point P

JJ) —Ax
coincides with the Point Z ſought. Whence
8 | We

of FLUXI1ONS. 57
we get y= — And putting this latter
Part of the Equation for y in the Equation of
the Curve x* +} =a xy, and there will ariſe
AE (x) = 7 I Being ſuch that the Ordi-

nate E D will be a maximum, or the greateſt

of any other Ordinate PM to the Diameter
AB.

EXAMPLE UII.

| — .

en „-a = NA ,, expreſs the F16. 33.

Nature of the Curve MD M. This
—2 23
thrown into Fluxions will be j ===<XEZ

»
37/4 — K

which J firſt make equal to o; but becauſe
this Suppoſition gives us — 2 x 3/a=0, from
which the Value of AE cannot be known, I
wy

afterwards make <<
7a —x

3 M =0. 1 AE (x) Da.

infinite; and ſo

EXAMPLE III.

| Ba AMF be a Semicycloid, whoſe PI c. 35.

Baſe B is leſs than half the Circum-
ference ANB of the generating Circle, and
Centre is C. It is required to Pad the Point
E in the Diameter 4B being ſuch, that the
Ordinate E D ſhall be a maximum, or the
greateſt poſſible.

Draw the Ordinate PM at pleaſure, inter-
ſecting the Semicircle in N, and at the Points
M, N of the Ordinate conceive the little Tri-
angles MR m, Nn; and calling the indeter-
minate Quantities 4 P, x; PN, z; the Arch

13 A N. 545

A Treatiſe \’

AN,u; So the ** Quantities AMB, a;
BF, * CJ or C Then from the Na-
ture of 9 PI ANB (a): BF(d): — N (2):

NM= Whence PM=z+, and

the Fluxion * E 25 = 0,

Fic. 35.

„hen it Point P coincides with F the Point
ſought. Now the right-angled Triangles
Nn, NPC are ſimilar. For if the common
Pat CNS be taken away from the right
Angles CNn, PNS the remaining Angles
Nn, PNC ſhall be equal. And therebore

CN: CP(C -:: Ny (d): Sn (Z) — =
Whenceputting this ValueforZ in 21305
and there will ariſe 22 o, and
ſo ye get x (which in this Caſe is LE) =
; + 5 *

Therefore aſſume CE towards B, a *

Proportional to the Semi · circumference An,

the Baſe BF, and the Radius CB, and the
Point E —— be that ſought.

EXAMPLE IV.

cut or divide the given right Line
AB in the Point E fo, that the Pro-
duct of the Square of AE, one of the Parts
into the other Part, be the greateſt of any
Product of the like nature.

Call the unknown Quantity & Ex; and

the given Quantity AB, a; then muſt AEXEB
Sax -& be a maximum, Now a Curve
MDM

of FLUX10NS.
MD M muſt be ſuppoſed ſuch, that, the Re-

lation of the Ordinate () to the corre-
ſpondent Abſciſs AP (x) is expreſſed by y =

Worms: and the Point E muſt be found

aa
ſuch, that the Ordinate E D be a maximum;

and fo j = — — from whence we
get AE () =g4.

And generally if you would have x” xa—x”
(where m and x expreſs any Numbers at plea-
ſure) to be a maximum, the Fluxion of it muſt
be made equal to o, or Infinity, Whence

my, — —
Mx Xx xa — 4 — 4 xXx = 03
which divided by x xa x,and there

comes out a m mr nx =0, and AE (x)
2
T Ir”
Tf , and #==—1 then will AE Da,
and the Problem may be thus laid down.
Continue ont the given Line AB (towards F 1 C. 37.
B) to the Point E, in ſuch manner that

the Quaniry 24 be a minimum, and not a
maximum; for the Equation of the Curve
MDM will be — =y, wherein if we
ſuppoſe x a, the Ordinate PM, which be-
comes BC, will be = that is, infinitez and
conceiving x infinite one ſhall. have y =x,
viz. the Ordinate will be alſo infinite.
If m=1, and z=—2 then will AE –

whence the Problem may be after this manner
ſtated. ‘ Corp

F. 38.

F 16. 39.

A Treatiſe
Continue out the given right Line AB (to-
wards A) to the Point E ſo, that the Quan-
AE xXTB | | |
tity be the greateſt: of any other

BE”
the like Quantity A} 75 — |

ExAMPLE V.

HE right Line AB being divided in-
F FT to Tu Parts AC, CEE To di-
vide or cut the middle Part CF in the Point
E being ſuch, that the Ratio of the Rectan-
le AEXEB to the Rectangle CE xx E F, be
eſs than any other Ratio formed in like manner.
Call the given Quantities AC, a; C;
CB, c; and the unknown Quantity CE, x:
then will AE Sa x, EB c- x, E F=
b—x, and therefore the Ratio of AE xx EB,
to C Ex EF will be A which
muſt be a minimum. Whence if a Curve
MD M be ſuppoſed ſuch, that the Relation
of the Ordinate PM (/) to the Abſciſs CP (x)
— 440+ Ax -A – ,

be . WN bl
the Problem will be brought to this, viz. to
find {uch a Value CE for x, that the Ordinate
ED be leſs than any other PM of like ſort.
Therefore throwing the ſaid Equation into
Fluxions, andafterwards dividing by 4 x, there
will come out cxx—axx—bxx+2acx—abc=0 3
and one Root of this Equation will ſolve the

Problem. .
If c=a-+5; then will x S5.
Ex AM-

of FLUXIONS,

EXAMPLE VI.

53 Oral the, Cones that can be inſcribed Pee. 40

within a given Sphere, to find that
whoſe convex Surface is the greateſt.

This Problem, in other Words, is to find the
Point E in A B the Diameter of the Simicircle
AFB ſuch, that drawing the Perpendicular EF,
and joining AF, the Rectangle 4FxFE be great-
er than any other Rectangle (ANN) like it.
For if the Semicircle AFB be conceived to
make an entire Revolution about the Diame-
ter AB, it is evident, that it ſhall deſcribe a
Sphere, and the right-angled Triangles AE F,
APN, will generate Cones inſcribed in the

Sphere; the convex Surfaces of which deſcri-

bed by the Chords AE, AN, will be to one
another as the Rectangles A FHF E, A Nx NP.
Now let the unknown Quantity AE x,
and the given one AB==a. Then from the
Nature of the Circle AF=y/ax, E F=
Vax xx; and therefore AAFEE = M aaxx—ax®
which muſt be a Maximum. Whence we muſt
conceive a Curve MD M to be ſuch, that the
Relation of the Ordinate PM ()) to the
correſpondent Abſciſs A P (x) is expreſſed by

Vaaxxrx ax =); and find the Point 75 ſo,

a
that the Ordinate E O be greater than any
other (PM) of like ſort. And making the

1 ä
Fluxion of the Equation — b, We
by 7,44 2 / aAXX—AX

get AE (x) a.

EXAMPLE

F rs. 41.

A Treatiſe
ExamyeLEe VII.

A MoNG all the Parallelepipedons equal to
| a given Cube 4, and having the given
right Line & for one of the Sides, to fir that
which has the leaſt Superficies.
Callone of the two Sides ſought *, _ the

other will 1 7 then aſſuming the alternate
Planes 65 1 of the Parallelepipedon, and

theirSun/ait. $44 — +2 will be half

of the Superficies, which “uſt A a Minimun.
And fo ( * all along) conceiving a Curve ex-

preſſed by ve += +F= =”, the Fluxion of
this Equation muſt be equi 20 #5 that js

x r

2 * * — an comes our

=, and eZ, ; Conſequently mie tluce
. of the Parlleleppedon required, will be
„= and Whence jou may obſerve

that two Sides are equal. 2 8

e Sg. . |

Awene all the Parallelepipedons that

are equal to a given Cube 2 to find

that which has the leaſt Superficies.
Call one of the unknown Sides x; then by
the laſt Example it is plain, that the two bo
ther

ther Sides will be each equal to-; andthere-
fore the Sum of the alternate Planes, which
is the half of the Superficics, will be-

2% which muſt be a Minimum: Whence
the Fluxion of this muſt be equal to , viz.

G & G 3 2
— mT, 05 and ſo x = az.” and” conlſe-
xX | rare

ently the two Sides ſhall alſo be c ual to. 43
Fa that the Cube itſelf ſolves the Probl

0 ‘ :
|

S 3 & %\

ExAMPLE IX.

Tu E Line AE being given in Poſition F 10. 414
on a Plane, together with two ſtable

Points C, F; and if two right Lines CP (u),

PF (z), be drawn from any Point P in it;

and if a Quantity be made up of theſe indeter-

minate ones 4 and ⁊, and other given right

Lines a, b, &c. at pleaſure. It is required to

find ſuch a Poſition of the right Lines CE,

E F, that the Quantity given, which is made

up of them, be greater or leſſer than that ſame
ben it is made up of the right Lines

C2, PE. | 21118 1 dad
Let us ſuppoſe the Lines CE, E F, to have

the requiſite Situation: Join C F, and conceive

the Curve DM to be ſuch, that drawing

P Ja at pleaſure perpendicular to C F, the

Ordinate QM may expreſs the Quantity gi-

ven. Now it is niet, that when the Point

P falls in the Point E, the Ordinate 9 M.

which then becomes O D, muiſt be the greateſt

or leaſt of all others of like ſort. Therefore

the Fluxion of it muſt be equal to o, or infi- |

nite : 1

— — ——
O— — 2
— —— — —
_ by

— — . ——

A T; reatiſe

nite; Whence if the given Quantity, for Ex-
„ is au xz Zz; then will ax 22 5 =0,
conſequently #: -: : 22: 2. Where⸗
fore we may already perceive that z muſt be
egative with reſpect to 4; that is, the right
Lines CE, EF, muſt have ſuch a Poſition
that z decreaſcs at the ſame time as u increa-

ſes. © |

Now if EG be drawn dicular to the
Line AE B, and from any Point & therein, the

Lines CL, GI, be likewiſe drawn perpendi-
culartoCE, EF, and the right Lines C Xe,

Fe H be drawn from the Point e infinitely

near E, and from the Centres C, F, be deſcri-
bed the ſmall Arches EA, E H; the right-an-
gled Triangles E LGand E Ke, E IG and EHE,
will be fimilar. For if the Angle L Ee be ta-
ken from the right Angles @Ee, LEA, the
remaining Angles LEG, X Be, ſhall be equal.
Whence EL: G:: KB (a): He (— ):: 2 z:
a. Therefore the Poſition of the right Lines
CE, EF, muſt be ſuch, that when E G is
drawn perpendicular to the Line AE B, the
Sine (GI) of the Angle GE C, is to the Sine
(EJ of the Angle GE F, as the Quantities
drawn into æ, to the Quantities drawn into .
Which was to be found. ee
Co ROL.

Te the right Line C E be given both in
TY Poſition and Magnitude, a0 EF only
in 8 and the Poſition thereof be re-

quired, it is evident that the Angle & E C be-
ing given, its Sine (CI) will be given like-

wiſe, and conſequently the Sine (G 7) of the

Angle (GEF) ſought. Therefore if a *
1 8

« *
io
oF

}

of FLUXIONS.
be deſcribed with EG as a Diameter, and the
Value of G be laid off in the Circumference
from & to J; the right Line E paſſing thro’
the Point J, will have the requiſite Poſition.
Let au be the given Quantity; then

will EI be 2 and ſo it appears, that

let EC and E F have what Length ſoever, the
Poſition of this latter ſhall be always the ſame,
fincc they do not come into the Value G1,
which conſequently does not vary. If a=,
it is plain that the Poſition of EF muſt be in
CE continued out from E; becauſe GLI,
when the Points C and F fall on each Side the

Line 43: But when they fall on the ſame

Side, the Angle FE & muſt be aſſumed equal
to the Angle CEG, .

EXAMPLE X.

Tus Circle AE being given in Poſi-

| tion, as allo the Points C and F with-
out the ſame: to find the Point E in the Pe-
riphery being ſuch, that the Sum of the right
Lines CE, EF may be a minimum.

Suppoſe the Point E to be that ſought, and
draw the Line OEG from the Centre 0;
which will be perpendicular to the Circum-
ference AEB; and fo * the Angles FEG,
CEG will be equal. Therefore if EH be
drawn fo, that the Angle E Z7O be equal to
the Angle CEO, and likewiſe EX fo, that
the Angle EXO be equal to the Angle FEO,
and the Parallels ED, EL to OF, OC; and
there will be formed the ſimilar Triangles

OCE and OEH, OF and O ER, HDE

and XLE; and calling the known Quantities
F *

F1c. 42.

Art. 37.

F 1G. 43.

1 Art. 56,

A Treatiſe

OE, OAorOB,a;z OC, l; Oe; and the
unknown ones OD or LE, «x; DE or OL, y.

Then will OH==, OK = _ and HD

(+—®):DE (): KI (» —=):LE(s).

W hence xx — =} — — and this

is an Equation ei to an Hyperbola,
Il

which may be eaſily conſtructed, and will cut
the Circle in the Point E ſought.

EXAMPLE XI.

A TRAVELLER ſetting out from a

Place C to go to a Place F, muſt croſs
two Countries divided from each other by the
right Line AEB. Now ſuppoſe him to go
the Length a in the time c in the Country ad-
Joining to C, and the Length & in the ſame
time c in the other Country adjoining to F: it
is required to find the Point E in the right

Line AEB thro’ which he is to paſs in the

ſhorteſt time poſſible from C to F.
Make a: CE(a)::c:<. Andb:EF(s)::
a

CZ CU

75 Then it is plain, that expreſſes the

Time of the Travellers going the Length CE, and

= the Time of his going the Length EF; fo

that . muſt be a minimum. Whence *
1

drawing EG perpendicular to AB, the Sine

of the Angle G EC muſt be to the Sine of the
Angle GEF, as a to b. >
| 8 is

of FLUXIONS.
This being premiſed, if the Circle C GH

be deſcribed with the Radius EC from the
Point ſought E as a Centre, and the Perpen-

diculars CA, HD, FB be drawn to the right
Line AEB, and the Perpendiculars GL, GI

to the right Lines CE, EF, we ſhall have
a: b:: CL: GI. But GA, and GE,

becauſe the right-angled Triangles GEL and

ECA, G EI and EV are equal and ſimilar,
as may be eaſily proved. Therefore if the un-
known Line AE be called x, we ſhall

have ED ===; and calling the known Lines
A B, 7; AC, 2; BF, h: from the Similari-

ty of the Triangles EB F, ED H, EB (f—x):
BF (Y): ED (= :DH= — . But

af—ax

becauſe of the right-angled Triangles E D H,

EAC, of equal Hypothenuſes E Ii EC,

ED +DH will be EA + A1C,, chat is,

bbxx , © bbbbxx | 1-4 %
aa aaf Zaafæ Þaaxx PTS

freeing the Equation from Fractions, and af-
terwards duly ordering the ſame, there will ariſe

aaxt— 2aafx*T-aaffxx—0afggxF-aaffgg=0.

—bb +2bbf bag

— bb
—bbbb

This Equation may be gotten after the fol-

lowing manner, without having recourſe to

the oth Example.
Having named the known Lines as be-

fore, viz. AB, ; AC, g; BF,; and the un-

known one AE, x. Make 4: C E(y/gg+xx)::c:

C 5 .
Ln = to the Time the Traveller 15 go-

F 2 ing

AX Treatiſe

ing the Length CE. And in like manner

b:EF F- ZF T TH): 2c:

. — * „ Tiime of his going

the Length E F: ſo that — . **

7 amines und cherer

** cx –

en ay/gg 7; xx T byff—2fx + xx ++ Þh

Whence dividing by cx, and freeing the E-

= O.

quation from Surds, we ſhall have the ſame

Fic. 34.

Equation as before; one Root of which will
expreſs AE the minimum ſought. AR

E x A * Y K. E XII.
60, L K F be a Pull hanging freely from

the End of a Cord CF faſten’d in C,
and let D be a Weight ſuſpended by the Cord

DF put over the Pulley F, which Cord is

faſten’d in B; ſo that the Points C and B lie
in the ſame horizontal Line CB. Nov if the
Pulley and Cords be ſuppoſed to have no
Weight, it is required to find in what Place

the Weight D, or Pulley E, will ſettle or

come to reſt. “PE
By the Principles of Mechanicks, it

is plain that the Weight D will deſcend

as far as poſſible below the horizontal
Line CB: therefore the Plumb Line D FE
will be a maximum. And therefore calling the
given Quantities CF, a; DFB,b; CB, e
and the unknown Quantity CE, x; there will
ariſe EF=y/aa —wx, EBV c z,

and DFE=b—yasa A c – 2c + e -x,
which muſt be a maximum. And fo the Flu-
An | xion

of FLUxX1oNSs.

Cx & *

xion of it will be ox

yaa Fre ——2ox Vaa — xx
So. Whence there ariſes 2cx* — 2ccxx —
aaxx+aacc=0o, and dividing by x -,
there comes out 2cxxX – aa – aa o, one

Root of which will expreſs CE ſuch, that

the Perpendicular E O paſſes by the Pulley F
and the Centre D of the Weight, when the
ſame is at reſt.
| Here follows another Solution of this Prob-
lem.

Call EFy; BF, z; then will þ—z+y=
maximum, and ſo j =z. Now it is evident,
that the Pulley F does deſcribe a Circle CFA
about the Point C as a Centre: and if FR be
drawn from the Point f, infinitely near to F,
parallel to CB, and F perpendicular to BF;
therefore will FR , and FS =z, Which
are conſequently equal to each other: and fo
the little right-angled Triangles FR, FS,
having the common Hypothenuſe Ff, are
equal and ſimilar : whence the Angle RF is
equal to the Angle SE, that is, the Point F
muſt be ſo ſituate in the Periphery FA, that
the Angles made by the right Lines EF, FB,
with the Tangents in F, be equal to each o-
ther: or elſe (which comes to the ſame) the
Angles B FC, D FC __ |
This being premiſed, if you draw FH ſo,
that the Angle FHC be equal to the Angle
CF or CFD; the Triangles CBF, CFH
will be ſimilar; as alſo the right-angled Tri-
angles E CF, EFA, ſince the Angle CFE is
equal to the Angle FH E, each of them be-
ing the Complement of the equal Angles EC,
CFO to two right less and conſequentl

= | 3 i

Fic. 34.

cf N + xx + bh

fore—

A Treatiſe

ing the Length CE. And in like manner

b: EF Y xx +b0)::
A — I = Time of his going

the Length E F: fo that * +

7 =a minimum: and there-

c & cr -t

agg A x T by ff fx +> wx ++ bb
Whence dividing by cx, and freeing the E-

= 0.

quation from Surds, we ſhall have the ſame

Equation. as before; one Root of which will
expreſs AE the minimum ſought.

Ex AMY LE XII.

L 1 F be a Pull hang ing they
the End of a Cord Ep in _
and let D be a Weight ſuſpended x the Cord

D FB pur over the Pulley F, which Cord is

faſten’d in B; ſo that the Points C and Blie
in the fame horizontal Line C B. Now if the
75 and Cords be ſuppoſed to have no
— it is required to find in what Place
the the Weight D, or Pulley F, will ſettle or
come to reſt.
. By the Principles of Mechanicks,
is plain that the Weight D will 2
as far as poſſible below the horizontal
Line CB: therefore the Plumb Line D FE
will be a maximum. And therefore calli N
given Quantities CF, a; D B, ; C
and the unknown own Quantity CE,x; there wil
ariſe EF=y/aa — ws, EB = = i,
and DFE=b— yaa c – c + yaa—x*,

which muſt be a maximum. And ſo the Flu-

xion

of FLUXIONS.
Aon Gl. it will be = nn
yaa ec —2x Vaa — xX
So. Whence there ariſes 2c * — 2c RR —
aaxx+aacc=o0o, and dividing by x—c,
there comes out 2:xX —aax—44c = 0, one

Root of which will expreſs CE ſuch, “that

the Perpendicular E D nes by the Pulley F
and the Centre D of he Weight, when the
ſame is at reſt.

Here follows another Solution of this Prob-
lem.

Call EF,y; BF, 2; then will b—z+y=
maximum, and ſo j Z. Now it is evident,
that the Pulley F does deſcribe a Circle C FA
about the Point Cas a Centre: and if FR be
drawn from the Point f, infinitely near to E,
parallel to CB, and FS perpendicular to BF,
therefore will FR = =, and FS =. Which
are conſequently equal to each other: and fo
the little right-angled Triangles FRV, FS,
having the common Hypothenuſe Ff, are
equal and ſimilar: whence the 3 RF is
equal to the Angle S E/, that * the Point F
muſt be ſo ſituate in the Periphery FA, that
the Angles made by the right — EF, FB,
with the Tangents in F, be equal to each o-
ther: or elſe (which comes to the ſame) the
Angles B FC, D FC equal.
This being ee if you draw FH ſo,
that the Angle FHC be equal to the An le
CFB or CD; the Triangles CBF,CFH
will be ſimilar; as alſo the right-angled Tri-
angles E CF, E FH, ſince the Angle CFE is
equal to the Angle FH E, each of them be-
ing the Complement of the equal Angles gh
CF D to two git _ 3 and PTY 275

ie
chend FO KFo) ::
EC (NY). Whence 4. — r = aa — iy from |

the Nature of the Circle: from whence ariſes
the ſame Equation as at firſt.

EXAMPLE XIII.

F. 45. 61. T* E Elevation of the Pole being given:
= To find the Day of the Year where-

| in the Twilight will be the ſhorteſt poſſible.
[| Let C be the Centre of the Sphere;
. APTOBHO® the Meridian; ZDM#O, the
Horizon; I the Crepuſcular Circle pa-
rallel to the Horizon; 4M NB the Equator
FED that Part of the Parallel to the Equa-
tor (the Sun deſcribes the Day wherein the
Twilight is ſhorteſt) contained between the
| Planes of the Horizon and the Crepuſcular
Circle; P the South Pole; PEM, PDM
Quadrants of Circles of Declination. Now
i the Arch Hg or OT of the Meridian com-
prehended between the Horizon and the Cre-
uſcular Circle, and the Arch O P of the E-
evation of the Pole are given: and conſe-
quently their Sines CF or FL or © X, and
V. Now we muſt find C X the Sine of the
Arch EMor DN of the Sun’s Declination,

| when he deſcribes the Parallel EWP.

If you ſuppoſe another Part fedg of a Pa-
rallel to the Equator infinitely near to FEDG,
and draw the Gates Pen, P4n: it is ma-
nifeſt, that the Time the Sun takes up in de-
| ſcribing the Arch ED muſt be a minimum, and
the Fluxion of the Arch M being the Mea-
ſure of it, and becoming #2» when E be-
15 _ comes

of FLUXIONS. 7I
comes ed, muſt be equal to nothing; whence
the ſmall Arches Mm, Nu, and conſequently
the little Arches Re, Sd, will be equal to
each other. Now the Arches RE, & D bein
contained between the ſame Parallels ED, e d,
are equal likewiſe, and the Angles at & and R
are right ones. Therefore the lictle right-
angled Triangles ERe, DS (conſider’d as
right * lin’d Triangles, on account of their“ 4r:. 3.
Sides being infinitely ſmall) will be equal and
ſimilar: and conſequently the Hypothenuſes
Ee, Dd ſhall be equal likewiſe.

This being premiſed, the right Lines DE,
EF, dg, ef, the common Sections of the Planes
FEDG, fedg, parallel to the Equator, and
the Horizon and Crepuſcular Circle, will be
perpendicular to the Diameters HO, Q, be-
cauſe the Planes of all theſe Circles are per-
pendicular to the Plane of the Meridian and

parallel. Therefore / Dd 2. 2

—2 —2 WF,
=V Ee F, or fe FE. Now it is plain
(from Art. 70.) if two Ordinates in a Semi-
circle be drawn infinitely near, the fmall Arch
intercepted between them will be to their Flu-
xion, as the Radius is to the Part or the Dia-
meter intercepted by the Centre and thoſe Or-
dinates. Whence (becauſe of the Circles
HO, SET) CO:CG:: Dd or Ee: DG—dg
or fe—FE::] F:: CO, or OX:
| CG TF or GL. But becauſe of the right-
| angled fimilar Triangles CFO, C KG, FLG:
| therefore CO: C G:: OV: GK. And GK:
G L:: CK: FL or 9X. Whence OV: CX
\ ;OX;X9:;X9: X Hby the Nature of the
8 | F 4 Circle:

„

Circle: that is, if © X be taken for the Ra-

the Angle H
the Arch 1. U 1 is ſuppoſed to

A Treatiſe

dius in the right- angle Triangle © X H, Lon
9 Degrees, (becauſe

be 18 De > and you make as the Radius
is to the ys Be of an Angle of 9 De-

rees, ſo is the Sine of the Elevation of the
Pole to a fourth Sine; this will be the Sine of
the Sun’s Southern Declination that Day of
the Year the Twilight is the ſhorteſt poſſible.
So that if you take o. 8002875 from the Lo-
garithm of the Sine of the Elevation of the
Pole, the Remainder will be the Logarithm
of the Sine ſought.

SECT.

of FTLVUxIONS. 73

SECT. .

Of the Uſe of Fluxions in finding
the Points of Inflexion and Retro-
greſſion of Curves.

BE. Aus E ſecond, third, 6c. Fluxions are
V uſed hereafter; before we go further, we
think it neceſſary to define them,

DEzrINITION I.

Tue infinitely finall Part 8 by the

continual increaſing or decreaſing of the
Fluxion of a variable Quantity, is called the *
Fluxion of the Fluxion of that Quantity, or
ſecond Fluxion. So if a third Ordinate 1 be F 1 C. 46.
ſuppoſed infinitely near the ſecond p, and
& be drawn parallel to AB, and mH to RS.
Vn is the Fluxion of the Fluxion Rm, or the
ſecond Fluxion of P M.
In like manner, if a fourth Ordinate of be

ſuppoſed infinitely near the third 29, and 7
be drawn parallel to AB, and L to S, the
Difference of the ſmall right Lines Vn, Lo, is
the Fluxjon of the ſecond Fluxion, or the third
Fluxion of PM. Underſtand the fame of

others.

| * See the Tranſlators Preface.

OB5sE R»

———6—jV— ——
PIR THEE ER __
1

7 ; .

2 A Treatiſe N

OBSERVATION.

Fun of Fluxions are denoted with Dots
over the Letters, being ſo many in Number
as the Order of the Fluxion is. For Example,
X is the ſecond Fluxion of x; & the third Hu-
ion of x; & the fourth Fluxion of x, 69c. Un-
derſtand the ſame of y, or any other variable
Quantity. And ſo if MP y, then will y ex-
preſs Hn; 5, Lo—Hn or Hn—Lo, &c.

CoRoLLARY I.

62.Jr each of the Abſciſſes AP, Ap, 44, Af,
be called x, and each of the correſpon-
dent Ordinates PM, pm, gu, fo, ); and every

of the Parts of the Curve AM, Am, An, Ao,

25 then it is plain that & will be the Fluxions

Pp, 24, / of the Abſciſſes; j the Fluxions

Rm, Fun, To, of the Ordinates; and # the
Fluxions Mm, mu, u, of the Parts of the

Curve AMD. Now in order (for Example)

to get the ſecond Fluxion Hun of the variable
Quantity PM, we muſt ſuppoſe two little
Parts Pp, pg, in the Axis, and two others
Mm, mn, in the Curve, to get the two Flu-
xions Rm, Sn; and therefore if the two ſmall
Parts Pp, p, be conceived. as equal to one
another, it is evident that x will be a ſtanding
Quantity with reſpect to q; and 4, becauſe Pp
which becomes pg, will not vary while Rm,
which becomes Su, and Mm which becomes
mn, does. If the little Parts Mm, mn, of the
Curve, be ſuppoſed to be equal to each other,
then will 4 be a ſtanding Quantity with re-
ſpect to x and ) And laſtly, if Rm and 1
a b up

of FLUX1ONS.

tity with regard to & and i, and the Fluxion
Hn (J) of it will be equal to nothing. |

In like manner, to get the third Fluxion of
P M, or the Fluxion of the ſecond Fluxion
Hn, we muſt imagine three little Parts Pp,
Pq, 55 in the Axis; three others Mm, mn,
n, in the Curve, and three others Rm, Sn,
To, in the Ordinates: Then will &, a, or 5,
be a conſtant or ſtanding Quantity, according
as the ſmall Parts Pp, pq, qf, or Mm, mn,
0, or Rm, Su, To, be ſuppoſed equal. The
ſame myſt be conceived of the fourth, fifth,
Sc. Fluxions.

All this is to be underſtood alſo in Curves, Pid. 47.

as AMD, whole Ordinates BM, Bm, Bn,
all iſſue from a ſtable Point B. As, for Ex-
ample, to get the ſecond Fluxion of B M, we
muſt conceive two other Ordinates Bm, Br,
making infinitely ſmall Angles M Bm, m Bn,

then if the ſmall Arches of Circles MR, mz S,

be deſcribed from the Centre B, the Difference
of the ſmall right Lines Rm, Su, ſhall be the
ſecond Fluxion of BM. And the ſmall Ar-
ches MR, #4S, or the ſmall Parts Mm, in u,
of the Curve, or, finally, the little right Lines
Rm, $n, may be taken as ſtanding Quantities.
Underſtand the ſame of third, fourth, &c. Flu-
xions of the Ordinate BM.

$SCHOLIUM.

== we are to obſerve, 15, That if Fre. 46,

there be ſeveral Orders of infinitely
ſmall Quantities: For Example, R M will be
infinitely ſmall with reſpect to PM, and infi-

nitely great with reſpect ro Ny. In like man-
1 NCT, 7

ſuppoſed equal, then will j be a ſtanding Quan-

F 1c. 48,

A T1 reatiſ, 7
ner, the Space MP pm is infinitely ſmall with
regard to the Space 4 PM, and infinitely great
with regard to the Triangle M Rm. *
2, That the whole Fluxion Pf is likewiſe

infinitely ſmall with regard to 4 P; becauſe
every Quantity which is the Sum of a finite
Number of infinitely ſmall Quantities, as Pp,
29, , with reſpe& to another A, will
be infinitely ſmall with regard to the ſaid Quan-
tity: And that in order for it to become of
the ſame Order, it is neceſſary for the Num-
ber of Quantities of the Order next below it,
and of which it conſiſts, to be infinite,

Conor, II.

Tur ſecond Fluxions, in all the Caſes
poſſible, may be repreſented thus:

1, In Curves whoſe Ordinates R, 1c,
are parallel, continue out the ſmall Side Mm
to interſect the Ordinate S in H, and deſcribe
the Arch n from the Centre m with the Di-
ſtance u, and draw the little right Lines 1,
li, ł cg, parallel to m & and & n. This being
done, if x be ſuppoſed a ſtanding Quantity,
viz. M Rm &, it is plain that the Triangle
mA is ſimilar and equal to the Triangle
MRm; and fo Hu is ); that is, the Diffe-
rence of Rm and Sun, and H= i. But if i be
conceived to be conſtant, viz. Mm mn or
m, then it is plain that the Triangle gk is
ſimilar and equal to the Triangle MR m; and
ſo kc=y, and Sg or cuy=#, Laſtly, if ) be
ſuppoſed invariable, viz. m Rn, then will
the Triangle mi be equal and ſimilar to the
*. MR m; and ſo i & o1 =, and
72 | 2

2% In

20, In Curves whoſe Ordinates BM, Bm, F 1 6. 50,
Bu, iſſue from a Point B. Deſcribe the Arches 51

MR, m, from the Centre B; theſe may be

looked * upon as ſmall right Lines perpendi- * Arr. 3. |

cular to B n, Bu. Continue out Mm to E,
and deſcribe the ſmall Arch »& from the Cen-
tre mn, with the Diſtance n; make the An-
le Em m Bn, and draw the little right
Lines 2 Ii, kcg parallel to & and Sn. This
being done, becauſe the Triangle B is right-
angled at S, the Angle B Sm BA, or +
Em Ha right Angle; and therefore the An-
gle BM E =a right Angle Sm H. It is allo
equal to the right Angle MRM + R Mm,
ſince it is external to the Triangle RM.
Therefore the Angle 8m = R Mm.

Hence, if you ſuppoſe & invariable, that
is, the little Arches MR, m &, equal, the Tri-
angle Sm A will be ſimilar and equal to the
Triangle R Mm, and fo Hn=y, and HAK i.
If A be ſuppoſed inyariable, the Triangle gmt
will be equal and ſimilar to the Triangle Rn;
and fo kc will expreſs j and Sg or cn, x. Laſt-
ly, if y be ſuppoſed invariable, the ‘Triangles
i ml, R Mm, will be equal and ſimilar; and ſo
iS or Inu , and Ie i.

by PRO. I.

65 7 find the Fluxion of a Quantity conſiſt-
ing of Fluxions.

Conſider any one of the Fluxions of the given

Expreſſion as 1nvariable, and proceed with the
others as variable Quantities, according to the

Rules laid down in Section the firſt. .
For

A Treatiſe ©
For Example, if + be ſup oſed invariable,
the Fluxion of 22 will be 27, and taking

Jas invariable, it will te ge In like
manner er, taking 5 4 25 invariable, the Fluxion

of . Wibe NN

vx x gl
and dividing it by 3 x, there will ariſe

zx3+z yp +29)
— and ſu poſin 7 inyariable

+) * F
ZXXX

i will te eke -r

the whole divided by xx, and * will be
e |

Ki wx OY

Taking & as conſtant, the Fluxion &f
—.— will be ) /x3+35— _
VII r yr

which divided by x. Jy, and then it will be

ax trite xY ; and ſuppoſing j to be in-

an it will be 2 —
7? 5 * *

In like manner, the Fluxion of

D, e ſuppoſing & ins

8 —
variable, will W TER 2 Mat xa 15 .

1 0

Here we muſt obſerve, „That in the laſt Ex-

by e, J cannot be ſuppoſed invariable; for if

e, its Fluxion y will be nothing and

. muſt not come into the Quan-
tity —

DEF t-

of FiUx10NS.
DETINITION II.

HEN a Curve AFX is partly convex, Fo. 52,
and partly concave, to a right Line AB 53, 54455:
or Point B; the Point F diffding the convex
from the concave Part, or the End of the one,
and the Beginning of the other, is called the
Point of [nflexion, when the Curve being come
to F, continues its Courſe towards the ſame
Parts; and the Point of Retrogreſſion, when the
Curve returns back again towards the Place
of its beginning. Ak-

PRO pp. II.
A General PR O L E M.

66 TH E Nature of the Curve AFK being
1 given, to determine the Point of Tufle-
*ion or Retrogreſſion F.

Firſt, Let us ſuppoſe the Curve AF to FIC. 52,

have the right Line 4B as a Diameter, and 53˙
the Ordinates P M, E F, Fe. parallel to each
other. Now if you draw the Ordinate FE
from the Point F, and the Tangent FL; and
another Ordinate MP from the Point M in
the concave Part AF of the Curve, as like-
wiſe a Tangent AT. Then it is evident,
14, InCurves that have a Point of Inflexion,
that while the Abſciſs A conſtantly increa-
ſes, the Part 4T of the Diameter, intercept-
ed between the Vertex of the Diameter 4and
the-Point ‘7, where the Tangent meets the
Diameter, does likewiſe increaſe until the Point

P falls in E, after which it continually decrea-

/ *

— K ——ů rr—

— —
— 8 — —

8 22 K OT ] . — – mo”
y—_ . _— —— — — . 2
SILENT a
0 »

rD — — — ———— — — — AI 4. ye

A Treatiſe

ſes: therefore 41 muſt become a maximum

AL, when the Point P falls in E the Point
ſought.

2%. That in thoſe Curves that have a Point of
Retrogreſſion, the Part AT continually in-

_ creaſes, and the-Abſciſs AP, till the Point T

Art. 47.

falls or coincides with L and afterwards it con-

ſtantly decreaſes; whence AE mult be a maxi-

mum, when the Point T coincides with L.
Now call AE, æ; EF); then will 41.

| Cha; and the Fluxion of this will be

95 Oh 12, A, ſuppoſing #invariable: which

(being divided by # the Fluxion of AE) muſt
be , or infinite: whence — is = 0
or Infinity: and fo multiplying by 55, and di-
viding by — 9, there comes out ) o, or infi-
nite. Now with this laſt Expreſſion, and the

2 Equation of a Curve, the Point of
nflexion or Retrogreſſion F may be found.

For the Nature of the Curve A FR being gi-

ven, we ſhall have a Value of j in x, and throw-
ing that Value into Fluxions, ſuppoſing x in-
variable, we ſhall get a Value of 5 in xx, which
being made equal to o, and afterwards to In-
finity : by means rhereof, in either of theſe
Suppoſitions, we may find AE fo expreſſed,
that its correſpondent Ordinate E ſhall inter-
ſe& the Curve in the Point of Inflexion or
Retrogreſſion F. 3
The Point 4 whereat the x* begin,
may be ſo fituate that 4 L = X —

SR inſtead of _ —x, and AL or AE a mi-

nimum inſtead of a maximum. But as the

Conle-

of FLUX1ONS. 81
Conſequence is always the ſame, and there is
no Difficulty ariſing from it, I ſhall ſay no
more thereof.
Here we muſt obſerve that 4 L can

never be = x + = for when the Point
T falls on the other Side the Point P, with
reſpect to A, the Value of 1 will be nega-
tive, (by Art. ro.) and conſequently the Va-
lue of — will be poſitive; ſo that in this

Caſe we have ſtill AE+ELor AL =x A

The fame things may be found after a diffe-
rent way. It is plain, taking x as invariable, Fi o. 48,
and ſuppoſing the Ordinate y to increaſe, that 49.
En is leſs than & Hor Rm in the concave Part
of the Curve, and greater in the convex Part:
therefore Hu (55) which is poſitive, muſt be-
come negative in the Point of Inflexion or
Retrogreſſion F; and conſequently * in that At. 47.
Point it muſt be nothing or infinite. |

Secondly, Let the Ordinates B M, B F, BMF 1c. 54, |
of the Curve A FX all meet in the ſame Point 55:
B. If any Ordinate B M be drawn, and the
Tangent MT meeting BT perpendicular to F. 56, i
BM in the Point 7; and if be taken infi- 57
nitely near M, and the Ordinate Bm, Tan-
gent 7, the Perpendicular Hr to B m meeting
MT in O, be all drawn: then ſuppoſing the
Ordinate B M, becoming Bm to mcreale, it
is plain, in the concave Part of the Curve,
that BY is greater than BO, and in the con-
vex Part the ſame is leſs. So that under the
Point of Inflexion or Retrogreſſion F, O:
muſt change from negative to poſitive. *

| is

A Treatiſe
Fic. 56. This being premiſed, about the Centre B

deſcribe the imall Arches MR, TH, and there

8 will be fortn’d the ſimilar Triangles m RM,
| MBT, THO, and the little ſimilar Sectors
BMR,BTH. Now if you make BM y;

| MR x. then will R (5): RM(5):: BM(9):

BT = E:: MR () : THS =: 7 XX

ny ea
HO ===. Now if you throw 272 ( 29
into Fluxions, ſuppoſing invariable, there

| comes out Bt—BTor Ht = =) .and

OS. 0
therefore OH Ht or Ot == . .

J
Whence multiplying by 5*, and dividing by
x, the 2 275 —yy mult wy no-
thing or infinite in the Point of Inflexion or
Retrogreſſion F. Now when the Nature of
the Curve (Fig. 54, 55.) AFK is given, we
ſhall have Values of 5 in x, and of 7 in *,
which being ſubſtituted in & + j*— yy, there
will ariſe a Quantity, which being made equal
firſt to o, and afterwards to Infinity; by
means thereof we may get ſuch a Value of
B F, that if a Circle be deſcribed from B with
the ſame as a Radius, it ſhall cut the Curve
FK in the Point of Inflexion or Retrogreſ-

ſion F.
F 16.509, Laſtiy, To find the ſame things moreover

another way, you muſt conſider that the An-

ole Bm E is greater than the Angle BN in

the concave Part of the Curve; and the ſame

| is leſs in the convex Part: and therefore the
F10. 5% Angle ByE—Bmn or Enn, that is, the
Arch En, being the Meaſure of it, coy

| | om

1
1
\

of FLUxXIONSS. 83
from poſitive to negative in the ſought Point
F. Now taking & as invariable, becauſe of

the right- angled ſimilar Triangles Hm S, Hnł,
we have Hm (): m & (*):: Hun (—5) : 14

Where you muſt obſerve, that 77s is

negative, becauſe as Bm () increaſes, R m (7
decreaſes. But becauſe of the ſimilar Sectors
Bm, mEk, therefore will B (0): S (::
mE (u): E = 7 and ſo Ek +knor En

N- ν
3 |

and dividing by x, it follows that # —y7 or F105: 54,

* + j* — muſt change from poſitive to ne- 53:

gative under the ſought Point F.

If y be ſuppoſed to become infinite, the
Terms & and 5* will be nothing with ref
to yy; and conſequently the Form * + ;*
—yj=0, or infinite, will be changed into
this — 5) o, or infinite; that is, dividing
by — 7, J will be =o, or infinite, which 1s
the Form of the firſt Caſe; and muſt be

ſo likewiſe, becauſe the Ordinates B M, BF,
EM do then become parallel.

Whence multiplying by 5

Coeur.

Wurx o, it is plain that the Flu- Fo. 52.
| xion of AL muſt be nothing with

regard to the Fluxion of AE; and therefore

the two infinitely near Tangents FL, FL, will

coincide and make but one right Line f FL.

But when y = Infinity, the Fluxion of AL F 1c. 53;

muſt be infinitely great in reſpect to that of

AE, or (which is the ſame thing) the Fluxi-

on of AE 1s infinitely little with reſpect to
| G 2 the

3 Na 3

A Treatiſe
the Fluxion of AL; and conſequently two
Tangents FL, Fi may be drawn from the

Point F, making an infinitely ſmall Angle

Fic. 56,
57+

FI c. 58.

LFI.

In like manner, when K + j* — 33 =o, it
is plain that Of muſt become nothing in re-
gard to MR; and ſo the two infinitely near
Tangents MT, mt, muſt coincide, when the
Point M becomes a Point of contrary Flexion
or Retrogreſſion; but on the contrary, when
** ＋5 — 93 Infinity, Oz muſt be infinite
with reſpect to MR, or (which is the ſame
thing) MR infinitely ſmall with regard to O::
and conſequently the Point #2 muſt coincide
with MN, that is, one may draw two Tangents
thro? the Point M, making an infimtely ſmall
Angle with each other, when that Point be-
comes a Point of Flexion, and contrary Re-
tro 10n.

ence it follows, that the Tangent to the
Point of Inflexion or Retrogreſſion I, conti-
nued out, does both touch and cut the Curve
A FK in that Point.

.
EJ AMP LE I.

L*. AFK be a Curve, the right Line

AB being a Diameter of it, and let
the Relation of any Abſciſs AE (x) to its cor-
reſpondent Ordinate EF (y) be expreſſed by

this Equation axx =xxy-+aay. It is re-

2 to find ſuch an Expreſſion for A E, that

the Ordinate EF ſhall interſe& the Curve

AF in the Point of Inflexion F.

The Equation of the Curve is y ===;

y
a

of FLUX10NS.

» 3 bot
and therefore j e, and finding the
xx + aa |
Fluxion of this Quantity, ſuppoſing x, to be
invariable, and making it equal to nothing, there

— v y ©
a.” 4 3.2 Sa Xxx +44
will ariſe 22 * ** ＋a – a’ xx + “WY

XX Tas £4
This multiply’d by xx + aa, and afterwards
divided by 24* x* xxx +aa, will be xx+aa—4xx
So. Whence AE (x) = <.

If zaa be put for its, Equal — in the Equa-

tion of the Curve y , we ſhall have
xx + as _ :

EF (OY) ia; fo that the Point of Inflexion

F may be determined without ſuppoſing the

Curve AFA to be deſcribed.

If AC be drawn parallel to the Ordinates
EF, and made equal to the given right Line
a, and C & be drawn parallel to AB, the ſame
will be an Aſymptote to the Curve AFK.
For if x be ſuppoſed infinite, we may take xx
for xx +44; and therefore the Equation of

axnx bo Fa
22 will be changed into

the Curve y =
this y .

EX AMS LE II.

69.F + T } —4 S — 4. Theny ==
and = 56 =—=2z ; Tuppo-

27 7
ſing + invariable. Now if the latter Expreſ-
ſion be made equal to o, there will come out
—6 x*=0. Which determines nothing;
therefore this laſt Expreſſion muſt be 2
infinitely great; and conſequently the Deno-

| G 3 nomina*

Fic. 59.

_ A Treatiſe
minator 25 = thereof infinitely ſmall, or

5 o. Whence the unknown Quantity AE
(x) a.

EXTAMPLE III.

70.F ET AFA be a Semi- cycloid, the Baſe
B K of which let be greater than the
Semi- circumference AD B of the generating
Circle, whoſe Centre is C. It is required to
find the Point E in the Diameter AB, from
which the Ordinate EF being drawn, will cut
the Cycloid in the Point of Inflexion F.
Call the known Quantities ADB, a; BA, 6b;
AB, 2c; and the unknown ones AE, x; ED, z;

the Arch 4D,u; EF,y. Then from the Na-
ture of the Cycloid y =z + 2 : and there-

fore) .. Now by the Nature of the

—

. Is — „ C > x
Circle we have 2 = Vacx— xx, == — 1
V 20x —XX

and i; (VT Z=) — — Whence ſub-
ſtituting for 2 and # their Equals, we have

J = Ez fr Fluxion of
AY 20x == XX :
which (ſuppoſing x invariable) will be
ber — acc — ec x

cd. From whence an-
Zen -N Xy/ RCX == XX. 8

ſes AE (*) A and CEN.
It is manifeſt, if there be a Point of In-

flexion F, that C muſt be greater than a; for

ik it be leſs, CE will be greater than CB.
I

Ex Au-

of FLUxions. 87
ExamPeLEe IV.

L* AFA be the Conchoid of Nico- F 1c. 60.
medes, and the Point P the Pole, and
the right Line B C the Af mptote. It is re-

to find the Point of Inflection F. Now
e Nature of this Curve is ſuch, that a right
ine PF drawn from any Point Fin it to the
le F meeting the Aſymptote BC in D; the

Liv F thereof is always equal to a ſtanding
ine 4

L
w PA perpendicular, and EF parallel
to BC, and ca he known Quantities 4B or
FDa; BP, b; and the — ones BE, x ;
EF, y; and draw DL parallel to BA. Then
becauſe of the ſimilar Triangles DLE, P EF,

DLG): LF LF (Vaa—xs):: PE (b+x):EF( 5
I — The Fluxion of this will be

———
—

5 e Now if you take the Fluxion

** al—xx
4 it again, and make the ſame equal to nothing,

n ——
aax*—XX Aaa—XxX

which may be brought down to x*+ 3bxx—
2aab=0; and one of the Roots Gut. wall |
be the Quantity of BE ſought.

If a=b, the Equation aforegoing will be-
come & 3axx—24*=0; which divided by
ra, and there comes out xx-2ax—244=0 3

and ſo BE (x) =— a+y3aa.

Otherwiſe.

Conceive the Lines PF iſſuing from the Pole
F as Ordinates, and uſe the Form yy=x * Art. 66,
+5?, where is ſuppoſed invariable. Nowif

G 4 you

== 0g

Centre P. Call the known Quantities

ou conceive another Ordinate PF making an
infinitely ſmall Angle FP with PF, and de-
ſcribe the ſmall Arches FG, DAH, $3) BR

| 543
BP, &; and the unknown ones PF, y; PD, z.
Then from the Nature of theConchoid y=z+ a+
and fo . Now from the right- angled Tri-
angle DBP, DB N, and becauſe of
the ſimilar Triangles DBP, dHD; PD H,

PFG, OE: B Chr: JH): HD

. bz

23D? and PP ): PF GEA) :: HD

hs 2 bzz+abz H
FG (x) =” . Whence we
Vu 605 Z/22—bb |
get S or EE » the Fluxion of
1 T4 BR |
which, ſuppoſing & invariable, will be j=

— —„ —

b2+ 24b22—abXzx | | bzT-2abz’ —ab*2Xx

7 —

a, c
ſubſtituting for æ what is equal to it. Now

At. 66. if in the general Form * yy =x+5

—

ou
put za for 5, and for y and 5, the V ues

already found in x and &, we ſhall have

2+-2a2— abbz X &* _ 2-þ2abbz+aabb x x
ö 2 r

Ta fa
which may be brought down to 22*—34bz—
abb o; and one of the Roots of this Plus a,
will expreſs P ſought. –

Ifa b, then will 22 —zaaa = , which

divided by za, will be 22222053

2
whence PF (z+a) = 1 7½ů = 2 2.

Ex ANMYLE

of FLUXx1oNs. 89

D a1 xv wa v.

L=z MF be a Conchoid of another Fc. 60.
Kind, the Nature whereof is ſuch,

that if a right Line PF be drawn from any

Point F taken in it to the Pole P, interſecting

the Aſymptote BC in D, the Redangle PD

x D F, will be always =to a conſtant Rectan-

gle PB BA. It is required to find the Point

of Inflexion F.

Call the unknown Quantities BE, x; EF,y;
and the known ones AB, a; B P,b; then will
PDxD F=ab: And becauſe of the Parallels
BD, EP, PD DF (ab): PB E (bx)::

PF “(bb+2bx+xx 451): PE (bb+2bx+xx).
Whence bbx+ 2Fxx+x*-l-yyx=abb+2abx+
405 ae
ar, or 72 — bbx- — A
and * la the

Flux ion of which willbey = eee

zx -x
And again, d this into Fluxions, there

Haar- 4abx X x* . #
will ariſe = — 0, which

ab.
may be brought down to x 2 =BE.
a
— rar 1 vue of
Z – xx
5 Do; then will K- Taxa, the two

nk @+yaa—dS4b and a—/Ta—Vvab

If you make

4
whereof, will be ſuch Expreſſions of B H and
BL, when 4 exceeds 84, that the Ordinate F 16. 64.
2 H M

F 16. 62.

F 16. 63.

A Treatiſe

HM ſhall be leſs than thoſe near it, and the
Ordinate LN greater, viz. the Tangents to the
Curve in Mand N, will be parallel to the Axis
AB; and then the Point E will fall between
Hand L. |

But when a==85, the Lines B H, BE, BL,
will be equal each to 2. And then the Tan-
gent in the Point of Inflexion F, will be pa-
rallel to the Axis 4B. Laſtly, when a is leſs
than 86, the two Roots are imaginary, and
ſo no Tangent can be parallel to the Axis.

‘This Problem may be ſolved likewiſe in ta-
king the Lines P F, Pf, iſſuing from the Pole
P as Ordinates, and uſing the Form y y =
x*+37, as in the laſt Example.

E x A M P L E VI.

73.L*r AED be a Circle, whoſe Centre is
the Point B; and let A FX be a Curve
of ſuch a Nature, that any Radius B FE be-
ing drawn, the Square of FE be equal to the
Rectangle under the Arch AE, and a given
right Line þ. It is required to find the Point
of Inflexion F of that Curve. 3
Make the Arch 4 E, the Radius BA
or BE Ha; and the Ordinate BF); then will
b;==aa—2ay+35y, and (throwing it into Flu-
.
xions) 3 S Ee. Now becauſe of

the ſimilar Sectors B Ee, B FG, B E (a): BF

2YY—24) N . r r
(3):: Re Ye
The Fluxion of this, ſuppoſing & invariable,
will be 4.5)—2a5+2y3)—2a3y=0; and there-

fore y5 —— If now for & and yy,

you

8
37
id
|
4
Ig
4
f
4
E
2
F

>
**

of FLUX1oONs. |
you put their Equals in , in the general
Form #* yj=#-þj*, there will ariſe .

Y—8

— 43%) == 8 ajy* + 4 09y)* + abby?

3 aabb
may be brought to 4zy%1 zA 1 2454.0

3aabby—2a*bb==0; and fo one of the Roots
of this will expreſs B F ſought.

It is manifeſt, that the Curve 4 FX, which
may be called a Parabolical Spiral, muſt have
a Point of Inflexion F. For becauſe the Cir-
cumference A E D does not at firſt ſenſibly
differ from the Tangent in A, it follows, from
the Nature of the Parabola, that it muſt at

firſt be concave towards that Tangent; and

afterwards, when the Curvature of the Cir-
cumference becomes ſenſible about that Cen-
tre, it muſt become concave towards the ſaid
Centre,

ExAMPLE VII.

Art. 66.

L which

71 L* AF be a Curve, whoſe Axis is Fi. 64.

the right Line AB; and let the Na-
ture of it be ſuch, that any Tangent FB be-
ing drawn meeting AB in the Point B, the
intercepted Part A B will be always to the
Tangent BF, in the given Ratio of M to n.
It is required to determine the Point of Re-
trogreſſion F.
all the variable Quantities A E, x; EF, yy,

then EB =— = (becauſe when x increaſes,
y decreaſes) and FB = . Now from
the Nature of the Curve, AE E ; or AB

Ir Din There

fore

AT. reatiſe

fore my/x*Þ3 — —X; and he Fluxion

will be . = ono
_— oi
poſing x invariable and negative; whence we

44 {OY hong
get 2 ww? _ Now making

myyy—nxy V

this laſt fractional Expreſſion equal to nothing,

and there will be had —yx—xj=o, from which

nothing can be determined. But if the ſaid

fractional Expreſſion be made infinite, viz. the

| Denominator equal to o, then we ſhall have

JEFF ==” becauſe of the E-

2
quation of the he Curves and conſequently x =
mnxxj—mmyyj. Now fquaring both Sides of
the Equation my) u , and we ſhall
WY MMYY——NNXX =

have likewiſe x =

mnxxj—mmy)).

nnxy.
there comes out j % = = 3, and fo
we get the following Conſtruction.

On the Diameter AD m, deſcribe a Se-
micirele 4 7 D, and take the Cord D [==z,
and draw the Line AT This will interſect
— Curve AF in the Point of Retrogreſſion

For if I be drawn perpendicular to AB,
becauſe of the ſimilar right-angled Triangles

DIA, TH 4, FE 4, DIG): IA (/mm—m)
::{H:HA::FE(y):EA H. And therefore

From whence, at length,

„yum ur nx.

It is manifeſt that N is parallel to D’7; be-
cauſe 4 B: B F:; AD (n: DIG. VC

of FLUXIONS.
the Angle AFB is a right Angle, and fo the
Lines AB, B F, BE, are continued Propor-
tionals. |

This Property will appear from other Prin-

cipals thus. Conceive * FB, Fh, two Tan- * Art. 67.

gents infinitely near each other in the Point F
of Retrogreſſion, making the infinitely ſmall
Angle B Fb. Then if the ſmall Arch B L be
deſcribed from the Centre E, there will be had
m:n::Ab:bF::AB:BF:: Ab—AB or Bb:
bF—BF or bL:: BF:BE. Becauſe of the
right-angled ſimilar Triangles BL, FBE;
whence, &c.

If n, it is plain that the right Line A
will become perpendicular to the Axis 4B
and ſo the Tangent FB will be parallel to the
ſame; which that it is ſo, is otherwiſe evi-
dent, becauſe in this Caſe the Curve AF muſt

be a Semicircle, a Diameter whereof is per-

pendicular to the Axis AB. But if m be leſs
than u, the Curve cannot have a Point of Re-

trogreſſion, becauſe then y /m n unn
is an impoſſible Equation.

s E Cr.

F 1s. 65.

A Treatiſe

w—

8 f 1 7 dd N (17D I \ 4 > 8 —
G
N 9 4 „, 4 — 8 1 ö

IN AS —

— CH” > Wn, © \

—

SECT. V.

Dſe of Fluxions in the Doctrine of
Evolute and Involute Curves.

DEFINITION.

r one End F of the Thread ABDF be
firſt fixed to the Point F in a Curve

BDF concave the fame way; and after-
wards the Thread be put about the ſaid
Curve, ſo as to touch it in every Part.
Then if the other End A of the Thread be
tightly moved in the ſame Plane with the
Curve, ſo as to confinually diſengage itſelf
from the Curve B DF, the ſaid Ene A of the
Thread, will deſcribe a Curve AHR, which
is called an Involute Curve or Figure.

And the Curve BD F is called the Evolute
of the Curve AHK, or the Evolute Curve.

And the ſtraight Parts AB, HD, XE, of
the Thread, are each called the Radius of
Evolution, or of the Evolute.

Co ROL. I.

Ecavst the Length of the Thread
B ABD F does . it is plain that
the Length of the Part B D of the Curve, is
equal to the Difference between the Radii HD
and AB iſſuing from the Ends thereof. *
e

of FLUXIONs. 95

like manner, the Length of the Part DF, will
be equal to the Difference between the Radii
FK, DH), and the Length of the whole Curve
B DF, to the Difference of the Radii FA, BA.
Whence if the Radius B A of the Curve be o,
or the End A of the Fhread falls in B, the
Origin of the Curve B DF; then will the
Radi of the Evolute DH, FA, be equal to the

Lengths of che Parts BD, BD F of the Curve ,
BDF.

Co Rol. II.

76 1 F the Curve BD F be conceived as a Po- F 1 c. 66. |
lygon BC DEF of an infinite Number |
of Sides, it is evident, that A the End of the 5
Thread ABCDEF will deſcribe the ſmall |
Arc AG, whole Centre is the Point C, until |
the ſaid Radius CG coincides with the Conti- |
nuation of the little Side CD adjoining to CB;
and then it will deſcribe the ſmall Arch GH, |
having the Point D as a Centre, until BH |
coincides with the Continuation of the little |
Side DE; and fo on, until the Thread be |
quite diſengaged from the Curve BCDE F. |
Therefore the Curve AHA muſt be confider’d |
as the Aſſemblage of an infinite Number of |
ſmall circular Arches 4G, G H, HI, IX, &c. |
|
|

having the Points C, D, E, E, &c. as Centres. |
Whence, |
1*, The Radii of Evolution touch the Curve |

continually, viz. DH in D, KF in K, &c. For |
DH, for Example, is perpendicular to the ‘
ſmall Arches & H and H, becauſe it paſſes q
thro* their Centres D, E. Whence we may |
obſerve, 15, That the evolute Curve BDF, FI d. 65.
terminates the Space wherein all the *

| iculars

Fic. 66.

diculars to the Curve AHK fall. 25, That if

A Treatiſe

any Radius VD be continued out, interſecting
the Radius A B in R, until it meets any other
Radius XF in S. We can draw always from
all the Points in the Part RS, two Perpendi-
culars to the Curve AHA, except from the
Point of Contact D; from whence but one
Tangent DH can be drawn. For it is evi-
dent that &, the Interſection of the Radu AB,
D, runs thro’ all the Points in the Part & &,
while the End A, of the Radius AB, deſcribes

Line AHK, to which it is continually
perpendicular: And that the Radii AB, HD,
do not coincide, but when the Interſection R
falls in the Point of Contact D.

2% If the little Arches be continued out,
viz. HG tol, IA to n, XI ton, &c. to-
wards A. Every little Arch, as J½, will out-
wardly touch Z@ the littlę Arch next to it,
becauſe the Radu CA, DG, EH, FI, in-
creaſe ſo much the more as the ſmall Arches,
the Curve AHA conſiſts of, are farther from
the Point A. In like manner, if you conti-
nue out the little Arches AG to o, GH to p,
HT to , towards contrary Parts from the Point
A; every little Arch, as H, will touch in-
wardly the {mall Arch 7X next to it. Now
becauſe the Points H and Z,, D and E, on ac-
count of the infinitely Smallneſs of the Arch
Hand the Side DE, may be confider’d as
coinciding: Therefore, if from any interme-
diate Point D of the evolute BD EF, as a Cen-
tre, with the Radius DH, you deſcribe a
Circle Hp, it will outwardly touch the Part
HA, which will entirely fall within that Cir-
cle, and inwardly the other Part HA, which
will fall quite without the ſaid Circle: That

| is,

of FLUXIONS.
is, it will both touch and cut the Curve AR
in the Point H; juſt as the Tangent in the

Point of Inflection does cut the Curve in that |

Pointe br]

3 Becauſe the Radius HD of the little
Arch HG, differs from rhe Radu CG, EH, of
the Arches G A, Hl, next to it, only by the
infinitely ſmall Quantity CD or DE; there-
fore if the Radius HD be leflen’d never fo lit-
tle, it ſhall be leſs than C &; and ſo its Circle
will inwardly touch the Part HA; and, con-
trariwiſe, if it be never fo little increaſed, it
will be greater than HE, and ſo the Circle
thereof will touch the Part H&K outwardly :
Whence the Circle H is the leaſt of all
thoſe that touch the Part HA outwardls,
and the greateſt of all thoſe that touch the
Part HA inwardly ; that is, no Cirele can be
drawn between this and the Curve. LY

4, Becauſe the Curvature of Circles increas
ſes in the ſame Proportion as their Radii de-
creaſe; therefore the Curvature of the ſmall
Arch HF, will be to the Curvature of the
ſmall Arch 4G reciprocally as the Radius BA
or CA of this latter, to its Radius DHor EH.
That is, the Curyature in ¶ of the Curve
AA, will be to the Curyature in A, as the
Radius BA to the Radius DH; and in like
manner, the Curvature in K is to the Curva-
ture in H, as the Radius D H is to the Radius
FR. Whence it is manifeft, that the Curva-
ture of the Line AHA, decreaſes ſo much the
more as the Radius of the evolute Curve BD
is greater; ſo that in the Point A, the begin-
ning of the Curve, it will be a Maximum, and
at the Point K, where it is ſuppoſed to end, a

Minimum. 85

lh, ea. 4 1a

FC. 67.

A Treatiſe

„ That the Points of the evolute Curve,
are only the Interſections of the Perpendicu-
lars drawn from the Ends of little Arches,
whereof the Curve AHK is ſuppoſed to con-
fiſt. For Example, the Point D or E is the
Interſection of the * Perpendiculars HD, IE,
to the ſmall Arch H; ſo that if the Curve
AHA be given rogether with the Poſition of
one of the Perpendiculars H to it, and it
be required to find the Point D or E, wherein
the Perpendicular touches the evolute Curve;
the Buſineſs is only to find the Point wherein
the Perpendiculars HD, ZE, infinitely near each
other do meet: And this is the Work of the fol-
lowing general Problem.

P R O p. I.

* E Nature of the Curve A M D Be-

ing given, together with the Poſition of

Perpendicular MC to it; to find the Length

if the Radius MC of the _— Curve; or,

which is the ſame thing, to find the Point

wherein the Perpendiculars ‘M C, mC, infinitely
near each other, meet.

Firſt, Let the Ordinates P M of the Curve
AMD, be perpendicular to the Axis 4B;
and letthe Ordinate mp be infinitely near MP,
becauſe the Point m is ſuppoſed infinitely near
M. From the Point of Interſection C, draw
CE parallel to the Axis AB, meeting the Or-
dinates M, mp, in the Points E, e. Laſtly,
draw MR parallel to AB, and then the right-
angled Triangles M Rm, ME C, will be ſimi-
lar: Since the Angles EMR, CMm, being

right Angles, andthe Angle CM R being com-

mon

« ® Theſe Perpendiculars are called the Radii of the Cur-
vature.

of FLUX10NS. 99

mon to them, the Angle E MC ſhall be equal
to the Angle R Mm. _ 28

Now call the given Quantities 4 P, x;
PM, y; and the unknown one ME, z. Then
will Ee, or Pp, or MR==x, RN A,
Mm=y/3*Fy’; and MR : Mn MH U
SEO. =. Now ſince the

Point C is the Centre of the ſmall Arch Mm,
the Radius C M thereof, which becomes Cm
while EM increaſed by its Fluxion Rm con-
tinues invariable; therefore the Fluxion of it
will then be nothing; and ſo making x inva-

table, , Whence we get

„ pe rob
ME (z)= —— — _ by ſubſtitu-
ting y for z.

2», Let the Ordinates B M, Bm, all iſſue Fo. 68,
from the ſame Point B. Draw the Perpendi-
culars C E, Ce, from the ſought Point C to the
Ordinates, which ſuppoſe to be infinitely near
each other, and defcribe the ſmall Arch MR
from the CentreB; then the right-angledTrian-
les RMmand EMC, BMR, BEGand Ce& will
ſimilar. And making B M==y, ME Dx,
M R==x, and we ſhall have Rm, Mm =

TT. 7 1 2/xÞy*
Vi, CEor Ce and A
3 | F&L 2848 £**. 8
And, as before, we ſhall get 2
Now B M ( C0 (2) : MR (v):

and me ME or Rn . |

DW. Whence ſubſtituting this for 4 i
9 H 2 |

100

pendicular to the Axis,

A Treatiſe

and ME (z) will be ET
If y be ſuppoſed infinite, the Terms x* and
5*, will be nothing in reſpect of yy; and con-
ſequently this laſt Form will fall into that
of the Caſe aforegoing. This muſt be ſo, be-
cauſe the Ordinates then do become parallel,
and the Arch MR a right Line perpendicular
to the Ordinates. |

Now if the Nature of the Curve 4M D be
given, we may get the Values of ) and y in x?,
or of xand 7 in 5; which being ſubſtituted
in the precedent Forms, and there will a-
riſe an Expreſſion for ME freed from Flu-
xions. And then drawing EC perpendicular
to ME, it ſhall cut MC perpendicular to the
Curve, in the ſought Point C. |

Coror. I.
Ecavss of the right-angled ſimilar
Triangles MRm and MEC, in the
x* +3 /x I
former Caſe MC = — , and in

the latter MC =! £9 ts

SCHOLIUM.

Pnrrt are other ways of finding the
T Radii of Evolution ſome of which
I ſhall mention for the ſake of thoſe who have

n_ been acquainted with Inveſtigations of this

Cass I. In Curves whoſe Ordinates are per-
if Way,

of FLUXIONS.

101

1 Way. Continue out MR to G, wherein F 1 6. 67.

it interſects the Perpendicular C. Becauſe
the Angles Rn, Mm & are right ones, RG

will be=2 and ſo MG = “45 Now

ſince the angles MRm, MP 9 the Points
©, q being the Interſections of the infinitely
near Perpendiculays, with the Axis xis AB) are ſi-

milar, there 4. Mg = EET . „P

4 & Ot – (7, 1
22 and dre A 9= => ＋ 2, the Flu

xion whereof (ſuppoſing 4 x ovarible) brings
out 9g =x + . — FL ? LEEDS “oy becauſe of the ſt.

milar Triangles CMG, C295 MG — — 27
(=>): MG ( ＋ 2 1 ANI 9
c. 1

x)
2d Way. From the Centre C, cſeribe *
ſmall Arch O0; then the little ri kal,
Triangles „2881 „Men, will be

cauſe 90 and MR, 9g, are pale, |

Therefore “= (Ai) MR (x): : 2.4
‘% +; =); 20 == Now
becauſe of the ſimilar Sectors C Mm, C 90,

Mn—20 (S) Mm 33):
WE 2 ) AC

34 Way. If the Tangents MT, mt, be
drawn infinitely near each other, Then will
\ H 3 PT

A Treatiſe
PT —AP or AT= %=—agheFluxion where:

of will bert =, and if the ſmall Arch

| THbe deſcribed from the Centre n, the right-

| angled Triangle HT will be ſimilar to the
i | 5 t· angled riangle Rm M; for the Angles
| lf Tt, R Mm or PTM are equal, ſince their
| lf Difference is the Angle Tt, which is infi-
| nitely ſmall. Whence Mm (y/z*+5″) : mR
1 T. (YH SL Now

0):7 (22): .

[il] the Sectors T DA M n, are ſimilar; for the
= Angles TM A m Done right Angle, and
i! ‘ the Angles Mm CT MCN are likewiſe equal

1 to one as Angle, becaule the Triangle C Mm
is conſider’d as r1 ht-angled at M. Therefore

— x) TA |

|| 7H 7 : Mm G): Tu -lig

| ru . ) 7 2 *+ 3 Vit
AG

1 Fs. 69. 4th Way, Taking # as invariable, the ſe⸗

5 * Art. 64. cond Flyxions any, be denoted *, and becauſe
| of the right-angled fimilar Triangles, Hm S,

6 Huk, Hm or Mm V ) or MR (6)
1 | £ : Hy (—9): nh=—— Now the An-

Un gle kms i is equal to that pn by.the Tangetit
[| at the Points M, n, and therefore equal to the
1 8 Angle MC m; whence the Sectors amt, MCm,

| are ſimilar; and ſo ( e) : M or
| Mp (FF 95 C L Note,

102

| Att. 2 | 1 4
ö 5 e mH

of FLUX1ONS. 103
M Hor Mu is taken for mk, becauſe their
Difference is only the ſhort Line Hi infinitely

leſs than either of them. In like manner, as
Hn is infinitely leſs than R m or Sn.

CAsE II. I» Curves whoſe Ordinates iſſue
all from a given Point.

1, Way. Draw the Perpendiculars BF,Bf, F 16. 68.
from the Point B to the infinitely-near Radii
CM, Cn; then becauſe the ri hi-angled Tri-
angles m MR, BM, are ſimilar (ſince if the
ſame Angle YR be added to the Angles
m MR, B MF, each of the Sums will be equal

to a right Angle) MFor M H=— 2 —

VI
and B F= hs the Fluxion whereof is

ILA
Bf —BF, or Hf –
Bf T =p Suppo
ſing x invariable. Now 3 the Sectors C Mm,
C Hf, are ſimilar, therefore Mm Hf: Mm

: MH: MC; and fo A ——

e
2d Way. Denote * the ſecond Fluxions in * 4rt- 64-
_— x 1nvariable; then becauſe of the ſi- F c. 70.

|
|
|

) milar Sectors BS, m EA RM (YS (::

g E 53 Now becauſe

˖ of the right-angled Similar Triangles Hm &,

Hnk, Hm or Mm Gan m S Or MR 5

:: Hun pit . And there-
fore En — f

— F 3 and finding a third

Proportional to En, Em or Mm, by means of
81 H 4 the

104 £1 A Treatiſe
the ſimilar Sectors E mn, MC m, the fame Ex-
preſſion: for MC, as before, will be had.

If you make Mm (V/3$5) i; and be
taken inſtead of x TY ne In the for-

mer Caſe, MC = == and in the other MC=

N * * wm „if 4 be foppobed i inva-

riable, there 2170 out in the former Caſe
NC = £ = or 25 (becauſe the oc of

* 1s $5+55=03 and ſo £ 2 * 2 )

0 FEY SL Wks

Eon 3
1 85. Ser ME or MC has been found to
have but one Value, therefore the in-
e Curve 4M D has but one evolute
BCG:

o – 4 1 .
-O—__ — = * «
7 $

. III.

„. bes (9 – (ggg)

2 * —53
“© poſitive, the Point E muſt be taken on the
fame Side the Axis A B or Point B, as it was
ſuppoſed in the Operation aforegoing. And
ſo the Curve in that Caſe will be concave to-
wards the Axis, or that Point. But if ME
be negative, the Point E muſt be taken onthe
contrary Side, and ſo the Curve will be then
: convex towards the Axis, or that Point. There-

bon it is plain, * in the Point of Inflexion
or

. ͤ— . .

of FLUXIONS.
or Retrogreſſion, which ſeparates the concave
from the convex Part, ME from being poſi-
tive will become negative. W hence the con-
tiguous or infinitely- near Perpendiculars from
converging become 8 Which can
happen but two ways only: for as they go on
increaſing ſtill the more, as they accede to the
Point of Inflexion or ſerragrelfion, they muſt
become at laſt parallel, that is, the Radius of

105

Evolution will be infinite: and where they

conſtantly continue decreaſing, they muſt at

laſt coincide, that is, the Radius of Evolution
will be nothing. All this is correſpondent to

what has been demonſtrated in the foregoing

Section. |

SCHOLIUM.

B E CAUSE hitherto the Radius of Evo-

lution has been conſider’d as infinitely

great in the Point of Inflexion; I ſhall here

ew, that in numberleſs Species of Curves

the Radius of Evolution in the Point of In-
flexion is equal to nothing. and that there is but

one Species where the ſaid Radius is infinite.

Let B AC be a Curve, whoſe Radius of E- PI d. 51.

volution in the Point of Inflexion A is infi-
nite. | |

Now if the Parts B A, AC be conſider’d as

evolute Curves, the Point 4 being 1

their Beginning, and the Curve D AE as an

involute Curve formed from them: it is plain
that this latter Curve will have a Point of In-

flexion in A, but the Radius of Evolution in

that Point will be equal to nothing. And if

a third involute Curve be formed from the ſe-
cond D AE as an Evolute, and a fourth Invo-
= | lute

106

F 10. 72.

A Treatiſe

lute from the third as an Evolute, and ſo on;

it is manifeſt that the Radius of Evolution in

the Point of Inflexion A in every of thoſe
Curves, will always be equal to nothing.

| Whence, Ce.

Prop. II.

L T AMD be an involute Curve, whoſe
Axis AB is at right Angles to the Tan-
gent in A: to find the Point B wherein the ſaid

Axis touches the Evolute BCG. |

If the Point M be ſuppoſed to become in-

finitely near the Vertex A, it is plain that the

Perpendicular MQ will interſect the Axis in

the Point B ſought. Whence if you ſeek the

F1c, 72.

Value in general of P 9, (Z) in x or, and
| 7 “24>

afterwards you make x or y =o, we may de-
termine the Point P, which is to coincide
with 4, and the Point © which coincides
with the Point B ſought; that is, P © will
then become equal to AB ſought. This will

be more plain by the following Examples.

EXAMPLE I.

Tyr the Involute A MD be a Parabola,
— whoſe Parameter is a right Line, ſup-
poſe a. The Equation of the Curve is ax =3y,

the Fluxion whereof is j = = T5
| 1 2 Va x

and throwing this laſt Equation again into Flu-
xions, by making x invariable, there ariſes 3 =

— Now ſubſtituting theſe Values for
4x v ax 3 and

of FLUXIONS.
5 and 5 in the general Form E—,andrhen

107

From hence ariſes the following Conſtru-
ction. |

From the Point 7, wherein the Tangent
M interſects the Axis, draw TE parallel to
MC: I fay, this ſhall meet MP continued
out in the Point E fought. For becauſe of
the right Angles MP7, MTE, MP (/as) :
PT(23)::PT (28): E —=#&vax ;
AN | y ax 4

and conſequently MP | PE = /az + 2.

Again, becauſe of the right-angled Triangles
MP, MEC, therefore PM (Var): PQ

Ga): ME ( 22225 EC or H

za 2. And therefore QR 2x. From
whence we get this new Conſtruction.

Take QA the Double of AP, or (which
is the ſame) take PX=T 9, and draw KC-
parallel to PM. This will meet the Perpen-
picular MC in the Point C, which will be in
the Evolute B CG.

Otherwiſe. yy Sax, and 2 yy Sax (and takin
x as invariable) zj / =0; whence there ariſes |

RT – And putting this Value in the Form

1 1
— 2 „there will ariſe * ME = 12 — 3 * Art. 77.
_ » |

and therefore EC or PX ===” x. * +?
=P9 + PTorTg, And
| be

x Y is
ſo we get the
i. fame

108

A Treatiſe

| ſame Conſtructions as before. For MP: PT

29:4: : P D. E= =
Now to find 7 * Point B, 3 the Axis
A touches the Evolute B 2 G: wehave P 2

2 =. And ſince this is invariable,

ans the Point M is aſſumed; therefore
when M coincides with the Vertex A, we
ſtill ſhall have P ©, which in this Caſe be-

comes ABS Ta.

To find the Nature of the 1 CG
according to Deſcartes’s way. Call the Ab-
ſciſs B A, 45 and the Ordinate KC or PE, t;

then will CA (7) = and AP, K

—AB(u)=3x. Now 2 ‘ # for its E-

FI c. 73.

Fic. 72.

qual x in the Equation 12 2 and there

ws come out 27att = 1647 selling the Re-
lation of BA to KC. Whence if the Invo-
lute be a common Parabola, the Evolute BCG
is a ſecond cubical Parabola, whoſe Parameter
a equal to x; of the Parameter of rhe Invo-
ute. |

It is evident, that if the Idvokite by the
whole Parabola MAM, the Evolute CBC
will conſiſt of two Parts CB, BC, having

contrary Concavities; ſo that B will beaPoint
of Retrogreſſion.

DEFINITION.

« BY Y Geometrical Carves, as AMD, BCG, 7
underſtand thoſe, whereof the Relation of

the Abſeiſſes AP, BK to the correſpondent Ordi-
nates

of FLUXIONS. 109
nates PM, K C, being all right Lines, can be
expreſſed by a finite Equation Fo from Fluxions.
And whatever is effetted by means of thoſe Lines,
is ſaid to be Geometrical.

Conde LM

W HEN the given Involute Curve AMD

is a Geometrical one, it is plain that
we can (as above) always find an Equation ex-
preſſing the Nature of the Evolute BCG;
and fo the Evolute will be likewiſe a Geome-
trical Curve. I fay moreover that it is to be
rectified, viz. we can find geometrically ſtraight
Lines equal to the Length of any Part BC of
it. For * by means of the Involute AMD, Art. 75.
which is a Geometrical Curve, we can deter-
mine the Point M in the Tangent C M to the
Part BC, ſuch that the ſtraight Line CM dif-

fers from the Part BC only by a given right
Line AB.

E Xx AM y LE II.

Tur the Involute MD M be an Hyper- F. 74:
L bola within its Aſymptotes. The Na-

ture of this is 4@ =xy. “ag
W aa

— =x, , and ſuppoſing
D
invariable — 2 — Er * Art. 77.

we get) == „and putting this Value in 4} 5

*2 28 4
there comes out ME == : fothat EC Art. 77.

or PR E. Hence the following
2% 29

Conſtructions are derived. Thro’

110

Ari. 78.

A Treatiſe |
Thro’ the Point T wherein the Tangent

MT interſects the Aſymptote AB, draw TS
to MC meeting MP continued out in

F; and aſſume ME cqual to; MS on the

other {ide the Aſymptote (which is here taken
as the Axis) becauſe the Value of it is nega-
tive; or elſe aſſume PK equal to; 72, on
the ſame ſide as the Point T 1s: then, I’fay, if
EC be drawn parallel, or KC perpendicular
to the Axis, they will interſect MC in the
Point C ſought: for it is plain that AS

LE — and 22

From an Inſpection of 5 Figure of the
Hyperbola MD M, it will not be difficult to
5 that the Evolute CLC muſt have a

oint L of Retrogreſſion, as the Evolute in
the laſt Example has. To determine which,
we muſt obſerve that the Radius of Evolution
DL is leſs than any other Radius MC; ſo
that the Fluxion of the Expreſſion thereof, |

ut ® 2o-

=
ans or infinite. And ſo taking x as

invariable, the ſecond e will be

3 —

r i

nite; whence ; by 2 +3*7 and after-
wards wage bas 45, there ariſes & +
3j —3 s, or infinite; by which ey
find ſuch an Expreſſion as 4 Z for x, that by
drawing the Ordinate H and the Radius of
Evolution DL, the Point L will be the Point
of * fought.

o, or infi-

— » = Lun. io — Y

— “— — — ea —— WS »*

of FLUXIONS.

— aa

In this Example y = 5 7 75

— ” . Whence putting
the latter Members of every of theſe Equati-
ons for the former ones in the Equation afore-
going, and there comes out AH (x) .
Therefore the Point D is the Vertex of the
Hyperbola, and the Lines AD, D coincide
with AL. Which is the Axis of the Curve.

EX AMP L E III.

111

T ETH generally expreſs the Na- F 6. 72,
ture of all Parabola’s, the Exponent 74:

m repreſenting a poſitive whole Number or
Fraction, and all Hyperbola’s, when the ſame
Exponent is a negative whole Number or
Fraction.

Now my” – y x, and the Fluxion of
this again, taking # as invariable, will be
mm —m 3j +my>—’j=0; and divi-
ding by my* , there comes out — y =
m— 1)

3 whence ſubſtituting this latter Ex-

preſſion in Bs , we * get ME . — 9

7 a mM—1y
and therefore EC or PK = I + .

M—Ix M—I)

From whence the following Conſtructions are

gained.

From T’ the Interſection of the Axis AP,
with the Tangent MT draw the Line 78 pa-
rallel to MC meeting MP continued out in
the Point &: Aſſume ME = : – MS, or

3 * elſe

TE +7 Treatiſe Q
elſe take PK = ——T’9; then if a patal-
3 lel to the Axis be drawn thro’ E, or a Per-
pendicular thro’ X; theſe ſhall interſe& MC
3 7 C ſought. A ko
Fc. fn be negative, as in the Hyperbola’s,

7+ the Value of 77 E will be negative; and ſo
the Curves will be convex towards their Axis,
FI c. 75, Which will be an Aſymptote then. But in
| the Parabola’s where is poſitive, there may
happen two Caſes: for when m is leſs than 1,
they will be convex next to the Axis, which
will be a Tangent to the Vertex; and when
m exceeds t, then they will be concave next
to the Axis, which will be perpendicular in
the Vertex. | | 2
Now in this latter Caſe, to find the Point
| B where the Axis AB touches the Evolute.

[ P 2 (2) MB; from whence ariſes

Fre. 72.

three Caſes. For when 2, as in the com-
mon Parabola, then the Exponent of y being no-
thing, that unknown Quantity vaniſhes; and

conſequently 4B =4z, viz. to one half the
Parameter. When z is leſs than 2, then the
E !xponent of y being poſitive, it ſhall be found
in the Numerator, and the Fraction will be-

4-4. 83. come nothing by making * it equal to o; that
is, the Point B in this Cafe will coincide with
A, as in the ſecond Cubick Parabola axx =.
‘F: 6. 56, Or laſtly, when m is greater than 2, and then
| the Exponent of y being negative, it will be
in the Denominator; ſo that when it becomes
nothing, the Fraction will be infinite: viz.
the Point B is infinitely diſtant from A, or
“v4 Which is the ſame) the Axis 4 B is an Aſym-
. ptote of the Evolute, as in the firſt hy

of EroUxoNnSs, 113

Parabola W In this latter Caſe; we
may obſerve that the Involute C Lo, when Fic. 77.
tha, Evolute AD M is a Semi- parabola, will
have à Point of Retrogreſſion; ſo that tho?
the Part LO of the Evolute be infinite, yet
the Part. DA of the Involute formed by it

will be determinate or finite; and the other
Part LC of the Exolute being infinite, will
notwithſtanding deſcribe the infinite Part DM |

of the Inxolute.
Now the Point L ma 1255 determined ar

the ſame way as in the Hyperbola. For Ex-
np; Leroy, or 7 3 then ill

which Vale being Aa Ss Bar!
tion 3*3+ 57 e; and chere comes

„ee ö e | T 5 Ae. 86)

Se Mm
than 1; and ſo the

HEN is greater

there are ſeveral Caſes. For if the Numera-
tor of the Fraction denoted hy be eyen, and

the Denominator odd, all the Parabola’s will P18. 7g
fall on each Side of their Axis juſt as the com

mon Parabola does. Bur if the Numerator

and Denominator be each odd, they have an
inverted Poſition on each Side heir Axis, ſo

that the Vertex A is a Point of Inflexion 3 as”

che firſt Cubical Parabola x =»?, or aax= -Fre. A

Laſtly, if the Numetator be odd, and the
Denominator even, theſe Curves have an i-
rene eden on bes ſame Side their 82 Fig. 5

[

Parabola’s . next to the Axis, 9

—_—

71 1
ſo chat the Vertex A is 4 Pont of 8

| Value. : Whence i it cyidently follows, gu *
Fro. 77. 1% That in the Point of Intex lon , (the
Evolution ma if be inflageely |
2 V ot infinitely ſmall Nl, pal as

Pre. 76. 25 Tbat in. the Point of Retro 1
the jus of Eyolution, * be fe
ite, as in a i; ot nothing, as if! pr Ar

-N That becauſe the Radius of Evolution

fite ar o, it does not from hente-follow

? ike Curves tave then a Point of Inflexion

| don: for in a*x = it is Infinite,

| TEEN and yet theſe Parabo-

£3 1. ‘s have the ſame Poſition with regard to their
“9 as the common Patabola.

ExAmyre IV.

Freatiſt

Fis. 73.

Pic. 78, v5: 1 the Involte AMD be an H

hola or Elli ; whole Axis is 2 .
. Parameter 1)

Iden from ce Nature of theſe .

M .
om — — ee ee
at. 4% LEE oo SOUL

ee. 2 —— Now if theſe Va-
Hab I qabrxx/aahs Tabus

| hes be introduced into —— . 5 , che ge-
N. 58. dan Exprefiion for * MC; pf will MC =

Sail.

— —— D 8 EST 4

m2 1 1 72 7 ; 245 bb

| =WE, becauſe on ck Side C LED

. Il

Line. 5

of EFLUXI10NS.

ran 242 + gw

—TY HTTY I FEY F? — — —

2
Hence we get the following Conſtruction,
which will ferve for the Parabola alſo.

Aſſume MC equal to four times a fourth

continual Proportional between the Parameter
AF, and the Perpendicular A{ 9, bounded by

the Axis: then will the Point C fall in the

Syolute.

If you make x =o, then will * AB =. * Art. $3.
And if in the 3 you make #=7 4, then F d. 79.

will DG be , 1. equal to half the

2b ? 1
TE of the conjugate Axis. And ſo it
ears that the Evolute BC G, when the In-
191 54 is an Elli pſis, termihates in the Point

9 of the conjugate Axis DO; wherein is
formed a Point of Retrogrefſion ; but in the

Parabola and Hyperbola ic runs out ad inſini-

Zum.

If 42 2 in the Ellipſis, there comes out
Meta; from whence it follows, that tlie

Radii of Evolution are all equal to one ano-
ther; and ſo the Evolute will become here a

Point; chat is, the Involute will be a Circle,
and the Evolute the Centre of it. Which is

an eſtabliſh” Truth from he? ren ry

4

EAM IIZ V.

of ſuch a Nature, that if from any

| Point Min the ſame be drawn the Perpendi-
cular MP to the Afymptote & Hand the Tangent

MT; the Subrangent | be PT a e

A New

% 0

Et AMD be a Logatithmick Curve FIC. 80.

|
|
Þ
|
|

A Treatiſe
3 N vr = a whence we get 3 = 2

„and the Fluxion of this again, ſuppoſing

—— will be y = 2 2 sand putting

Jet. 7 theſe Values in Ax, Fr there will ariſe *

E = — 2 2. and therefore E C or P K

8 . F rom whence ariſes this Con-

fru8ion. |

Aſſume PX 27 Qon the ime Side 7, be-
cauſe the Value of it is negative; and draw
KC parallel to PM: I fay the fame will inter-
ſe& the Perpendicular MC in the Point C

fought. For 7. A 2 L.

If you have a mind on the Point N be in
the eſt Part auen of the Curvature,

the general Expreſſion & Y ＋ 5 350,
(ef An. 86.) muſt be uſed: then 1 2772 =

be put for 5, 5, 3 „ PM (y) will be =ay+.
Now when x is invariable, it is plain that
the A y are to each other as their Flu-

xions 5 or! — *; from whence it follows, that

they will be in a geometrical Pro Peek Alſo.
For if the Aſymprote, or Axis PX, be ſuppo-
{ed to be divided into an infinite Number of
ſmall equal Parts Pp or MR, pf or m, fg
or A, &c. intercepted by the Ordinates PM,
pm h &c. then will PM: pm:: Rm: Sn
::Rm::P 3 or ꝓm: pm n or 1 5

2 n

of FTLuxlos.
In like manner we prove that Am: fu: : Fu: go,

117

and ſo on. Therefore the Ordinates PM, pm,

Fu, go, &c. are in a geometrical Progreſſion.

5 ExAMr LE VI.

L* AMD be a Logarithmetical Spi- Fu. 3.

ral, of ſuch a Nature, that if you
draw a right Line AA from any Point Min it, to

the Point A being the Centre, and the Tan-

ent MT; the Angle AMT is of a given
Quantity; viz. always the ſame.
Becauſe the Angle A MT or-4m Mis in-

variable, the Ratio of m RO) to RM (x) will

be alſo invariable; therefore the Fluxion of

2 muſt be nothing; ſo that (ſuppoſing & inya-

x WP | |
riable) 5 =o. Therefore if the Term y) be
ſtruck out of = 2 —.— the general Expreſ-

ſion * for ME, when the Ordinates iſſue from “ Art. 77.

the ſame Point, ME will be =», viz. ME =
“AM. From whence comes the following
. Conſtruction.

Draw AC perpendicular to 4M, meeting
MC the Perpendicular to the Curve in C,
which will be a Point in the Evolute 4 CB.

The Angles AMT, ACM are equal, be-
cauſe by adding AMC to each of them, the
. wholes will be right Angles. Therefore the
Evolute AC & will be a Logarithmick Spiral
84 only from the Involute { M in Po-
ſition. |

If the Point C in the Evolute 4 C G be gi-
ven, and it be requir’d to find the Length of

0 M the Radius of the Evolute in that Point,
which is * equal to the Part AC of the Spiral » 4,,, 55.
| | 3 making

118

A Treatiſe

making an infinite Nienber of Revolutions
before its Acceſſion to 4; it is manifeſt, that
AN need only be drawn perpendicular to AC.
So that if AT be drawn perpendicular to AM,
the Tangent M will be equal likewiſe to the

Part 4M of the Legprithmick Spiral 45 D

given.

If you imagine an infinite Nutaber of Or-
dinates A M, Am, An, 40, &c- making infi-
nitely ſmall equal Angles; it is manifeſt, that
the Triang les MAm, m An, n Ao, &c. will
be — becauſe. the Angles at A are equal,
as likewiſe are the Angles at n, u, 0, & 1

the Nature of the Spiral. Therefore AM:

Am:: Am: An; and Am: An:: An: Ao, &c.

Whence the Ordinates AM, An, An, Ab, &c.
ae in a geometrical Progreſſion, if they make

ee Angles with each other.

EXAMPLE VII.

F16. N * r AMD be one Spiral (of: an infinite

k >

Number) formed in — Sector BAD,
“of ſuch a Nature, that any Radius AMP be-
ing drawn, and calling the whole Arch BPD, &;
the Part BP, æ; the Radius AB or AP, a;
and the Part IN, y; there is always the fol-
Towing Proportion, 5: *:: an: yn.
Pa Equarion of the Spiral AMD is 22 =

7 0 thrown into Fluxions is my”

7 += _ BY Now — of the fimilar Sectors
TA AE 15 AM (5) : 4P(a) : MR (3):

; ® p (= . | Now putting this Value for &
. the Equation aloe found, and 19 =

2 —

of FT los. 119
ELLE; which again being thrown into Flu-

1 making & invariable, and mm 1 “+
my”j=0z whence divi ling by 2:9”
N, e 3 and therefore * NE. eater 77

2 >, = So that the

“IF” Ar

Gllowing Conſtruction 7 out thus.
| Thro A draw T A e to AM
meeting the Tangent N in Taadthe ne Perpendi-
cular M in 9; make Wn 142:
T:: MA: ME. Then if E C be drawn pa-
rallel to T9, it will meet MY in a Point, as
, which will be in the Evolute.
For becauſe MR G, TAY are parallel,

Re (Hue (444)
2 FIAT e ME =

x*+ y
Pm) my
1 Ex AM LE VIII.

9 L* T AMD * {a og whoſe Baſe F 1 d. 83.

BD is eq the Circumference
BE A of the pena > Errcle |

Call AP, xz PM, y; the Arch AE, u; and
the Diameter AB, 24. Then PE =\/1ax—xx,
from the nature of the Circle, and y = +
24x xx, from the nature of the Cycloid;
which laſt Equation thrown into Fluxions will
be j = # + AX—XX 24 — XX

1 Or
ere VZzax – xx

ox W putting ——o; and ſuppoſing
| Viax— xx

14 * inva-

AA 7 reatiſe |

— and ſubſtitu-
— XX

ting theſe Expreſſions in IZLE, and
— — ö

there comes & out M C e chat

is, 2 BE or 2 MG.
If you make #=0; then will I N=

be the Radius of Evolution in the Vertex 7

But if „za, the Radius of Evolution in D
will become nothing. From whence we are
aſſured, that O is the Beginning of the Evo-
. and the Point N the End, fo that BN =
B |

Now to find the Nature of this Involute,
“you need only compleat the Rectangle BS,

daeſcribe the Semicircle DIS on DS, as a

Diameter, and draw D af – parallel to M C or

BE. This being done, ts pin that the
Angle BD Tis 2 to ro the Angle 7 e EBD; and
conſequently the Arches D 7, ual to

one another; whence the Chords 57 5 BE or
GC are likewiſe equal. And therefore if IC
be drawn, it will be equal and parallel to DS,
which by the Generation of the Cycloid is
equal to “he Arch BE or DI Fee} conſe-
quently the Evolute DCN is a Semi-cycloid,
having the right Line N’S for its Baſe, being
equal to j the Circumference DS of the ge-
nerating Circle: that. is, it will be the ſame

| 10> ang, AMDB having an inverted Si-
f W |

;

—

K Conor,

of FLUXIONS.

Conor.

121

Tris * evident that the Part DC of the · 4re. 75.

Cycloid is the Double of its Tangent
[e, or N Chord D 7; and the
Semi- cycloid DCN, is the Double of the
N BN or DS of the generating Cir-
cle. | „

ANOTHER SOLUTION. _

or PHE Length of the Radius MC maybe
= T found jor any Calculus ** 2
Conceive the Perpendicular C to be infinitely
near the former one, another Parallel me, ano-
ther Chord Be, and from the Centres C, B,
deſcribe the ſmall Arches GH, EF; then

will the right-angled Triangles & Hg, E Fe,
be equal and ſimilar; for Gg Ee, ſince B

or ME is equal to the Arch AE, and in like
manner, Bg or me is equal to the Arch Ae;
moreover Hg or g- MG Fe or Be—B E;
whence GA will be equal to EF. Now ſince
the Perpendiculars MC, C, are parallel to

the Chords E B, e B, the Angle Mn ſhall
be equal to the Angle E Be. Therefore fince
the Arches GH, E, that are the Meaſures of
thoſe Angles, are equal, it follows that the
Radii CG, BE ſhall be equal likewiſe; and fo

MC muſt be aſſumed the Double of AMG or

LEMMA.

96 77 there be any Number of Quantities a, b,
c, d, e, &c. either ſinite or infinite, the

Sum of their Differences, viz. a—b+b—c+ .

-d +d—e, Cc. is equal to the greateſt Quan-
-. —_

122

- 1
  «
  » N

B. oausf the Sectors C Mn, CG Hare

8 t A * *

” * 2 1 5
1 1*
829 — « 14 4 15

.
tity a Minus the leaſt e, or elſe equal to the grea-
teſt, when the leaſt is gf N Re

: 10 R. a 4 : 14 pg . « 2 * 1 we
0 «+
1

Umilar, Mum is =2 CH=2 EF; and
ſince this is always ſo, be the Point M where
it will; therefore the Sum of all the little-Ar-

ches Mm, that is, the Part Am of the Semi-

P88 AM_D, is the Double of the Sum of

\ll the little Arches EF. But the little Arch
EV being a Part of the Chord AE perpen-

dicular to BE, is the Difference of the Chords

AE, Ae, ſince the ſmall right Line e per-
endicular to Ae, may be conſider’d as a ſmall
ch deſcribed about the Centre A. And

therefore the Sum of all the little Arches EF
in the Arch 42 E, will be the Sum of the
Differences of all the Chords AE, Ae, &c.

in that Arch; that is (by the Lemma above )

equal to the Chord AE. Therefore it is evi-
dent, that the Part AM of the Semi-Cycloid

AMD, is the Double of the correſpondent

Chord AE.

Act. 2.

58 Tu E SAcRN Ggm®, or the Trape-

zum MG Hm Dad Tx MG

EFX BE; that is, it is the Triple of the

the Triple of the circular

Triangle E B For E Be; whence the Space
ace BE

MGE 4, or Sum of all the ſaid 9575 1, Is
J be-
.

ing the Sum of all the ſaid Triangles.

IT r

Foz F . 5 F .

| of Puvikions, ;

JE I r III.

123

CA. BP, 2; the Atch AZE or EM 8

or BG, u; and the Radiue X A, a;
then will the Parallelogram MGBE be uz.
Now the cycloidal Space MG B A==3BEZ A
=3EKB+jan; and therefore the Space
AME’B contained under the Part 4 M of the
. Cycloid, the Parallel ME, the Chord BE,
and the Diameter A B,’ is == EKB+1au—
E. Whence if you allume BP (z) za, the
Space A ME B will be the Triple of the cor-
reſpondent Triangle EXB; and ſo we have
the Quadrature thereof independent of the
Quadrature of the Circle; (which Mr. Hugens
firſt obſerved) as likewiſe the Quadrature of
the following Space.

If the Segment BEZA be taken away from

the Space AMEB, there will remain the Space

AZEM=2BK Ta 12-1 Z; therefore when
the Point Peoineides with &, the Space AZ EM
will be then equal to the Square of the Radius.

It is plain, that among all he Spaces AMEB,
AEN, there are but theſe two only that can

be ſquared independent of the Quidiratizre of
the irele. |

EXAMPLE IX.

tr AMD be a Semi- cycloid deſeri-
| bed by the Rotation of the Semi-
circle A E B about or along another Circle
BG at reſt. It is requir d to find that Point
in the Perpendicular MG of a given Poſition
That touches the Evolute.

Becauſe

F1c. $4.

— — — —
— – Err —

BMA A Treatiſe {0

Becauſe in the Ute of general Expreſſions or
Formules, we muſt firſt take right Lines per-
pendicular to the Axis AO, as Ordinates to
the Curve AMD, and then find an Equation
expreſſing the Relation of the Ordinates to the
Abſciſſes, or of their Fluxions, which often is
a very operoſe Performance; therefore in Oc-
currences of. this Kind,. it. is much better to
endeavour at the Solution from the Genera-
tion itſelf. 1181 |
When the Semi-circle AE B is come to
the Situation GB, wherein it touches the
Baſe DB in G, and the deſcribent Point A,
falls on the Point Mof the Semi- cycloid AMD,
it is plain, .
10 That the Arch GM is equal to the Arch
GD; as alſo the Arch GB of the moveable
Circle, to the Arch GB. of the immoveable
® 47.43, 2JT hat M is * perpendicular to the Curve:
For if you conſider the Semi-circumference

MGBor AEB, and the Baſe BG D, as the
Aſſemblage of an infinite Number of little e-
qual right Lines, each equal to its Correſpon-
= it is manifeſt, that the Semi-cycloid ZMD
will be the Aſſemblage of an infinite Number
of little Arches, whoſe Centres are ſucceſſively
the Contact Points &, each being deſcribed

thro? the ſame Point Mor A.

3 That if from O the Centre of the im-
moveable Circle, the Arch M E be deſcri-
bed, then the Arches MG, EB, of the

moveable Circle, will be equal to one ano-
ther, as well as their Chords MG, EB, and
the Angles O@M, OBE. For the Lines
OX, OK, joining the Centres of the Circles
are equal, “ar Sooy they paſs thro’ the Points
of Contact B, G. Therefore drawing the _
. 11

of PFLUxTONS. 125
dii O M, O E, and & E, the Triangles OLM,
O KE, will be equal and ſimilar. Whence
becauſe the Angle ON M is equal to the Ang gle |
OKE, the Arches MG, BE, of the equal
Semi-circles MGB, BE A, being the Mea –
ſures of theſe An les, ſhall be N as like –
wiſe their Chords MG, EB; 41 ſo the An-
gles O GM, OBE, are equal likewiſe.

This being laid down, let mC be another F 10. 85;
Perpendicular infinitely near the firſt, we ano-
ther concentrick Arch, and Be another Chord;
and from the Centres C and B, deſcribe the
ſmall Arches GH, EF. Now the right-an-
gled Triangles GH g, E Fe, are equal and ſi-
milar; for Gg or Dg – DGS Ee, or the
Arch Be—the Arch BE: Moreover, Hg or
mpg—MG=Feor=Be—BE. Whence the
little Arch G FH will be equal to the little Arch
EF; and ſo the Angle & CH, is to the Angle
EB E, as BE to CG. And therefore t
whole Difficulty is brought to this, viz: to
find the Relation between thoſe Angles ;

which may be effected thus:

Draw the Radii OG, Og, XE, Ke, and
call OG or OB, ; KE or KB, or KA, a;
then it is manifeſt; that the Angle EBe=
OBe—OBE=O0Ogm—OGM=(drawi

ly the Angle GOg= — EBF, and & CH
4 —.— EBF. Whine GCH: EBF or BE

5 Y | a

7 _—

4 7 hs >

er ks Tt And 6 the R

Pic. 86. — —.5 if — = OA (2944) 0

| 0 15 A 8. the Point. ene je te
Evolute. .

It is wo al 155 That the Evolute be gin

442 +1 7-8tho- Point D, in which Fugen tha

, becauſe the Arch GM becomes infi-
pitely ſmall in that Point. 27, And that it
— in che Ppint V; ſo that OA: OB: 4B
25 N:: OA AB or OB: OB BN or ON;
that is, G A4, OB, ON, are continual Propor-
tidnals, ,3% H the Cirele VS. Q be deſeribed
0 then, I ſay; the Evolute
Di generated by the Rotation of the
mopesble ECS, haying GS or EN as
«Diameter, alor the immoveable one NS
That is, the; aid Evolute is a $cmi-cycloid,
{becauſe the Diameters AB, BN, of the move-
. able Circles, are to one another, as the Radii
OB, ON, of the immoveable one) having an
inverted Situation with reſpect to that of the
ther, the Vertex being in .
To prove this, let us ſuppoſe the Diameter
of the moveable Circles to fall in che right
Line O drawn at pleaſure from the Centre
O; this ſhall paſs chro the Points of Contact
S, Es and A or TG: BN or GS: :
1: EC, the Point C will be in the Evolute,
as likewiſe in the Citeumference of the Circle
6; for the Angle & MT being a right An-
zle, the Angle GCF ſhall be fo Itkewile. But
ater br: | Angles AH, CU, the
Arch TMor GB is to the Arch CS; as the
er G to the Diameter G S:: OG: 155 1

”

of FLUX10NsS. 127

2:03: NS; 2 z and therefore the Arches Ga *

are 1 hence, Ge.

Con. L.

101, _ is * plain that the Part DC of the Art. 75.
Cy cod j is equal to the right Line CM.

and ee wn DC is to the Tangent CG: 4 B

+BN:BN:: OB ON: ON; that is, as the

Sum of the Diameters of e genergring Cir- * 087

cles, or of the moveable and immoyeableon
is to the Radius of the immoveable Circle!
1 Truth of this W be ſhe wn otherwiſe,

Becauſe of che Similarity of the Triangles F : c. 85.
CM, CGH, Mm:GHorEF:: MC GC: 0.7
+0 B. (a2) :O. Whence (as in It.

the Part 4M of the Cycloid is to the cor-
— — *. Chord AE, as: The Sum of the Di-
ameters of the Semi nerating Circle, and the Baſe
is to the Semi en of the Baſe. f

Conor I

3 —— — wats

— ——

40! | Te- Trees hash cl Fer a 10. * —

XM. Pew e e)
2 e *
cn is 6H: 27 2

en STIs dena CH=BE,

“and 3 Hu ſhall be 424

EFx EB. That is the Dueperiam AG Hin
2 7 3 to r eames
Whence

- — ——U— — — — — — —Pm . — — — — — ——— — —

3 ˖KX·— —
way —

128

e
Whence the Space MG B A contained un-
der MG, AB, Perpendiculars to the Cycloidy

the Arch BG, and the Part MA of the Curve;
is to the correfpondent Segment B EZ A of

the Circle, as 2a+3þ is to b.

Coror. III.

Fic. 87. 10 3- r is evident that the Quadrature of any

Part of the Cycloid depends on the
Quadrature of the Circle; but if O © be ta-
ken as a mean Proportional between O K, OA,
and the Arch © EM being deſcribed with that

as a Semi-diameter.. I ſay, the Space ABEM,

contained under the Diameter AB, the Chord
BE, the Arch EM, and the Part A M of the
Cycloid, is to the Triangle EK B:: 2a+36:6.
For call the Arch AE or GB, a; and the
Radius O ©, z; then will OB (0): O (g)::

GB (i): R Qor M ==>. And therefore

the Space R or MG E, that is,
1TB + DD, Be Now the Space

I 102 #MCBA=E B EZ A=2IL; BxB

EW XE Z 4 (22). Now if the Space
aforeſaid be taken e this, chere ſhall remain

„ „% 2aau+-3abu+bbu—zzu 24-＋-30
ABEM H

ETB AU EEB, becauſe by Conſtru-

ction 22=24243ab+bb. Whence the ſaid
Space is the only one, among others like it,

Whoſe Quadrature is independent of that of

the Circle.

Moreover,

as.
of FLUX Ns.
Moreover, we may have another mixted
lined Space, whoſe: Quadrature is independent
of that of the Circle. For if from the Space
ABEM;: _— ent BEZ A au- EKB)

be. taken away; will remain AZ ENR |
Le en ee thi — x E K B=
—

e ata 75 8 Aas

That is, if the Semi: eircumference be biſected
in the Point E, the Space AZ EM ſhall be to
the Double of the Triangle EX B, viz. to the
Square of the Semidiameter:: OX n :
03 AY

omg: e e 3007,

104 I N the Wesch Circle 4E re- 7. 10. 885
volves within the immoveable one
G D, the Diameter 4 B thereof, which be-
fore was poſitive, is here negative; and there-
fore the Signs of the Terms affected with it
_ its Dimenſions are odd, muſt be chang-
‘ Whence,
2 If you draw M G at pleaſure perpendicu-
ki to the Cycloid, and make O A (b—2a):
OB (b): : M: GC; the Point C will be * in * 4 1003
the oute DCN deſeribed by the Rota-
tion bf the Circle (having B N as a Diameter)
BD… the Circumference NS concentrick to
29% If the Arch M E be deſcribed from the
Centre O, the Part AM of the Cycloid ſhall
= * 7 the Chord GB 4 2—24: C6. Art. 101.
e Space MG B A 1s * co the Segment Art. io.

TER: Lr
43 If we aſſume O r
vis a mean Proportional between OK, OA;
8 K then

130 » 4) [Treatiſe 0
dam reg 1B EA comprehended under
the Part HM of the Vid the ME,
the Chord dig, and the Diameter AB, ſhall

4rt. 103. be to ne EMH ifi ;þ. But
ik ve make 0 , or \OE=z\ 204-=2abÞ bb,

that is, the equal to a Quadrant
the ons ace AZE » 750 omprehended Aude e

Part 1 M of the deid, — — ches AM
. % A lde ee de dene
——— ——— of thy
— FI D * * nigh ole

203 G AY Agne T 05 10 ide 190
(AO 605 T7 oh v. art da 9716p
A)? a 4

ro. r the Radius B of the immoveable
2 4 Te Circle be . to become infinite,

2 Win che Areh B become arttmght
1 — 4MD will be 92 |
mon Cycloid. Mow nee, in this Caſe; —
the-Diameter of the moveable Circle, is
—— wich velpect i, Tha of the immoyeable
| Dhefetorey ic rc 2c nag!
„ MG:GC: 47 b, becauſe þ4: 281, this
is, ee. and conſequently, if you aſ⸗
fume BVA B, and draw the righeLineNS |
$621 4. parallel to B D, the Evolute DV ſhall be
| generated by by the: Rotation of the Circle,
ving BN as a Ben vr alen the
Baſe NS.

83 2% The part 4 M of the 2 leid i is tothe

correſpondent (Chord A : : +4: b,’; And the
Nr BEA
4 . YO. 9117 * * hag
De. ps 2 3% Becauſe BY, or + ©. OB, Mich I
call x, is =Þ-b+ /24a4- 3ab++bb,’there comes
Mite Tan, aan and æ a ſince
the Terms affected with b vaniſnh, they be-
ing

of F.1vX1oNs. |
ing infinitely lefs than the other Terms. Con-
quently if in the common Cycloid you aſſume

“BP=$ AB; and draw PAM parallel to the Baſe F 10. 83.

D; the Space AM EB ſhall be the Triple
-of dne Triangle EK B. In like manner, yon
will dind that When P coincides with the Cen-
tre X, che Space AZ E M contained under the
Part AA of the Cycloid, the right Line ME,

:und:the Arch 4 E; Shall be equal to che Squire

of the Radius; Which. on uu A
nnn cre 999:

{. 3 $1IGOhDY 0 | 167 T1910 7 Wn

$EnoLk vu.

A Ways to each other; the Angle
DOG hall likewiſe be. always to the Angle
M:: GK: OG. Therefore hen we have
given O the beginning df the Cycloid DILL,
che Radii OG, GA, of the generating Circles,
and the Point bf; Contact G, 11 you have a ind
to determine the Point M in this Poſition
which deſoribes the Cycloid, it is only draw-
ing the Radius KM mr that tbe Angle GN
be et en Angle DOG: OA RN. NW
this may be done . always, when
the Relation ofithe ſaid Radii can he exp
in Numbers; and ſo the Gyeloxd DM wall
Then be a trical Curve.
For let, for Example, 08 G
I is manifeſt that the Angle MAG — *
25908 che given Angle. And ſo the whole
Matter conſiſts in dividing; the Angle DOG
into five ęqual Pants.) Bit every Ceometri-
cian knows, that a given Angle or Arch may
be divided into any Number of egual Parts ge-
nnen, In always comes nit

— adam Andie 6, Gil, why . 6. $44

132 * A Treatiſe | W

ation containing ſtraight Lines on]
9 Sc. | FE 5 4
I fay,’ moreover, that the Cycloid D AM
is a mechanical Curve; or, which is the ſame

thing, that the Points M in it cannot be 80
metrically determined, when the Ratio of OG

to & & can be expreſſed only by Surds.

Fic. 89. For every Line, whether mechanical or ge-
ometrical, returns into itſelf, or elſe goes on
(or is extended) ad inſinitum; ſince the Gene-
ration thereof may be continued at pleaſure.
Now when the Point A, in the moveable Cir-
cle ABC, in one Revolution, has deſcribed
the Cycloid A DE, this will be but the firſt
Part of the Carve; ſince as the Circle rolls
on, there will be deſcribed a ſecond Patt EF,
and a third G HI, &c. until the deſcribent
Point A, after ſcveral Revolutions, returns
again to the ſame Point in the immoveable
Gele from whence it went. So that if you
again revolve the Circle AB Cas before, the
ſame Curve Line ADE FG 77 will again
be deſcribed by the Point 4. Now when the
Radii of the generating Circles are-incommen-
ſurable, their Circumferences are ſo likewiſe
and co A y the deſcribing Point A in the
moveable Circle AB C, can never return again
to the Point A in the moveable Circle from

whence it went, be the Number of Revolu-
tions never ſo many; therefore there may be
— an infinite Number of Cycloids that all toge-
ther make up but 8 ADBFGHL
&cc. Now if an indefinite right Line be drawn
_thro’ the immoveable Circle, it is plain that it
ſhall cut the Curve continued ad inſinitum in
2 an infinite Number of Points. But becauſe
| | 5 25 Equation N the Nature of a geo-
SE metricat

of FLUX1ONS. 133

metrical Line, muſt at leaſt be of as many Di-
menſions as is the Number of Points that a
ſtraight. Line can cut the Curve in. There-
fore the Equation that expreſſes the Nature of
the Curve I D EFG Hl, &c. mult be of an
infinite Number of Dimenſions: Which be-
ing impoſſible, it is plain that the ſaid Curve
is a mechanical or tranſcendent one.

c II.

107 * find any Number of Involutes AM, F 16. 90.
BN, EFO, to the given Evolute BF C.

K is manifeſt that the Points A, B, F, of the
Thread AB FC of the Evolute BF 2 will de-
ſcribe the Curves AM, B N, FO, to which
the given Curve BFC ws the common Evo-
lute. But becauſe the Curve FO is deſcribed
by the Evolution of the Part FC, it does not
begin in F. And in order to find where it
does begin, the Part B F remaining, muſt be
taken as the Evolute, the Point being that
to which the Thread is fixed; and beginnin
at F, the Part EF of the Inv hte EFO mu |
be deſcribed, which begins in E, and is the
Involute to the whole Curve B Cc. |
If you have a mind to find the Points M,
N, 0, without the Thread AB FC, you need
only aſſume the Parts C, CN, CO, equal
to. AB FC, BFC, 765 in any Ta CM
“SINE P A. |

Conor.

H= EN OE it S 15 „That the bnd
Curves AM, BN, ‘EFO, differ very.

| ney from each other. I mean as t6 their Na-
K 3 ture;

134

Ireatiſe

fincein the Vertex / of the Curve AN;
che on th of Evolatiortis equal ro 4 B, wheres |
as that of the Catve BN’is nothing. It is like –
wiſe evideny from the very 6 5 87 the Curve
5 762 7 7 it differs very much from the Curyes
© 2% That the Curves AM; BN, Eo, are
geometrical ones ofily, when the given Evo-
ute B FC, is a geometrical Curve, and recti-
fiable : For if the ſame be not a geometrical
Curve, when BA is aſſumed for an Abſciſs,
the Ordinate X C cannot be determined geo-
metrically; And if it be not rectiſſable, when
og Tangent CM is drawn, the Points M,N,
cannot be determined in the Curves AM,
25 90 EF O, becauſe ſtraight Lines carmot be
fd e qual to the 180 BF C, wa the Pings
B Br F ; erco,

Lads.

8 8 UM.

F. 91. 0 115 Evolute BAC has in

flexion in A; then from the Evolution

of the Part B A D thereof, beginnitig at rhe
Point D (not the Point of Inflexion) there
will be formed the Part BE of the Invo-

Nr and from the Evolution of the Part DC,

Will Io? the temaiting Part D Gt the

Invol.

E, the Cancavities of DE, EF, lie
555 – Now. it the 225 oregol 5 rhe

P —— — — EEOC EI 7s

So that the whole involute Curve
.formet! from the Evolution of the Curve BAC
will be FEDG. Now it x evident that this
Curve has two Points D and E of Rettogref-
ſion, with this Difference, that at D the Parts
DE, DG, have oppoſite Convexities _ at

etermundtion | of Pohtits of Retrogreſſion

191 01 1 A 6f

of HFrurloNs. 135
of the ſame Nature with D was handled,
therefore ‘ E ſhall bere ſhew the Manner of

| finding the Pois E, which may be called
Points of Retrogreſſion of the tecond Kind;
and this is what no body, that I know of,
has hitherto” conſider e.
In order to which, draw at pleaſure two
Perpendicuſars MN, in n, to the Part YE ter-
Leiner br the Points N. u, of the Evolute;
from Which draw two other Perpendiculars
NH, to NM, un; then the ſmall Sectors
MNm, N Hu, will be ſimilar, becauſe the
Angles MNn, Min, are equal. Therefore Nu:
Mm::NH:NM. Now in the Point A of
Inflexion, the Radius VH becomes * infinite 47. 81.
or nothing; and the Radius MN, which be-
comes AE, continues finite. Therefore in the
Point E of Retrogreſſion of the ſecond Kind,
the Ratio of Nu, the Fluxion of the Radius
of Evolution M, ta- Mm the Fluxion of
the Curve, miſt becottie finitely great or
or infinitely ſid; Ang becauſe * Nn = Art. 86.
E O.IOAT AH

—E +} +5}

A—

5
5 *25 + * TM * +2
Vx +)”, therefore =; 145 =2, or In-

finity z and multiplying by &), we have the
following gene Expretfon x*3 +) 3—335″
Do, or Infinity, for determining the Points

of Retrogreſſion of the ſecond Kind.

If an Involute DEF, or HDE FC, has aF 1c. 92,
Point of Retrogreſſion of the ſecond Kind, the 93.
Evolute B AC may have a Point of Retrogreſ-

ſion of the ſecond Kind alſo; ſo that the ſe-

cond Point A of Retrogreſſion anſwers to the

ſecond Point E, viz. both lie in the Radius of
XK4 Evolution

» and Mm =

„

—

9 5
F P ⁵˙T nk od. ated

136

this Su

Evolution iſſuing from the Point E. Naw in

ppoſition it is evident, that the Radius
of Evolution EA, will be always a Minimum
an Maxima. And cherefore the F Iuxion of

455 76. 2+ 1 “he general Expreſſion * for the Ra

—. Toa

dit of Kvalution; muſt beo, or e in the
Point E ſought; from whence we get the ſame
general Expreſſion as before; which muſt be
uſed to inveſtigate the Pas ah Wegen
of Is ſecond Ba 1. ;

r
a „ 1 31 ‘ ? * ,
4 8 4
\ £ —
; FS 4 * ah»
4 4 , 1 . 9
2 * P
” 1 : ” — — 4

311 1 R911 1 * 7711 : 1

. 4 ” k & 3 = £ SZ? LiS inf

% PR * od .
. F © * * . iy , «# 4
4 4459814. e 3 1 © : ‘3 +
1 —
6 £
. 4 * i 4 – . 4
7 1 – 4
g l | 4
Fg v4 4 | *

” : + 5 *

. 1 4
4 p
b .
6 e
_ -. *

5 *
& * Þ mY
%
—
–
wen. — * *
:
- “W205
  . *
  Fn 7
  “4
  „ ow
  : FP
1 1
. s * 2
4 7+ 5 ,, 4 * –
, 1 . a
4 2 “Y m 4 * Y 0 138 4
e * » 8 > Ca#3p4®S & 1} * 4
%
N , 8 F.
1 1 \
= 140 £ : 5 1 19 1 1
.
172 28 wi Ju 1
LIT ESI IE 14 U 10
f »
x .
n .

Q wk bw 4 bd $+4% £# Þ\ .

E. © F 1
w – * 4
10 f 1 Ae v ö 8
„ 4 wk >
| * =
{1 C4434 4.424 1 1 7 :

* – |
1 le . M

8 8 —_ | 4 2 4 8 N 4, S. > B»
The Uſe of Fluxions in finding of *
Cauſtics by Reflection.

.
N

19 b. DerINIT ION.

T an infinite Number of Rays B A, B M, BD, FC. 94.
=» iſſuing from a luminous Point , be reflect- 95.
ed by the Curve AMD in ſuch manner, that

the Angles of Reflexion be equal to the Angles

of Incidence; the Line Z FN. that touches

the reflected Rays, or their Continuations. AH,

MF, DN, is called a Cauſtick by Reflexion. –

Corot. I.

I 10.JF HA be continued out to V, ſo that FG. 94.
IAA B, and the Cauſtick HFV be
taken as an Evolute, and IA as the firſt Ra-
dius of Evolution, then will the Involute /LK
to the ſame, be of ſuch a Nature, that the |
Tangent FL. ſhall be“ conſtantly equal to the ,,;. 7.
Part FH of the Cauſtick Plus the right Line
HI. And if Bm, m, be ſuppoſed two re-

flected Rays infinitely near B M, M; and if

Fm be continued out to /, and the little Ar-

ches MO, MR, be deſcribed from the Cen-
tres F, B; the ſmall right-angled Triangles
M On, MRm, will be equal and * fe

| | mee

I 138 „2 A reutiſe \ I

| | Since the Angle OmM= Fm D=RmM, and
[i the Hypothenuſc Aim is common ; ig that On:
|

j Nn. Now becauſe Om is the Fluxion’otf
[|| a and Nw the Fon of F , dad Kis
ll F alFays 10, wherever the Pon M be taken:
i Therefore I. IA, or AH+ HF— ME,
| Art. 96. the Sum of all * the Fluxions Om in the Part
| AM of the Curve, is B M=B 4, the Sum
j Art. 96. U of all the Fluxions R in the ſame Part
1 AM; and conſequently the Part HF of the
1 Cauſtick FN, will. Pe Gaal to B M—B A
i 4M F—A H.
iſt There may happen:ſeveral Caſes, according
i zz the incident Ray B A is greater or leſs than
14 BM; and the reflected Ray HA, as/a Radius
[| bu of Evolution diſengages irſelf, from the Patt
N of the Curve, to become MF; But we
can always prove, as we have already, that the
Difference of the Radii of Incidence, is equal
to- the Difference of the reflected Rays Plus
the Part of the Cauſtick taken as an Evolute
that one of the Rays diſengages itſelf from, be-
5 fore it coincides with the other. For Exam-
Fr 6.95. ple, B M—B A=M F+FH—A H; and con-
v2 – – | Tequently H= M==BA+A HMT.
Fro. 94, If the Arch A P be deſcribed from the Cen-
rence of the incident Rays B A, B A. And
if the luminous Point B becomes infinitely di-
Fi e. 96. ſtant from the Curve 4D, the incident
on Rays BA, B M will become parallel, and the
Arch A a ſtraight Line perpendicular to
thoſe Rays. 5 R 65 f 75 NE / , | *
. 912911 13 -Q o R O L. II. $3019 2

— r

2 a * 44 1 3 _ —]— —

” — — — —
Ia a —_ —— — – £ – — —_ — “
— 8 2

an
——
-_

= . a —

—
— 2
— Om
4

P16. il I= che Bigure BA MD be fuppoledtobo
I mene en he fame Plane, fo hay tho
VT” Point

of FTU Ns.

Point, B coincides: with H and the Tangent to
the Curve A D in its firſt Situation, ſtill
touches it in this lattet one; and if rhe: Curve
42 MA — on —— or fel; ſothan
che P arts 4 A 9 4 $ = I –
by this Motion the — — 4
kind of Cycloit II. A, which will be am In-
volute to the Cauſtick H 5 taken as an E-
volute. P

For from the Generationi it ors. 1 That

the Line LM drawn from che deſeribem Pt
L, to the Point of Contact M, will be“ per- Art. 43,

icular to the Cuive TL A. 22, That La
or L 4, and L’Mz=BM. ze, That the
Angles made by the right” — ML, BM,
with the common Tangent in M are *
And therefore if LM he continued out to I.
ae be de Incident Ray Þ 5 reflected!
Whence the Perpendiculat LF touches the
Cauſtick HF N; and fince this is {6 always,
fer the Point L be taken where it will, it is
phat that J LX is the Involute to the Cauſtick
AEN, the right Line Hi being the Radius
of Evolution.

Hence it follows, that the Part FH or F,,

HT F=B M4-M F—B A—A4H. Which is
what has been otherwiſe demonſtrated in n the
Corollary aforegoing.

Conor. III.

112 Wes the Tattgem DN is inbuitel
; near the Tangent FM, it is mant-
feſt that the Point of Contact N, and the
Point V of Iiiterfe&ion , will both coincide
with V the other Point of Contact: So that
the Point F, way the” reflected Ray MF
5 touches

140

Fic. 55

N89

Art. 110.

214 Treatiſe

touches the Cauſtick HN is determit d, with
only ſeeking the Concurrence of the infinitely
near reflected Rays AA E m7. Conſequently
if we ſuppoſe an infinite Number of infinitely
near incident Rays, the Interſections of theſe
Rays reflected; will forma Polygon of an in-
„ ATI —_— = the en HEN.

ner. os oy

2 .
2 ” Dy

AE Nation ef ube Cure AMD, the
. Iuminons Point B, and the incident Ray
B M being, given: To. find the Point F in the re-

flefted’ Ray MF cw in 1 18 mie it

Aber the Cauſtist.

Pino (by Section 8 M the 5
of the Radius of Evolution in the Point
AV aſſume the infinitely ſmall Arch Mm, draw
the right Lines Bm, Cn, m, from the Cen-
tres B, I, deſcribe the ſmall Arches MR, MO,
draw C E, Ce, CG, Cg, perpendicular to the
incident and reflected Rays; and laſtly, call

the given Quantities BM, y; ME or MG, 4.

Now we. prove (as in * Coroll. 1 .) that the

Triangles M Rm, MO m, are equal and fimi-
lar; and ſo M R=M O. But becauſe: of the

Equality of the Angles of Incidence and Re-
flexion, we have alſo CE=CG, Ce=Cg; and
therefore CEC e or E ma — — Cg or SG.
Whence becauſe of the ſimilar Triangles BMR,

BEA; FMO, FGS,BM+BE(2y—9):
BM(9): : MR+E9 or MO+6GS:MR or

1 Of: MG (s): MF===

2479 —
If the luminous Point B falls on the other

| – Side the Point E, with reſpect to the Point M,
£11990 * $5 or

of FLVUxToRs.
or (which is the ſame thing) if the Curve AMD

be convex next to the luminous Point B; then
will y be negative, and conſequently F =

141

” * . *
lr 008 “1 811 7 011
7 —— . EE : * 1 : ” 1
YA 2 ＋4 401 ary

If be infinite, that i is, if che point B he at F c. 96.
an infinite Diſtance from the Curve 4M,
then the incident Rays will be parallel, and
MP — I mg art is ys, ien

> ry TAs Av

4 wt wb | * 1 541 # +»

Cone. bob mul

2 n v 41 27 208 |
1102 zcivse MF has but one Value, Fc. 94,

wherein is the Radius of Evolution; 95-
therefore the Curve: 4 M can have gnly; one
Cauſtick by Reflexion H _ —_ 8 is by Aer. 80.
dare one Evolute to tt . NA 1
4 I. 5 £ 3 ws wid 7: Mr <0
Conor. II.

Ly: Ani the, lovolute AMD is, a geo- F 6. 97.
metrical Curve, it is manifeſt * the . 4. $5.

Evolute is one likewiſe z, that is, any Point C

in it may be determined geometrically; and

conſequently any Point F in the Cauſtick there-

of, may be determined geometrically, or (which

is the fame) the Cauſtick NN will be a geo- F 1 ©. 94,

metrical Curve. I fay, moreover, that the ſaid 95:

Cauſtick is always re&ifiable ; becauſe by means

of the Curve AMD, which is ſuppoſed to be

a geometfical one, it is * manifeſt that ſtraight · 47. 110.

1 me be ound ber to any Part ef,

4 abe cee.

| pep
11 WW — bee the reflected Ray is

141 2 A Treatife \ \o

‘TY 1 mA uod 141 Alice Wo 6} 5 {404 4 10

(1 de ex ly 1 HE * 25vun05 or

ne ee 2d x Up

F 16. 97. 116. Ir the Curve AMD i next to
. the luminous Point 3 the es; of

“EC rn) bl Aves he poi; and

the Point # \auſtif.conſequoneebe-allaned,on
cheiſame:Side-che Point M as che Point G, a8
we — ſuppoſed in the —— grH-
going. Therefore the 3 reflected
ays do diverge. ; .
Bur if the Curve AMD be concave next

40 — 10 the lumzino bas Point , z MEN wal
1 — hers is greater than $a, nega-
cd v. ® doc when he {ame \is leſs, and infinite when
it is A. Whence if a Circle beideferibeil
with a Diameter equal to 1 MC the Radius of
Evolution, the infinitoly near reflected Rays
will converge, when the luminous Point g falls
8 9 1 7 the Oireumference thereof], dt di-
-24 n verge when the fame fille wirhin; ald finall

5 ai be p ar: n fon it falls in the: Circumfe-
rence. N 5 : . Ani 9 (ig! Nl A
Ai WE £2113 A 23am 10% yok up {0
7 II 1 SE 15. > 0 * 14 IV. Wy fact E 3
Ne. 11 nn 9) Nin MIA 420116} 33 (rot 21

20 1 5 ob meidene Ray! B u.. touches the
Er —.—— AMD in the Point , then

5 0j end therefore 14 —

is then in the

Direction 6f the ineident Ray, and it is the
Nature of the Cauſtick to = all the refle-
Fed Rays; therefore it ſhall touch likewiſe
the incident Ray BM in the Point M; that
185

_ A OO PISS *

au. Maas T e aca us RD 2

of FLUX LIONS.
is, the Cauſtick and che given Curve will have ä
the ſame Tangent in the common Point MN.
If the Radius CM of Evolution be nothing,
then fill will ME (a) o; and conſequentiy
MF=0, Therefore the given Curve, and 8
the Cauſtick, make an Angle with each other
in the common Point M, equa) to the Angle
of Tncidence.
If the Radius of Evolution c M be infinite;
2 Arch Am will become ſtraigt
e, and MF N hecauſe ME ſa) being
infinite, Y will be nothing with regard tb,
Now ſince this Expreſſion os Value: i neg.
tive when the Points B, C, are both on- the
Gme’Side the Line AMD; amd poſitivewher

one is on one; ſide, and the other on the other;
Therefore the infinitely near reflected Rays

5 Ae lien en! 184 mh

any 99

00 811
211 { “Ve OE At. n un 5918
, 7 Pg “©.
; Oc 0 * 0 L. N al lech in
9

*
: * to ;

I 118 415 T is 233 chat When any two of the
Points B, C C, F, are given, the third is
eaſily found. ;

16, Let the 1 hve AM D be a ci Fic. 98.
rabola; and the luminous Point B the ——_
Then (from the Conick Elements) i
niſeſt, chat all the reflected Rays ill —
rallel to the Axis; and. ſq in — Poſition of
the Point M, MF will he infinite; and con ·
ſequently 4==29, ‘ When if you aſſume ME
HA, and dra the Perpendicular 26,
this will intorſedt the Perpendiqular AC to
the Curve AM, :utithe Fim C niht
rere a af 5 ne At

AX ih 21 c hllszeg avrents 4221

ii.“

—— — — 73k — ———
9

144

Fic. 99.

Art. 113.

Fr o. 110.

FIN os

F in the reflected Ra

Art. 113.

2 A Treatiſe \q

52N Let AMD be an Ellipfis, and the lumi-
nous Point B one of the Foci. Then it is evi-
dent, that all the reflected Rays MF will co-
incide in the other Focus F. And if you call

MF, 25 then will Are $ fork whence
comes a ME LP = THE Which, Was

— 5K.

But if the Curve AM D be an A pre,
the Focus F will fall the contrary Way; or

vithout the Curve; and therefore M F(z) will

become negative; and conſequently A4 Ea)

kT w_ — . Whence ariſes the following

7 2

e Which will ſerye for che Hu-
O. e 4 > . 1

Aſſume ME a fourth Proportional vo half
the tranſverſe Axis, and the incident and refle-
Cited Rays, and draw the Perpendicular EC:
This ſhall interſect the Line MC perpendicu-
lar to the Curve in the Point 2 when dl
be in e ann | |

1411

* 15 v4 4

ExAMPLE I.
tp ef { :

ng» ET AMD be a e ha let 1
incident Rays PM fall perpendicular
to the Axis. It is required to find the Points
1 * wherein they
ER 0

touch the Cauſtick-4 n
If we draw M the Radius of Evolution,
. Line CG icular to the refle-
cted Ray MA F, it is manifeſt ® that we muſt
aſſume HF = MHC. But this Conſtruction
ma be ſhorten’d, if you conſider that *
Nis drawn parall to the Axis, AP, and

the

of FLUX10NS.

the right Line ML to the Focus L, the An-
gles LMP, EM ſhall be equal, fince from
the Nature of the Parabola LM 9 = © MN,
and by the Suppoſition PM © =D MF. It
then the common Angle PM F be added to
both, the Angle LA 4 ſhall be equal to the
Angle PM, that is, a right Angle. Now it
has been demonſtrated (Art. 118. Num. 1.)
that L H perpendicular to ML ſhall interſect
the Radius of Evolution MC in ¶ the middle
thereof. Therefore if MV be drawn equal
and parallel to LI, it will be one of the re-
flected Rays, and will touch the Cauſtick
AFK in the Point F. Which was to be
found. | 25

If the reflected Ray MF be ſuppoſed paral-
lel to the Axis A, it is maniteſt that the
Point F of the Cauſtick will be at the great-
eſt Diſtance poſſible from AP, becauſe the
Tangent to that Point will be parallel to the
Axis. And conſequently in order to determine
that Point in all the Cauſticks, as AFA, form-
ed by incident Rays parallel to the Axis of the
gs Curve, we need only conſider that then

9 .

Now let ax =yy. Then) —— ,
and fo 4 P(x) a: that is, when the Point
P falls in the Focus L, the reflected Ray MF
will be parallel to the Axis; which is a known
Truth otherwiſe eſtabliſh’d ; ſince in this Caſe
MP coinciding with LM, M F muſt alſo co-
incide with MN, and LH with L. Whence
MF will then be equal to ML; and therefore
if FR be drawn perpendicular to the Axis, AR

or AL + ME will 3 a. We may ob-

leryc;

& muſt be equal to , And therefore
x.

145

A Treatiſe 7
ſerve, that in this Caſe the Part AF of the
1 Cauſtick is equal to the Parameter, ſince it is
N Art. 110. * always equal to PM MF.

i: To determine the Point & wherein the Cau-
14 ſtick AFK interſects the Axis JP, the Value
hd of MO muſt be found, and made equal to
| that of M; for it is plain, when the Point
N | F falls in K, the Lines MF, MO become equal
0 1 to each other. For when the Point Fcoincides
ol | with K, it is manifeſt that the Lines MF, MO
will become equal to one another. Therefore
if the unknown Quantity MO be called t; from
the Biſection of the Angle PM O by M perpen-
dicular to the Curve, we ſhall have MP (9):
MO (i) :: P 2 10 09. = 2. And

146

therefore OP = et) 2 = FF — 9), becauſe

of the right-angled Triangle MP O; and di-
viding both Sides by t +, there comes out

1
— 75 I5 from whence we get MO (7) =

Art. 77. LEI =MF(o) => ſince * ME

„

(a) ., And ſo by means of —2.y5 +#, the Point P may be determined ſo, that
drawing the incident Ray PM, and the re-
flected Ray MF, this latter will touch the
Cauſtick AFK in the Point K, wherein it in-

terſects the Axis AP.

— .

Now in the Parabola ) =x*, ix 77,

ji, and ſubſtituting theſe Ex-
preſſions in the foregoing Equation, there will
Fre come

of FLUXIONS.
come out by * bs * —=x*; and ſo we

ſhall get A P(x) of the Parameter.
Now. to find the Nature of the Cauſtick

(ESE or —2)y :3—Y::MP (3): MS
—— * ;
—

et IE; 29%) ).
0 ni bp: 2257 ):8PorPR

(a — x) 3 Therefore with 2z = y +

LE „and 4 =x + 2 together with the
2

—2
Equation of the given Curve, we get a new
Equation freed from æ and y, which expreſſes
the Relation of A R (u) to FR (⁊).
When the Curve AMD is a Parabola, as
3

in the Example, we ſhall get z x – or
({quaring each Side) 2x G + 4x* = zz, and

=3x. Whence we get the Equation
ſought 822 =4#*—jauu + 3aau expreſſing
the Nature of the Cauſtick A FR. Here we
may obſerve, that PR is always the Double
of AP, becauſe A R(#) == x. From whence
we get moreover another way of determining

the {ought Point F in the reflected Ray MF.

L 2 E X A M-

130

» A Treatiſe

EXAMPLE II.

“FORD © | Ba AMD be a Semicircle, the Line
| AD a Diameter, and C the Centre,
and let the incident Rays PM be perpendicu-
lar to AD.

Becauſe the Evolute of the Circle is a Point,
viz. the Centre of it, therefore if CM be bi-
ſected in H, and HF be drawn perpendicular

Art. 113. to the reflected Ray MF, it ſhall & cut the
ſaid Ray in the Point F wherein it touches the
Cauſtick 4 FK. Now becauſe the reflected
Ray MF is equal to + of the incident Ray
PM; therefore when the Point P coincides
with C, it is plain the Point F will coincide
with & the middle of CB. And the Part
F is the triple of MF, and the Cauſtick
4A the triple of BK. We may likewiſe
obſerve, that when the Angle 40 M is made
one half a right Angle, the reflected Ray MF
ſhall be parallel to 4C; and therefore the
Point F will be the higheſt Point of the Cau-
ſtick above the Diameter AD.

The Circle whoſe Diameter is the Line
MH paſſes thro’ the Point F; becauſe the
Angle ZFM is a right Angle. And if from
the Centre C, with the Radius CK or CH
the half of CM, the Circle XN be deſcri-
bed; the Arch H ſhall be equal to the Arch
HK: for ſince the Angle CM is equal to
CMP or HCA, the Arches x HF and HX
that meaſure the ſaid Angles in the Circles
MFH, K HG ſhall be to each other as the
Radu 1 MAH, HC of theſe Circles. Whence
it appears, that the Cauſticx A FX is a Cy-
£loid generated by the Rotation of the *
4 7x Sg –

148

of FLUXIONSS. 149

cle MF H along the immoveable Circle Le,
the Beginning thereof being X, and the Ver-
tex A. fit i

ExaMPLEe III.

| Ba VOD be a Cirele, the Line AD Fa. 103.
the Diameter, and the Point C the
Centre; and let A, one End of that Diameter,
be the luminous Point from whence all the

Rays of Incidence A M flue.

From the Centre C draw C E perpendicular
to the incident Ray AM; then from the Na-
ture of the Circle, it is manifeſt that the Point
E biſects the Chord AM; and ſo ME (a) =

1). Whence MF ( 0 : that is, we

muſt take the reflected Ray M F= AMthe in-
cident Ray. Conſequently DR AD, CK=3
CD, and the * Cauſtick A FK =4 AD, like Art. 110.
as its Part A F AM. If you aſſume AM
= AC, the reflected Ray M F will be parallel
to the Diameter AD; and conſequently the
Point F will be the higheſt Point poſſible
above the ſaid Diameter. 1

If you take C=3CM, and draw H
perpendicular to MF; the Point F ſhall be in
the Cauſtick: for drawing H perpendicular
to AM, it is manifeſt that ML ==3 ME =
FAM, becauſe M HIC M. Therefore the
Circle having MH as a Diameter, ſhall paſs
thro’ the Point F of the Cauſtick; and if from
the Centre C with the Radius C X or CH ano-
ther Circle XH be deſcribed, it ſhall be
equal to the former one, and the Arch HRK
ſhall be equal to the Arch HF: For in the
Iſoſceles Triangle C MA the external Angle

L 3 | KCH

150 A Treatiſe ©
KCH=2CMA=AMF, and therefore the
Arches HK, H r. the Meaſures of thoſe
Angles in the equal Circles, ſhall likewiſe be
equal. Whence it follows, that the Cauſtick
AFK is likewiſe here a Cycloid generated by
the Rotation of the Circle MFA along the
immoveable one KH, the Beginning being
K, and the Vertex A. | a
This may be demonſtrated otherwiſe thus.
If a Cycloid be deſcribed by the Revolution
of a Circle equal to the Circle AMD about
this ſame, beginning at 4, we have demon-
ſtrated (Art. 111.) that the Cauſtick {FX
will be the Evolute to the ſaid Cycloid taken
Art. 110. as an Involute. But *the ſaid Evolute is a Cycloid
of the fame Kind, viz. the Diameters of the
generating Circles of them ſhall be equal; and
the Point X will be determined by taking CX
a third Proportional to CD DA and CD,
viz. equal to CD. Whence, Sc.

E x A N IV.

Fre. 104. 122. Er the Curve 4M be one half of
na common Cycloid deſcribed by the
Rotation of the Semicircle VCM along
the right Line B D, whole Vertex is A, and
the Beginning thereof D; and let the incident
Rays X M be parallel to the Axis A8.
Fre. 10% Becauſe * MG is equal to the half of the
Art. 95. Radius of Evolution; therefore * if G be
Art. 113. drawn” perpendicular to the reflected Ra

MF, the Point F-ſhall be in the Cauſtic
DFB. And ſo MF muſt be aſſumed equal

of FLUX IONS. 151

If from I the Centre of the generating
Circle MGM to the Point of Contact &, and
to the deſcribent Point M be drawn the Ra-
di HG, H; it is evident that HG will be
perpendicular to B D, and the Angle C M H=
MGH=GMX£: whence the reflected Ray
MH paſſes thro’ the Centre H. Now the
Circle: having GH as a Diameter, paſſes thro?
the Point Flikewiſe ; becauſe the Angle G FH
is a right Angle. Therefore the Arches GM,

GF the Meaſures of the fame Angle GN
ſhall be to one another, as the Diameters MN,
GH of their Circles. And conſequently the
Arch GFS GN GB. Whence it is evi-
dent, that the Cauſticx DFB is a Cycloid
deſcribed by the entire Rotation of the Circle
GFH along the right Line BD.

EXAMPLE V.

123.Þ-E7 the Curve AMD be {till the half p;c. 103.
of a common Cycloid, the Baſe BD
whereof 1s equal to one half the Circumfe-
rence ANB of the generating Circle; and
F let the incident Rays PM be parallel to the
Baſe B D.

If & © be drawn perpendicular to PM, the
right-angled Triangles G. A, BP will be
equal and ſimilar; and therefore M = P N.
Whence it follows, * that M mult be aſſu- * 4re. gs,
med equal to the correſpondent Ordinate PN II.
in the generating Circle A NB. –
WM hen the Point Fis at the greateſt Diſtance
poſſible from the Axis AB, the Tangent MF
in that Point muſt be parallel to the Axis.
And conſequently the Angle PMF will be a
right Angle, and PM or PV half a right

L 4 Angle.

KY y Y pn v7

152 Vl Treatiſe
Angle. Whence the Point P falls in the Cen-
tre of the Circle AND.

Here it is worth obſerving, ha as che Point

P afterwards accedes towards the Extremity
B of the Diameter, fo does the Point F like-
wife accede towards the Axis AB, until it
comes to a certain Point XK; after which it re-
cedes therefrom till it comes to D. So that the
Cauſtick AFKFD has a Point of Retrogreſ-
ſion in Kk

10, To determine which, we muſt obſerve * that

1j. the Part AF=PM+ ME, the Part AFK
HLA, and the Part XFof KFO is
=HL-+ LK— PM— M F: whence HEL +
LX muſt be a maximum. And if you call
AH; H,; the Arch A; then will
HL + LK =u+23, the Fluxion whereof is

eee and 7 20% Do, by ſubſtitu-

8 ung = — whence we getſa⸗ 4=

ER becauſe of the Cue:
100 Waser A 1 (* .

Conor. 5 ON

I 124. 14. Pan Space APMor AFX FAI con-
1 tained under the Parts of the Curves
2 or AFK R, 4M, and the reflected Ray
MF is equal to o the half of rhe circular Space
APN. For the Fluxion thereof, viz. the
Sector FMO is equal to che half of the Re-
ctangle Pp SN, the Fluxion of the Space
APN; fince ther rght-angled Triangles MOm,
E Ae kay een al, MO ſhall be
1 equal to MR or JP or Pp, and moreover

MF=PN, |

ExAMs

of FrUxXxIONS.

EXAMPLE VI.

Lr the Curve AMD be the half of FI. 106.

a Cycloid generated by the Rota-
tion of the Gircle MCN about AG K equal
to it, and let A be the beginning thereof, and
D the Vertex. And let the incident Rays
AM all iſſue from the Point A. The Line
B H joining the Centres of the generating Cir-
cles conſtantly paſſes thro’ the Point of Con-
tact &, and the Arches G M, G A, as allo their
Chords, are always equal; alſo the Angle HGH
=BG A, and the Angle &MA=CGAM. Now
the Angle HG M+B GA=GMA+GAM),
becauſe if the Angle AG be added to each
Side, there will be two right Angles made.
Whence the Angle HG M hail be always
equal to the Angle GNA; and fo likewiſe
to the Angle of Reflection G MF. Whence
it follows, that MF always paſſes thro I the
Centre of the moveable Circle.

Now if CE, GO, be drawn perpendicular
to the incident Ray AM, it is manifeſt that
MO=O 4, and O E=430 M; ſince the Point

C being in the Evolute, GC M; there- Art. 100.

fore ME =3 AM; that is, a =; and con-

| a N—. ;
fequently M F ( . Whence if
you draw & F perpendicular to MF, the Point

will be in the Cauftick A FX.

The Circle whereof G His a Diameter, does
pals thro’ the Point F; and ſince the Arches
GM, 16 F, the Meaſures of the Angle GHM
are to one another, as the Diameters MN, GH,
of their Circles; the Arch G ſhall be equal
to & M, and conſequently to the Arch G& A.

From

154

Art. 120.

Fr6. 107.

Ari. 91.

2 Treatiſe Jv
From lo } it is evident, that the Cauſtick

AFK is a Cycloid generated by the Rotation
of the moveable Circle HHG about on along

the immoveable one A 94.

Con Av.

1 F a Circle be deſtnbed about the Cen-

tre B, with a Radius equal to BH or
AK; and an infinite Number of right Lines
parallel to B D falls on the Circumference of

it: it is then manifeſt * that by their RefleQi-

on f will form the {me Cauſtick A FR.

ExanyLe VII.

127 1 * AMD be a Logarchmick e
| and let the incident Rays * ) all
iſſue from the Centre .

I the right Line 04 be drawn Kom C,
the End of the Radius of Evolution, pe
dicular to the incident Ray AM, it ſhall meet

the ſame in the Centre A. Th erefore A M
00 Sa; and conſequently N ( —)

Whence A ſhall be an Iſoſceles

Triangle; and ſince the Angles of Incidence

and Reflection AMT, FMS, are equal to one
another, the Angle A ‘F Mc is equal to the An-
gle AMT. Whence it is manifeſt that the
Cauſtick {FX will be a Logarithmick Spiral,
differing only ou “wg b one Fl D

in *

” Io 4-1

4 A : N 1 4 = al

% Simon: Bios,
” 5 :

of Froxloxs.

PRO p. II.

1557

I 28TH E Cauſtick HF by Reflefion, and the pio. —

luminous Point B being given, to find
an infinite Number of Curves, as AM, whereof
the ſame is the Canſtict by Refleftion.

Take the Point A at pleaſure in any Tangent
HA, for one of the Points of the Curve AM
ſought; and from the Centre B, with the Di-
ſtance B A, deſcribe the circular Arch AP;
and with any other Diſtance B M, another cir-
cular Arch. Then aſſume A HH E=BM
—BAor PM; and by the Evolution of the
Cauſtick Z F, beginning at E, deſcribe the

involute Curve, cutting the circular Arch de-

ſcribed with the Radius B M in the Point M,

which will be * in the-Curve AM. For by +» 4,; 8

Conſtruction PM MF= 4 H+ HF.

Or elſe take a Thread BME, and havi
fixed one End in B, and the other in F, firecch
the Thread by Means of a Pin at M, which
ſo move along, that the Part MFof the Thread
wraps about the Cauſtick HF: Then it is e-
vident that the Pin M in thus moving, de-
ſcribes the Curve MA ſought. |

Otherwiſe thus :

129 TR aw any Tangent (FM) excepti
D HA, * and in the 7
find the Point M fuch, that B M+M F=BA
+4 H+HF; which may be done thus:
Aſſume FK==B A4+A4 HA4HEF, and biſect-
ing BA in G, draw the Perpendicular 7 .

|
|
|
|
|
‘
|
|

A Treatiſe ©
This ſhall interſe& the Tangent FM in the
ſought Point M: for B M=MF.

If the Point B be ſuppoſed at an infinite
| Diſtance from the Curve AM; that i is, if the
incident Rays BA, B M, be parallel to a right
Line given in Poſition; the former Conſtru-
ction will ſerve likewiſe in this Cafe, in con-
ceiving the Arches deſcribed from the Centre
B, to become right Lines perpendicular to the
incident Ray. But the latter Conſtruction
will not do here; for which we ſhall lay down

| the following one.

Axt. 110.

Aſſume FR AHL EE And find the
Point M fuch, that MP (a Parallel to 4B)
| rpendicular to AP, be equal to MX; then

lain * that M will be the Point ſou btin
the urve AM, becauſe PMEMP=21+
H F. Now this is done thus.

Draw KG DN to Ap; and afli-
ming KO = XG, draw XP arallel to OE,
and p paralle] to EA: I fay the Point M
will be that required. For becauſe of the fi-
milar Triangles GK O, PMX P M will be

MK; ſinc KX O. If the Cauftick
HF degenerates i into a Point, the Curve AM
will become a Conick Section. 4

Conor. J.

H E N c E it is manifeſt, that the Curve,

which paſſes thro’ All the Points A,
SETS, by Hr Evolution of the Curye
beginning at A, and that its Nature va-

ries according as the Situation of the Point A
in the Ta “A H yaries. Wherefore be-
“cauſe the Curves (AM) are all generated by

the ſame geometrical Conſtruction from _—
| ſai

of FLUXIONS. 157
ſaid Curves: It is manifeſt * that the Nature Art. 108.
of them differ, and they are only geometrical

when the Cauſtick Hꝰ is a geometrical Curve
and reCtifiable.

Coror II.

z. ACun vx D N, together with the lu» FIC. 110.
minous Point C being given, to find
any Number of Lines (as AM) being ſuch,
that they may cauſe all the Rays DA, NM,
reflected from them, to unite in a given Point

If the Curve HF be ſuppoſed to be the Cau-
ſtick of the given Curve DN, formed by the
luminous Point C; it is manifeſt that the ſaid
Line HF muſt likewiſe be the Cauſtick of the
Curve AM, having the given Point B as a
luminous Point: So that FK=B A+ A H+
AF, and NK=B A+ A H+ HF+FN=BA
+AD+DC—CN, ſince HD+D C=HF Att. 110.
2 N+NC., Which gives this Conſtru-
ion.

In any reflected Ray aſſume the Point A at
pleaſure, for one Point in the ſought Curve
AM); and in any ether reflected Ray MN, aſ-
ſume the Part VX=BA+AD+DC—CN,
and oy Point M will be found as above (Art.

217 —

3 % — one” air 13

F1c.111.

A7. reatiſe

SECT. VII.

The U/e of Fluxions in finding of
Cauſticks by Refraction.

DEFINITION.

y the Directions of an infinite Number of
Rays B A, B M, B D, all iſſuing from the
ſame luminous Point B, be alter’d after their
Concurrence with a Curve AMD, either nearer

or farther from the Perpendiculars MC to it:

And if the Law of Alteration be conſtantly

ſuch, that (C E) the Sine of the Angles of In-

F1c.112.

cidence (CME) be to (C G) the Sines of the
Angles (CMG) of Refraction, in the given
Ratio of m ton; the Curve HF I that touches
all the broken or refracted Rays, or the Con-
tinuations of them, A H, MF, DN, is called
the Cauſtict by Refraction.

CoRoOoLLAR x.

I 32.J r the Cauſtick HN be taken as an

Evolute, and the Involute A LK be
deſcribed from it beginning at A; the Line LF
Plus FH, a Part of the Cauſtick, will be always
equal to AH. And if you conceive another
Tangent Fm infinitely near to FMI, and
another Ray of Incidence B m; and from the
Centres F, B, be deſcribed the ſmall Arches

MO,

of FLUXIONS. 159
MO, MR; then the little right-angled Tri-
angles M R m, M O m, will be ſimilar, the
former to M E C, and the latter to & C;
becauſe if the Angle E Mm be taken from the
right Angles RME, C Mm, the Angles RMm,
EMC, remaining, ſhall be equal: and, in like
manner, if the Angle C Mm be taken away from
the right Angles @ MO, Cm, the remaining
AMm, GMC, will be equal. Therefore Rm
:Om::CE:CG::m:n. Now ſince Rm
is the Fluxion of BM, and Om the Fluxion |
of LM; B MBA the Sum of all the Flu- » ,,, 96.
xions (Rm) in the Part of the Curve AM, will
be * to ML or A H—M F–FH, the Sum of
all the Fluxions (Om) in the ſame Part, as n
to ». And therefore the Part TH AN

MF+ 34A — 2B Ms
2 mM

There are ſeyeral Caſes, according as the
incident Ray BA is greater or leſs than B M,
and the refracted Ray A V wraps about, or
unwraps ( itſelf from) the Part HF: But we
prove, as already, that the Difference between
the incident Rays, is always to the Difference
of the refracted Rays, (Plus the Part of the
Cauſtick that one of theſe Rays diſengages it-
ſelf from before it falls on the other) as to n.
For Example, BA—BM: AH—MF—FH p,,.
: m: n. From whence we get FH=A4H— |

MF+ ZZ BM—EBA.
Mm M

If the Arch AP be deſcribed from the Cen- RIO. 111.
tre B, it is manifeſt that PM will be the Dif-
ference of the incident Rays B M, B A. And
if the luminous Point B becomes infinitely di-
ſtant from the Curve AMD, the incident
Rays BA, B M, will be parallel, and the *
4 P

160

A. Treatiſe

AP will become a right Line per icular
to the ſaid Rays. a dare

P 0 P. I.

133J “HE Nature of the Curve AMD, the

luminous Point B, and the Ray of Iu-
cidence B M, being given, to find the Point F in
the refracted Ray MF given in Poſition, where-
in the ſame touches the Cauſtic by Refraction.

Finsr, find * the Length MC of the Ra-

dius of Evolution at the given Point MA,
aſſume the infinitely ſmall Arch Mm, draw
the right Lines Bm, Cm, Fm, from the Cen-
tres B, F, deſcribe the little Arches MR, MO,
draw CE, Ce, CG, Cg, perpendicular to the
incident and refracted Rays; and call the given
Quantities B M, ); ME, az MG, b; and the
lirtle Arch MR, x. This done, |

Becauſe of the right-angled ſimilar Triangles
ME C and MRm, MGC and MOm, BMR

and B Ne, ME a): MEG G) :: MR(s): MO

=_— – And BM): B Q or BE (H Ha) ::
MR (): ee. Now from the Na-

ture of Refraction Ce: Cg: : CE: CG: min.

And therefore n: n:: Ce CE or De (= — 2 )

+$Cg—GCors gun W herefore be-
_ cauſeof the right- angled ſimilar Triangles FMO

and FSg, MO—S’g (= 1 ———) ;

0
1

of FLUXI1oNs.
MO CE): :MSor MG (6b): M F=
— {2 From whence comes out

bmy=—any—aan

the following Conſtruction.

Towards C M make the Angle E C Hh Fio.ngl

GCM, and towards B take MK===. Then
if you make HX: :HE:: MG: : ME I fay

the Point F will be in the Cauſtick by Refra- |

Ction! ‘- ;
For Sund of the Similarity of the Trians
angles CE, CEH,CG:CE::n: n.: MG

(): EH=Z. Whence HE—ME or HM=
— „H- Kor u- Er aan

And: dend 1 * 2 wer = pa), HE

\::a16 N. Ae e

2
It is manifeſt, that when the Value of HK
18 negative, that of MF ſhall be fo likewiſe:
Whence it follows, that when the Point M
falls between the Points C, , the Point H
falls between the Points K, E.

If the luminous Point B be next to E. or

(which is the ſame thing) if the Curve 4H
be concave next to B, then will y be changed
from poſitive to negative; and conſequently,

N Vm
m Tam -in, J 22
And the Conſtruction will be the ſame. .
If be ſuppoſed infinite; that is, when the

ä luminous Point Bis at an infinite Diſtance from |

the Curve AM D, the incident Rays will be
M then

-c

|
|

„ A Treaſe \

chen parallel, and A1 F -g, {inte thi

Term aan is o With reſpect ro bmy, any; and

becauſe ux(s) then vaniſhes, We. N
e 1K. |

4 N

Con. 1

we tray bedemonſtracd IR the a *

manner, as in the Cauſticks by Refle-
Sion, chat the Curve z M D has but one
Cauttick by Refraftion, the Ratio of n to 1
von; which Cauſtick is always a geo-
metrical Curve and rectifiable, when the pro-

poſed Curve 4 Al Mig erde G Curve.

— 2.

® fre. 114. 11

wm. 4

2 nner. II. AST

135 Jo. the Point E fills on the omen Side

cular AC with regard to

vant G, and C E be=CG;.it is evident
That the Cauſtick by Refraction will be-
come 2 Cauftick by an. For MF

; becauſe * =#,

(G 2 – wok
And a is here ative and equal to ö. And
this with what has been before demon-
Ae i th Se Section aforegoing.
| be infinite with reſpect to u; (id is evi-
| | * 3 2 or 3 Ray MP will
i} 5 e Ferpendieular C A. So that
the Cauſtick by Refraction will be the Evo-
lute. For MF wil be = which in this Caſe
becomes MC: That is, the Point F will co-

Incide with the OPS which is in the E-
volute.

Coo.

of F LUXTONS.

. +601 vo C 0 R o * III IP
ei nie

136 15 che convex Side of the Curve AMD
be next to, the luminous Point B, and

che Value of MF(; A.) be poli:

anf
tive; it is evident that the Point F muſt be
aſſumed on the ſame Side of M as G is, which
was ſuppoſed in the Operation of the Problem;
and on the contrary, if it be negative, the
ſame muſt be aſſuned on the contrary Side.
The ſame is to be underſtood when the con-
cave Part of the Curve AMD is towards the
Toint B. But we muſt obſerve that then AF

= = ä From whence i it follows,
_— the (infinitely near refrafed or broken
do converge: when the Value of MF: is

— * ive in the rſt Cale, and ne negative. in the
Reads andon the contrary, they iverge when
it is negative in the firſt Caſe, and poſitive in
the fecond. This being premiſed, It is evi
dent,

155 That if the convex Side of the 222
AMD he next to the luminous Point B, and
m leſs than n; or if the concave Part thereof
be next to the: ſaid luminous Point, and #2
greater than ; then the infinitely near refra-
er or broken Rays will always diverge.
25, If the convex Side of MD be turned
towards the luminous Point B, and m exceeds
1 or the concave Side, and m he leſs thay n;

the infinitely near reſracted Rays do con-

p TE MELT 1K leſs than N

163

which becomes —

“A Treatiſe

Now ſince MK o, when the Rays of In-

cidence are parallel, it follows that in this Caſe

the infinitely near refracted — do
converge. |

Send. W.

1. he Ray of Tackleane B M — *
  the Curve AMD in the Point M,
  then will ME (a) So; and therefore MF==b.
  And ſo the Point F will coincide with the.
  Point G.

If the Ray of Incidence B be perpendi-
cular to the Curve 4M D, the right Lines

ME (a) and M (ö) will be each equal to the

Radius CM of Evolution; becauſe they co-
| incide with it. Therefore M F # my

== Y
= when the Rays of In-

cidence are parallel “ag one angther,
If the refracted Ray MF touches the Curve

AMD in the Point M; then will MG (6)

=0. And conſequently the Cauſtick does

then touch the given Curve in the Point M.
If CM the Radius of Evolution be nothing,

all ſtraight Lines ME (a), MG (6) ſhall like-

wile be equal to nothing; and ſo the Terms
42, bbmy are nothing with reſpect to the
Terms my, any. Whence it follows that
“MF=0; and ſo the Point M is both in the

-Cauſtick and given Curye.
If CM the Radius of Evolution Va infinite,
the right Lincs ME (a), MG (6) ſhall 7
2 de

this Quantity is * vat [hed the Point F · 4 133.
falls not on the ſame ſide the Line AMD as
B; and on the contrary is poſitive, when F
and Bare both on the ſame gde AM D; there-
fore the Point F muſt be * aſſumed next to Art. 136.
the Point B, that is, the infinitely refracted
Rays are diverging. It is evident that the
ſmall Arch Mm then becomes a right Line,
and the Conſtruction above will not do here.
And fo we ſhall lay down the following one,
which determines Points 1n Cauſticks by Re-
fraction when AMD is a right Line.
Draw B O perpendicular to the Ray of In- F1e. 114.
cidence B V interſecting the right Line MC
icular to AD in the Point O; likewiſe
draw O L perpendicular to the refracted Ray
MG, make the Angle BO H equal to the An-
le LO M, and allo BM: BH: : ML: MF.
ſay, the Point F will be in the Cauſtick by
Refraction.
: For let MC be of wwhnie Magnitude: you
| pleaſe, the right-angled Triangles ME C and

BO, MO and MLO will always be fi- –. +

| milar: 4nd therefore when it becomes-infinite,
we ſtill have ME (a): MG(b):: BM(j): ML

=, And becauſe of the ſimilar Triangles
OE OB AH we have Ar O L: O
b (1 m): . 0 H= 2 7 Whence

BAM (9): 2 ( ): L (N. MF. 9.5
55220 n 126013 Ni
as M 3 CoroL.

of FLUXIONS. 165

finite alſo; and conſequently the Terms vm,
any ſhall be-nothing va 2 to aan, bbmy :

So that MF will be = _ Now fince

— . —— HE EINE — . —

— * —

— — es ”

Fi. 115.

Art. 1 37+

Art. 137.

Art. 132.

_—

A Treatiſe
Weser

: 38. HEN any two of the three Points
W B, C, Fare given, it is manifeft that
the rg may be — found.

EAN I l.

95 4

Lor AMD bea Quadrant oba Cc;
whereof the Point C is the Centre;
and let the Rays of Incidence B A, BM, BD
be parallel to each other, and perpendicular to
CD; ud finally let the Ratio of to 1 be as
3 to. 2 which is the Ratio between the Sine
of the Angle of Incidence and Refraction in

the Paſſage of the Rays of Light from Air

into Glaſs. Now becauſe the Eyolute of the
Circle HMD is the Centre C thereof if a

Semicircle ME C be deſcribed; with ithe Ra-

dius GNA, as a Diameter, and you aſſume the
Chord C g E, it follows char the Line
MG will be therefrified R # in which the

Point Fmay be determined as above (Art, 1 33

To find the Point V, wherein the Ray of
Inridence BA perpendicular to AMD touches
the Cauftick. ty Refraction, we have * AH

e And if a Setnicirele

CND be deſcribed: ith ‘D as a Radius, and

you take the Chord C =3CD; it is *-ma-
niſtſt that the Point N Will bei in the Cauſtick
by Refraction, becauſe the Radius of; Inci-
dence BD tauches the Crete AMD in the

7 oint DB, 2 “I
. WAP be rawn p 90 CD, it is * ma-

nifeſt that the Part FH AH Fe; : TIA

2 * „
o 4 — C — * 0

* –

. COR TS gene Two Ts

SO” ov

of FLUXIONs. 467

ſo that the whole Cauſtick HFN =3C A—

DN=LZXECA.

If the acute Part of the Circle AMD ***: i
be turned next to the incident Rays BM, and
the Ratio of m to be as 2 to 33 then on the
Semi-circumference C EM having CM as a

Diameter, aſſume the Chord CG CZ; and 1

draw the refratted Ray MG, in which the
Point M may be determined according to the
general Conſtruction (Art. 133.) $7
We ſhall have * 4Z (= Y = — 2 b, Art. 137.
that is, A H will be * next to the convex Part Art. 136.
of the Quadrant AMD, and the Double of
the Radius 4C. And if we ſuppoſe C G or
10 ECM, it is manifeſt that the refracted
Ray MF will touch the Circle AMD in M,
becauſe then the Point & will coincide with
the Point M. Whence’if you take CE =
3 CD, the Point M will coincide with the
Poim N, wherein the Cauſtick ZH FN touches

the Circle AM D: but when C is greater Art. 137.

than 3 CD, the Rays of Incidence ( can
be no more refracted, that is, out of Glaſs
into Air; becauſe it is impoſſible for C &, per-
pendicular to the refracted Ray MG, to be

eater than CM: ſo that all the Rays that
fall on the Part D N will be reflected. 4h
If AP be drawn parallel to CD, it is ma-
nifeſt * that the Part FH = AH— MF+} 137.
PM: fo that drawing VX parallel to G ,
the whole Cauſtick HFN=2CA+Z3AK=

4 1

Vu 44

—— — — Ay

168 5 A Treatiſe \

oo.

EXAMPLE II.

646 E T the Curve 4 MD be a Logarith-

Fro. ar L mical Spiral, the Centre thereof be-
ing the Point 4, from which let all the inci-
dent Rays (AM) iſſue.

with A, viz. a. Therefore if you ſubſti-

Art. 113. ture F for a in * | the Value * of
MF when the Concavity of th the Curve is pert
the luminous Point; we ſhall have MF =
Whence the Point F coincides with G.
If you draw the right Line AG, and the
Tangent MT, the Angle 46 0, the Comple-
ment of the Angle AGM to two right An-
lcd, will be equal to the Angle 4 MT. For
e the Circle, whereof C M is the Diame-
ter, paſſes thro’ the Points A and G, the half
of the Arch A M is the Meaſure of each of the
Angles 460, AMT. Therefore it is evi-
\ dent that the Cauſtick AGM is a Logarith-
mical Spiral, the ſame as the given one, differ-
ing from it any in Poſition,

E =
|

ene — w;) II.

Fic. m 141. E hats HF, the bee Point
act B, and the Ratio of m to n, being gi-
ven: To ind any Number of Curves (as AM)
whereof it may = the Cauſtict by RefraZtion.

Tate the Point A at pleaſure i inany Tangent
for one Point in the Curve A M, and from the
Centre B, with the Interval BA, deſcribe the
” Un h.4P, and — Rien with any *

f FLUX1ONS.
terval. Then take AE PA. and with the
Cauſtick HF as an Evolute, deſcribe the In-
volute EM, which will interſect the Arch de-

ſcribed with the Diſtance B M in the Point
M; and this will be in the Curve ſought:

169

For PM: AE or ML: n: u. Arr. 132.

Otherwiſ.

Tx any Tangent FM excepting HA
5 I find the Point M with this Condition,

that ZF+FM+ —BM=HA+— BA.

And then if you take FR B A+ A H—
FH, and find the Point M ſuch, in the Line
FR, that MK=— BM, the ſaid Point will

be * that ſought. Now this may be effected . . 132.

in deſcribing a Curye GM of ſuch a Nature,

that the right Lines MB, MK, drawn from F

any Point in it to the given Points B and X,
may be to each other always in the conſtant
Ratio of m to x. Now this Curve may be thus
found.

Draw MR perpendicular to B , and call
the given Quantity BA, a; and the indeter-
minate Quantities B R, x; RM, y. Then be-
cauſe of the right – angled Triangles B RM,
XR M, we ſhall have B M=y/xx+y, and

K M = aa—2ax+xx+yy. So that to ful-

fil the Conditions of the Problem y/xx+1y .:

vVaa—lax+txx+yy::m:n. Whence yy=
2AMMX=AAMM Ns | j ;
5 D —— xx, Which is a Locus ad Cir-

culum;

ans * may bach thus deſcribed.
a
Aſſume BG= — and B 2 and

| with G 9. as a Diameter, deſcribe the Semi-
| circle GMO. I fay, this will be the m_

_. fought. For ſince N or BY-BR= —.—

| ET
| —, and RG or B R—B . from
|

the Nature of the Circle, where © Rx RG

E. we have yy EET,
= Pie. x20… AIFXDE Rays of Incidence B A, BM, be pa-
4 rallel to a right Line given in Poſition, the
former Solution will ſtill take place: But the

latter one is of no Effect, and inſtead of! it we

ll wo A Treatiſe
1 RY
ll

| 7 5 may uſe the following one.

I! | Aſſume FL=4 H—F, draw I G parallel
to AB, and perpendicular to AP; aſſume
|

Lo α e, and draw LP parallel to GO,

and P.M parallel to & L. Then it is manifeſt
Art. 132, * that the Point M will be that ſought. For

1 tt Bona ©
ſince LO==—LG, ML==PM. |

If the Cauſtick by Refraction FH becomes
‘a Point, the Curves (4M) will then be the
famous OVALE of Deſcartes. | |

Conor. 1.

Ar TER the ſame manner as in cane
by Reflexion, we demonſtrate * that

os . the Cares AM are of a Nature very ert
from one another; and that they are —.

metrick Curves but when the Cauſtick by] 4

fraction

of FLUXIONS,
fraction HF is a en ! polite
fiable Alſo.

S II.

A Curve AM, the luminous Point , Fi6. 121.

and the Ratio of m to » being gi-
ven; To find any Number of Lines DN of
ſuch a Nature, that they may again refract the
refracted Rays MN, ſo as to unite them at
length in a given Point c
f we imagine the Curve HF to be the
Cauſtick by Refraction of the given Curve
AM, formed by means of the luminous Point
B; it is manifeſt that the ſaid Line HF muſt
likewiſe be the Cauſtick by Refraction of the
Curve DM ſought, having the given lumi-

nous Point C. Therefore _ BAS 4 H= *Art.132.

E-BM+ MP+ FH, and L Fl
NC=HD— c, ; andtherefore=BA4–

Ah B MN+HD 2 D C
—N C. And by the uſual TranſpoſitionZ

BA— BNA DC+AD=MN+

—N C. Which gives the following Con-

ſtruction. |
Firſt aſſume the Poiat D in any refracted

Ray AH, as one Point in the fous ht Curve
DN); then on any other refracted Ray ME,

take the Part MK => 3A *

172 24 Treatiſe 8
DCA; and having found the Point N
Art. 142. (as above *) of ſuch Condition, that NX =

Art. 32.—N C, it is evident * that the ſame ſhall be
in the Curve DN.
| A GENERAL CoROLLARY
For the three laſt $eftions.

Art. 80, 145. Tr is manifeſt * that a Curve can have
$5, 107, but one Evolute, one Cauſtick by Re-
108, x 19. flection, and one Cauſtick by Refraction, when
12, 134, the luminous Point, the Ratio of the Sines of
143- the Angles of Incidence and Refraction is
given; which Lines are always geometrical
and rectifiable, when that Curve is geometri-
cal. Whereas the ſame Curve may be an E-
volute or a Cauſtick by Reflection or Refra-
ction (in a given Ratio of the Sines, and Po-
ſition of the luminous Point) to an infinite
Number of Lines of different Natures, which
will not be geometrical, but when that Curve

is geometrical, and withal rectifiable.

—

of FLUXIONS.

SECT. VL

The Dſe of Fluxions in finding the

Points of Curves touching an in-
finite Number of Curves, or Right
| Lines given in Poſition.

PRroOP. 1.—

LE T7 AMB be any given Curve, whereof Fic. 122.

the right Line AP is the Axis; and let
us conceive an infinite Number of Parabola’s
AMC, AmC, all to paſs thro the Point A,
and to have the Ordinates PM, pm, as Axes.
It is required to find the Curve touching all
the ſaid Parabola’s.

It is manifeſt that the Point of Contact of

every Parabola 4 MHC, is C the Point wherein
the Parabola Am C, which is infinitely near it,
interſects the ſame. This being ſuppoſed,
draw CK parallel to AP call the given Quan-
tities AP, x; PM, y; and the unknown Quan-
tities AA, u; KC, z. Then from the Nature

of the Parabola 4 P (xx) : P K (un—2ux+xs)
i:MP(y); MP—CK(z=z). Whence zxx
=24x)—14y ; Which is an Equation. common
to all the Parabola’s as AMC. Here I ob-
ſerve that the unknown Quantities do not va-
ry, while the given ones 4 P (x), PM (0) do,
viz. become Ap and pm; and KC (z) is ne-
Ts

e — CY * * 2

. „

AFTreatiſe
ver invariable, but when the Point C is the
Interſection aforeſaid : For it is manifeſt that
every where elſe the right Line KC interſects
the Parabols $ AMC, An C, in two different
Points, and conſequently it will have two dif-
ferent Values correſpondent to the ſame Value
of AK. Therefore if the Equation above
found be thrown into Fluxions, with 0 and
25 taken as: conſtant Quantities, the Point C
will be determined to be that of the Interſe-
ion aforeſaid. Whence 22xx==24uxy-|-: 209%

-,: And fo A K(«) == —4 by fub-

ſtituting _ pr 2; and the Nature of

the Curve AM B being given, we may find a

Value of 5 in ; which being put in the Va-

þ © of AN, the faid unknown Quantity will

gth be expreſſed in known Terms freed

from uxions. Which Was propoſed to be
nnn

If “Pl Curves or rights Lines of a determi-

nate Poſition, be propoſed inſtead of the Pa-

rabola’s (AMC), the Solution of the Problem
will be much the ons. as will “PF Wl the
en an e. * 7

EA EE p 10 22 |

147.] E * 407 expreſs the N ature
of the Curve AMB: This will be
half an Ellipſis, whoſe: conjugate Axis is AB
equal to a, and perpendicular to AP, and
| tranſverſe Axis the Double of the Conj ugate.
Now SNN and therefore AK

(nx = Whence if AK

iyi

J be

of FLUXIONS.
be aſſumed a fourth Proportional to M, PA,
AB, and KC be drawn perpendicular to 4A;
it ſhall interſect the Parabola 4M C inthe Point
C ſought. Again; to find the Nature of the
Curve touching all the Parabola’s, or which paſ-
ſes thro” all the Points C thus found, we muſt find
an Equation expreſſing the Relation of A K
(2) to KC [z) after this manner. Subſti-
tute = for 4 its Equal in zxx==2uxy—uuy,
aa

d we get y=—®*—: and th fore x o
and We get Y ind therefore x or-

. Now if theſe Values be put for «
and in eee e, there will ariſe uu.
4aa— 4a, wherein x and y are got out, and
which expreffes the Relation 25 4K to K C.
Therefore it is manifeſt, that the Curve ſought
is a Parabola, whereof the Line BA is the
Axis, the Point B the Vertex, the Point A
the Focus, and conſequently the Parameter is
four times 48.

3 Ye COS. Ee
KC(z) ns But ſince this Expreſſion
zs poſitive when 2 is greater than a,, negative

when it is leſs, and nothing when it is Jos 5
therefore in the firſt Caſe the Point of Con-
ta&t C falls above AP, as it was ſuppoſed to
do in the Inveſtigation; in the ſecond, it falls

below; and in the laſt Caſe it falls in A P.
If the right Line AC be drawn to interſect

M in G; I ay ME BN, and the Point

E is the Focus of the Parabola A4 MC. For,

q
7
[
1

— ——

— — 7
. ³˙ m Z ˙¹¹¹% ‘; ũů! War wo Wo a_ no —_— ̃—»— ̃⁰˙ A een}.

— 2,

176 – A Treatiſe

— — Ä — —

PG Si-. And therefore MG A-
3.2. 2, The Parameter of the Parabola AMC,
is 44 — 45, putting 4425 – 49 for xx; and
therefore MG =a— is the fourth Part of
the Parameter. Whence it is evident, that
the Point & is the Focus of the Parabola; and
conſequently the Angle BAC muſt be biſected
by a Tangent to the Curve in 4.
Hence the Parameter of the Parabola A NHC
is four times B ©; and when the Vertex M
falls in 4, the Parameter will be four times
\ AB; conſequeutly the Parabola, whereof the
Point A is the Vertex, is Aſymtotick to that
paſſing thro! all the Points C. %
Becauſe the Parabola B C touches all the
Parabola’s (as AMC), it is manifeſt that all
the ſaid Parabola’s will cut the determinate
Line AC in Points, which will all be nearer
to A than the Point C. Now it is ſhewn in
the Doctrine of Projecties, that (AX being a
Horizontal Line) all the Parabola’s, as A MC;
will be the Paths of Bombs thrown out of a
Mortar wich a given Force placed in A, hav-
ing all poſſible Elevations. And conſequently
if a right Line be drawn biſecting the Angle
34 the fame will be the Poſition of the
Axis of the Mortar; ſo that a Bomb thrown
out of it, will fall on the Plane AC given in

Poſition to a Point C, which will be farther
from the Mortar than when it has any other
Elevation. | v9 |

PR o r. II.

Fi. 123. 148. | ET any Curve AM be given, whereof

AP is its Axis: To find another Curve
BC of ſuch a Nature, that if any Ordinate P *
95 | £

72
A

- NN
  4
  ö 4
  : =
  21
  7
  I
  bus *
  Es
  7p
9
7 *
25
% p
eb” :
% 7 I

of FLUXIONS. 177
ze draum, and the Perpendicular P C to the
ſought Curve; theſe Lines PM, PC, may be

always equal to one another. =

If an infinite Number of circular Arches be
ſuppoſed to be deſcribed from the Centres P, p,
with the Radu PC, pC, equal to PM, pm.
It is evident that the Curve BC ſoughr, muſt
touch all the ſaid Circles; and that C, the Point
of Contact of every Circle, is the Point where-
in the Circle infinitely near it cuts it. This
being premiſed, draw CA perpendicular to
A P, and call the given and variable Quanti-
ties AP,x; PM or PC, y; and the unknown .
and conſtant Quantities AK, 4; KC z. Then

from the Nature of the Circle C PK —

KC, VIZ. yy =xx—24x+uu+z2, which
is the Equation common to all the ſaid Circles.
This thrown into Fluxions, will be 2y5=2xx

‘ —24×3.and ſo we get PA Er ===>.
From whence comes the following general
Conſtruction. _

Draw N Q perpendicular to the Curve A M,.
take PK=P ©, and draw KC parallel ro PM.
I fay, this will meet the Circle deſcribed from
the Centre P, with a Radius PC PM in
the Point C, wherein it touches the Qurye
CB ſought. This is evident, becauſe Pd =
29 PS

The Value of PK may be found otherwiſe
thus: | .
Draw PO perpendicular to Cy; then the
right-angled Triangles pO P, PKC, will be

| “> 08 ſitmilar.

[|

|
|
|

Fro. 123.

Fic. 124.

A Treatiſe
ſimilat. Therefore Pp): Op (Y:: PC O:
F oh

When PS =P M, it is manifeſt that the

Circle deſcribed with the Radius PC, will
touch KC in the Point K; ſo that the Point
C will then coincide with A, and conſequent-
ly will fall in the Axis.

But when 5 ©. 1s, greater than PM, the
Circle deſcribed. with the Radius P C, cannot
touch the Curve BC; becauſe it cannot at
all interſe& the right Line Cc.

10329 £04
1 EXAMPLE.

L 1 the given Curve AM be a Para-
17515 bola, the Equation whereof is a x =
Y. Now P 9 or PA (x—s)=343/ and con-
ſequently x fa, and yy aaf, be-
cauſe of the right – angled Triangle PKC. If
theſe Values be put in ax =yy, we ſhall have
aa EA Haaf AZR, of aa TA ,

expreſſing the Nature of the Curve B C.

Whence we may perceive that the ſaid Curve

zs alſo a common Parabola as well as AM, be.

cauſe they have both the fame Parameter a,
and the Vertex B is diſtant from the Vertex

A by the Diſtance BA e.

P R O p. III.

yo. 7 ET A M be any given Curve, the
L right Line AP being a Diameter, and

the Ordinates PM, pm, parallel to a right Line

AQ given in Poſition; and having drawn MQ,
m q; parallel to AP, then draw the right Lines

PQC, pqC. HU is required to find the wy

—Eü

of FLUX1ONS.
AC of ſuch a Nature, that all theſe laſt menti-
oned Lines may be Tangents too; or, which is the

ſame thing, to find the Point of Contact C in
every right Line P QC.

Conceive another Tan ent pgC to be infi-
nitely near P OC, and draw CA parallel to
AQ; then call the given and variable Quan-
tities AP, x PM or A 23 the unknown
and variable Quantities A A, a; KC, z. Now

becauſe of the ſimilar Triangles P A9. PKC,

AP 2 AQ (9): :PK (u): KC 2 4
2 22 Which is an Equation common to all

the right Lines as KC. The Fluxion thereof
us 2 hou

is 7 ＋ o. From whence ariſes. AK
( 2 55 and ſo che ee general
Conſtrudion.

Draw the Tangent M 7, in which 8
AK a third Proportional to AT, AP: Then
if XC be drawn parallel to A 9, it ſhall cut
the right Line P in the Point C fought.

! AP(ﬆ)::4P(ﬆ): AK
2
g

EXAMPLE I.

ET the given Curve AM be a Para- Fie. 124

bola; ſo that the Equation thereof
will bear . . Then AT=AP; whence AK (u)
=x3; that is, the Point & falls in T. Now
to have an Equation expreſſing the Relation
of AK (u) to KC (z)z we ſhall have KC (Z)
2 becauſe PK is the Double of 4 P.

N 2 Now

180 A Treatiſe
aa putting 1 and 1 z for their Equals x and
4 in ax, and then will 442 =22:
Wbence we may perceive that the Curve 4C
is a Parabola, whoſe Vertex is A, and Para-

meter a Line equal to four times the Parame-
ter of the Parabola AM. ; |

EXAMPLE II.

Fi6. 125. 175 zT the given Curve AM be a Qua-

b drant B MD, whoſe Centre is the

Point 4, and Semidiameter the Line AB or

AD, which I call a. It is manifeſt that P.

is always equal to I Mor AB, viz. is an =
variable Quantity. So that the Ends P, Q.

be ſuppoſed to ſlide or move alo 8 Sides

BA, AD of the right Angle BAD. Now, AK (a)

willbe==, becauſe AT ===; and on ac-
count of Fon Parallels K C, A 9; AP (x): PD

x = =

(a): AR ()) c= r. Whence we

ma yk perceive that to a the Point of
tact C; we need only aſſume QC a third

. Proportional to P and AP. If you ſeek the
uation of the Curve BCD, the fame will

de found this here, u%—3aau* + 3a u . – o.

| +322 +214a22F zπ

Conor. J. 0 1.1

173 J- hs Relation of DC the Part of the

Curve BCD to its Tangent CP be
required, you muſt firſt conceive another Tan-
gent c 7 u ere near rc P; then —

of FLUXIONS.
the little Arch PO from the Centre C, and
cp—CP or Op- Cc will be == which

is the Fluxion of CP =: whence
L there comes out Cc = Op-+ {£4 Now be-

a
cauſe of the right-angled ſimilar Triangles
PA, PpO, PD(a): AP(x)::Pp(x):Op
== And therefore Cc = =D —

De. Whence it is evident, that where – ever
the Point C be aſſumed, this Proportion will
always be had, viz. DC — Dc (3= :CP

—p (i ::3:2. And conſequently the

Sum of all the Fluxions DC- Oc correſpond-
ing to the ſame right Line PD, that is, * the * Ari. 95.
Part DC of the Curve (BCD) is to the Sum

of all the Fluxions CP —cp anſwerable to the

ſame right Line PD, viz. * to the Tangent » 4,4, 96.
CP::3:2. And allo the whole Curve BCD

is to its Tangent BA:: 3: 2.

Con ol. II.

1 the Curve BCD be taken as an Evolute,
the Involute DM formed from it begin-

ing at D, will be of ſuch a Nature, that CN: CP
: 3:2. Becauſe CN is always equal to DC
the Part of the Curve BCD. hence it
follows, that the ſimilar Sectors C Na, C PO,
are to one another ::9:4. And therefore the
Space DCN comprehended under the Curves
DC, DN, and the right Line CN, which is
Al 3 a Tan-

4
W
ö

“
,

:
9
14
1
1
ö

Art. 2.

» A Treatiſe ©

a Tangent in C, and perpendicular at N, is to
the Space D CP, contained under the Curve
DC and x the Tangent DP, C, as to 4.

Cox O. III.

T xx Centre of Gravity of the Sector

Nn muſt be in the Arch PO; be-
cauſe C’P ICN. And ſince the ſaid Arch
is infinitely fmall, it follows that the Centre of
Gravity muſt be in the right Line AD; and
therefore the Centre of Gravity of the Spaces
DCN, BD F, made up of all thoſe Sectors,
muſt be in the right Line AD. Conſe 2
ly if a Figure be deſcribed on the other ſide of
B D Fequal and ſimilar to BD F, the Centre
of Gravity of the entire Figure ſhall be in the
Fs A. |

Coro. IV.

156 Bega of the right-arjtlee. Kcbllar
Triangles pH 2A, pPO, PD (a):
AD or } ÞM ETD: Pp (x) 20 =*

— xx — And becauſe of the fimilar Sectors
CPO, PO, CNn, CP: C Nor 2:3: : PO

r e

223 — h erz Now the

Rectangle MP x Pp, viz, * the little circular
Space Mh =x4/aa—xx. Whence AB
x Nu==3MP pm: and conſequently the Part
ND of the _ DMF drawn into the Ra-
dius AB, is the Seſquialter of the circular

t ‘DMP, and the whole Curve DNF

is cqual to z of the Fd Quadrant BMD,
PRO.

of FLUXIONS. 183

PR Op. IV.

L T AM be any given Curve, whereof pic. 126.

the right Line AP is the Axis; and
let there be an infinite Number of Perpendiculars
MC m C drawn to the ſame. It is required to
find that Curve which all theſe Perpendiculars
are Tangents to; or, which is the ſame thing, to

find the Point of C mac Ci in every hd ea of
lar MC,

Firſt imagine another Perpendicular C in-
finitely near to MC; let M be an Ordinate,
and through the Point of Interſection C draw
the right TEE C perpendicular, and CE

— to the Axis: = call the given and

ds Quantities 4P,x; PM,); and the
unknown and invariable ones A A, 4; KC, .

This done, PQ will be ===, PK or CE=
u—x, ME =zy+ 2; and becauſe of the right-
angled ſimilar Triangles MP, MEC, MP
(9) : P2(L 22): :ME (HT): EC (ws)

3 2 7. Which is an Equation common

to all thi Perpendiculars as MC, and the Flu-
xion thereof (ſuppoſing x invariable) will be

Ay I fo e I; from whence comes out

ME Z TY _ Now the wess of

the Curve AM being given, we ſhall have
| Values of * and y in &, which being put in

. —_ 5 Will give a known Value of ME
N | N 4 freed

Fre. 127.

A Treatiſe ‘
freed from Fluxions; which is what was pro-
poſed. Fi |
It is manifeſt that the Curve paſſing thro’
all the Points C, is the Evolute of the Curve
AM: and becauſe in the fifth Section theſe
Curves are fully handled, it is needleſs to give
again new Examples of them here.

Pray, V.

>” apd Y two Lines AM, BN being given,
2 together with a right Line MN con-
rinuing always of the ſame Length. Now if the
Ends M, N of this Line continually move along
the two former Lines, it is required to find the
Curve that it always touches. py”

Firſt draw the Tangents MT, NT, and con-
ceive anothet right Line n infinitely near
MN, and which by conſequence cuts it in the
Point C, wherein the ſame touches the Curve
in which we are now determining Points.
Now it 1s plain, that while the right Line MN
is moving to the Situation u, the Ends there-
of will deſcribe, or continually be in the ſmall
Parts Mm, Nu of AM, BN, which on ac-
count of their being infinitely ſmall, are com-
mon to the Tangents 7 M, T N: So that while
the right Line M is moving to the infinite-
ly-near Situation u, the Ends thereof may
be concived as moving along the right Lines
TM, TN given in Poſition.
This being well underſtood, draw MP, CK
perpendicular to NT, and call the given and
Variable Quantities T P, x; PM, y; the un-
xnon and conftant Quantities TA, 2; K C, z;
2nd the ſtable Line MMa. Now germ?

f * 4#* 7
Sf „ n= + a

f Fiuxions.
of the right-angled Triangle MP N, PN
/aa—7); and becauſe of the ſimilar Trian-
gles NPM, NKC, NP (Vaa—yy): PM(y)
: NX a -* – Na = 9): XC (2)
Z -Y. And this thrown into Fluxi-
2 | | vb te
ons will be aauy — aa, -a,
aay—))Yy/aa—xx: and making /aa—yy
== for brevity’s ſake, there comes out PK

Tuns mw mms |
(-K) — 2 =—— by ſub-
ſtituting x) for its Equal y, becauſe of the
ſimilar Triangles RM, MPT; and there-

fore MC — ZZ — from whence comes

the following Conſtruction.

Draw TE perpendicular to MN, and aſ-
ſume MC NVE: I ſay, the Point C will be
that ſought. For ſince the right- angled Tri-
angles MM, TNE are ſimilar, ir N (a) :
NP (m):: N T ( : NE or MC =

mm nx

— |
Otherwiſe. Draw TE perpendicular to MN,
and deſcribe the ſmall Arches MS, NO, from
the Centre C, call the given Quantities NE, 1;
E T,; MN,a; and the unknown Quantity
CM, t. Then will Sn or Onu=z; and be-
cauſe of the 77 7 ſimilar Triangles
MET and M, NET and ON, CMS
and C VO, ME (-a): ETO) :: : SM

==. And NE G: ETOS :: O :O

n n
. And s NO ( L) Ms
7 r -r,

Fans &

ariſes the ſame Conſtruction as above, If

( L D (a): MC (0. From whence

185

186

x Fic. 1 28.

4 A Treatiſe

If AM, BN be po Ber right Lines at
right Angles to each other; it is manifeſt that
the Curve ſought is the ſame as that of Ar-
ticle 172. | | | |

Px o P. VI.

* 7 L, M, N be any three given Lines,

aud from every Point L, I in the Curve
L, let two Tangents LM and LN, Im and In
be conceived to iſſue, to the Curves M and N,

one to each. It is required to find a fourth Curve

C, to which all the right Lines MN, mn join-
ing the Points of Contact of the Curves M, N
may be Tangents. | “oF

Draw the Tangent LE, and thro’ any Point
E therein draw E F, E & perpendicular to the
two other Tangents ML, NL, let the Point
be ſuppoſed infinitely near L, draw the little
right Lines LA, L perpendicular to 1,1;
as likewiſe MP, m P, N 9, n 9 perpendicular
to the Tangents ML, ml, NL, ul, which
will interſect each other in the Points P and
©. Then the right-angled Triangles E FL
and LH}, EGL and LAI will be ſimilar;
as likewiſe the Triangles L MH and Mn,
Lu and N Ds right-angled at H and mm, K
and NV; becauſe cach of the Angles LA,
Mm added to the Angle 2 Mw, makes a

right Angle. And in like manner we prove
that the Angles LA, Nn are equal to one

another. ; | |

This _ premiſed, call the little ſide
Mm of the Polygon, the Curve M is con-

ceived to be, 4; and the given Quantities
EFn; EC, ; MN or un, a; ML or m5;

“+

of FLUX1ONS.

NL or ue; MP or P/ f; NE 1 53
(J here take the right Lines MP, N for
given ones, becauſe the Nature of the Curves
M, N being given by Suppoſition, they may

always * be found); then we ſhall have, 1*, MP * Art. 78.

(H: ML(b): : Mm D:LH= ® EF(m)
( LX Af 2 LN
or Luc): (8) :1 LK (Z) „Se

4*, (Drawing MR parallel to NL or 11) ml
(b):In();::mM(#):M === 5„ MR+

Na 7 ) R (5) MN):

3 a+
A which was to be found.

When the Tangent EL coincides with the
Tangent ML, it is manifeſt that EF (m) will
become nothing; and therefore the Point
ſought C will coincide with M. Likewiſe
when the Tangent EL coincides with the
Tangent LN; then E@ (i) will become no-
thing; and conſequently MC a. Whence
it is eyident, that the Point C ſought will co-
incide alſo with NM. Laſtly, if the Tangent
EL falls in the Angle GLI; in this Caſe E &
(n) will be negative: ſo that then MC =
ä — — z and the Point ſought C will fall
cen — bbon |
without the Points M and N.

E x A M-

187

188

4 Treatiſe

EXAMPLE 1.

rie. 129. ee gute; the Curves MN, tobeParts |

of the ſame Circle; then it is plain,
that 1 in this Caſe bc, and fg. So that MC

—.—5 3 whence it follows, that the Point
C ſought, is determined by dividing the right
Line MN into two Parts, being to one ano-
ther in the given Ratio of m to u, viz. ſo tha
MC; NC: : n: 1:

EXAMPL E II.

1618 UPPOSE the Curves M and & be any
Conick Section. Here the general
Conſtruction will be changed into a far more
ſimple one, from the Conſideration of a Pro-
rty of the Conick Sections demonſtrated in
reatiſes of thoſe Curves, viz. That if from
every Point L, I, of a right Line EL, be drawn
two Tangents LM and . N, Im, In, to any
Conick Section, all the Right Lines M N, m n,
joining the Points of Contact, will interſe3 one
another in oue and the ſame Point C, thro’ which
the Diameter A C, to Ordinates that are paral-

kl to the Right Line EL paſſes. For it fol-

lows from thence, that the Point C is deter-
mined by only drawing a Diameter whoſe Or-
dinates are parallel to the Tangent E L.

In the Cirele it is manifeſt that the IN
ter muſt be perpendicular to the Ta
that is, a Perpendicular 4 B drawn 7 **
Centre A to the Tangent, will interſe& the
right Line MN in the Point C ſought.

SCH os

—

of FLuxtons. 189
| SCHOLIUM. —Y

Fur Solution of the following Pro- pic. 228.
blem depending on the Method of

Tangents, will be had by means of the afore-
ſaid Problem.

The three Curves C, M, N, being given,
and a right Line M being continually mo-
ved about the Curve C fo as to always touch
it: And if from the Points M, N, wherein it
cuts the Curves M and N, be drawn the Tan-
gents ML, NL, interſecting each other in
the Point L, which by the Motion deſcribes
a fourth Curve LI. It is required to draw
L Ethe Tangent to this Curve, the right Lines
MN, MI, together with the Point of Con-
tact C being * 1
For it is manifeſt that this Problem is but
the Inverſe of the aforegoing one, and here
M is given: So that e are to find the Ratio
of EF to EG, which determines the Poſition
of the Tangent EL. Therefore if you call
the given Quantity MC, h, we ſhall have

acm 2
= 1 Whence comes out mm —

Ex and conſequently the Tangent LE
muſt be ſo ſituate in the given Angle ML G,
that. if from any Point E in it, be drawn E E,

EG, perpendicular to the Sides of that Angle,
they may be always to each other in a given
Ratio, viz. of bbgh to accf—ccfh. Now this

is done by drawing MD parallel to NL and

When

190 A Treatiſe.
4 F 0. 129. When the Curves M and N, are Parts of the

1 247i. 161. ſame Conick Section, it is * manifeſt that then
you need but draw the Tangent L E parallel

to the Ordinates to the Diameter N thro*
the Point C.

— — — —

— ä — —
rags i OI Gs ale “Cece
— RR ent BEA as 3 of 5

C

PPP

„ ——— ll. 41444

5 *
2

—

ly ec BOOT AY

— — — — —
2
Lp 2 be

2 –

— ogs Go
K rr ets „ . ox”
©

of FLUXIONS.

8E d r. IX
The Solution F ſome Problems depend-
ing on the Methods aforegoing.

P R 0 1. T:

163 2 AMD be a Curve (Apr, PM

=, AB=a) of ſuch a Nature, that
the Value of 2 Ordinate y is expreſſed by a Fra-
ction, the Numerator and Denominator of which,
do each of them become © when Da, Viz. when
| the Point P coincides with the given Point B.
Ii is required to find what will then be be 1 alus
f tbe Ordinate BD: |

Let ANB, COB, be rwo Curves (having
the Line AB as a common Axis) of ſuch a
Nature, that the Ordinate P N expreſſes the
Numerator, and the Ordinate PO the Deno-
minator of the general Fraction repreſenting

any Ordinate PM: So that PH — ZN “uf

Then it is manifeſt, that theſe, two- Curves
will meet one another in the Point B; ſince
by the Suppoſttion PN, O, do each become
o when the Point P falls in B. This being
ſuppoſed, if an Ordinate hd be imagined infi-
nitely near to B D, cutting the Curves ANB,

COB,

191

F16. 130.

| nn

Art. 2.

A Treatiſe

COB, in the Points f, g; then will 54 =
= 5 which will be equal to ̃ D. Now
our Buſineſs i is only to find the Relation of by
to bf. In order thereto it is manifeſt, when
the Abſciſs AP becomes AB, the Ordinates
PN, PO, will be o, and when A P becomes
Ab, they do become bf, bg. Whence it fol-
lows, that the ſaid Ordinates bf, bg, them-
ſelves, are the Fluxions of the Ordinates in B
and h, with regard to the Curves ANB, COB;
and conſequently, if the Fluxion of the Nu-
merator be found, and that be divided by the
Fluxion of the Denominator, after having

made x As or AB, we {ſhall have the

Value of the Ordinates 5d or BD ſought.
Which was to be found. 5

EXAMPLE I.

D 2/Gax

L. ‘T * 8 0
it is n when x a, that the Nu-
merator and Denominator of the Fraction will

Now

each be equal to o. Therefore we muſt aſ-

2X *
— the a of the er
. FT and divide it by the Fluxion of the

Denominator — — ; ==, after having made

a, viz. divide —jas by —jx; and there

comes out Za BD.

1 n Ex AM-

= Numens

A T II.

d. > == Then when a,

7 will be = 2 a2.

This Example may be ſolved without Flux-

ions thus.

Taking away the Surds, and then will aan

2aaxy——=2xyy — 2a a – Do,
which being divided by xa, will be brought
down to aax—43*2aay—ay}=0; and ſubſti-
tuting à for x, we have as before ) —2 4.

LEMMA.

193

E TBCG be any Curve, which the right Pio. 131;

Line A E touches in the Point B, in
which take two Points (having an in variable Po-
fition) A, E. Nou if this right Line moves a-
bout the Curve ſo as to touch it continnally, it is
evident that the flable Points A, E, by the. {aid
Motion will deſcribe two Curves AMD, ‘E NH.
Then if D L be draum parallel to AB, ard
which conſequently makes with DK (in which
the right Line A E is ſuppoſed to be when it
touches the Curve BCG in G) the Angle KD..
equal to the Angle AO D formed by the Tan-
gents in Band; and from the Centre D be
deſcribed any Arch K FL.

{jay DK: KF L:: AE: AM DEN H,
viz. + when the Point of Contact falls between
the deſcribent Points, and — when it does not.

For ſuppoſe the right Line A Ein its Mo-
tion about the Curve B C & to be come to the
Situations or Politions ACN, nn infinitely

near

© A Treatiſe ©
near each other, and draw the Radii DF, Dy,
nel to CM, Cm.” Then it is manifeſt
that the Sectors D Ff, C Mm, C Nn, are ſi-
milar; and ſo DF: Ff:: CM: Mm::CN:
Nn::CM+CN or AE: Mm+ Nun. Now
ſince this is always ſo, let the Point of Con-
ta& C be any where taken, – it follows that the

Radius DK is to the Arch KFL the Sum of

all the fmall Arches Ff: A E:: A MD+ENH
the Sum of all the little Arches Mm + Nn.
Which was to be demonſtrated.

Conor. I.

x isplainthat the Curves AMD, E N 7H,
are formed by the Evolution of the

| Curve BCG; and ſo the right Line AE is al-

ways perpendicular to thoſe two Curves in all
the various Poſitions of it; ſo that their Di-
ſtance is every where the ſame; which is the
Nature of parallel Lines. Whence it appears
that a Curve Line 4 M being given, we can
find an infinite Number of Points in the Curve
ENA without the uſe of the Evolute BCG,
by drawingan infinite Number of P icu-

rs to that᷑ Curve, and taking them all equal to

the right Line I K.

4 ©

Cor 0 L. II.
r. BC, CG, the Halves of the Curve

3 1 BCG; be ſimilar and equal, it is ma-

nifeſt’ that the Curves AMD, E NH, ſhall
be ſimilar and equal; fo that they only differ

in Potion. hence it follows, that the

Curve 4M D will. be to the circular Arch

KF L:zAE:DR: That is, in a given Ratio.

PRO.

of F LU XION Ss,

par. II.

195

_ there be any two Curves AE V, FIC. 152;

BCG, * with a third Curue
AMD of ſuch a ; Nature, that a Part of a Curve
E M being deſcribed from the Evolution of the
Curve BCG, the Relation of the Portions or
Parts AE, EM, to the Radii of Evolution EC,
MG, be exps efed by any given Equation. It is
required to draw the Tangent MT from the gi-
ven Point M in the — AMD.

Conceive another Part or Portion e of the
Curve infinitely near to E M, and the Radii of
Evolution Ce , Gm R, to be drawn. Then,
1, Let CH be perpendicular to CE, meeting EA the Tangent to the Curve AE Y in H. 2% ML parallel to CZ, meeting the Arch GL deſcribed from the Centre M with the Radius MG in the Point L. 3, GC be perpendicu-
lar 2 M meeting the ſought Tangent MT
in

This being done, make 4 E, EM,
CE=u, G M=z, CH, EH, the Arch
GL r. Then will Ee==x, Fe or Rm==#
Az; and becauſe of the right-angled ſimilar
Triangles e FE and E C H, CE (u): C HU::

Fe (2): On And CE (a): EH (7):

Fe (z): Ee 2 == Nov by the Lemma

NF ne —— ; and therefore RM(RF—me

l Im +54
Whence becauſe of the GemilatCrigngles mR M,
O 2 MCT,

FC. 133.

Fic.133

rt, 8.

MG (>): er

A Treatiſe
MGT, mR O KE +E+5 ): F

Ang 25 – but if 2 and

be put for their Equals | 1 2 x in the Fluxion
of the given Equation, we ſhall get LIVE of

3 in zz which being ſubſtituted in — there

will come out a known Value (freed from Flu-
ions) of the Subtangent & ſought. Which

was to be done.
If the Curve BCG degenerates into a Point

O; it is manifeſt that the Portion of the Curve

ME (Y) will then be an Arch of a Circle equal
to the Arch GL(r), and the Radii CE (u),
GM () of Evolution will be equal to each
other: So that &, which in wa Caſe be-

comes OT, will be Y +5 += 75

EXAMPLE.

L. TY ===, this thrown into Fluxions
will be 5 _ EE (v being taken negative

, becauſe while x and — 004 tf

2 5, by ſubſtituting = for its Equal ;

and therefore C = + 5+

— * —

„putting for its Equal

SCH 0-

of Fruxloxs. 197
55 So uh O LIV u.

171 Ir cke Point O falls in the Axis AB, and Fc. 134.
the Curve AE be a Semicircle, the
Curve AM D will be half a Cycloid genera-
ted by the Rotation of a Semicircle B & N a-
long an equal Arch B GN of a Circle deſcri-
bed from the Centre O, the generating Point
A falling without, within, or upon the Cir-
cumference of the moveable Circle BSN, ac-
cording as the given Quantity à is greater,
equal to, or leſs than OV. To prove his, and
withal determine the Point B,
Let us ſuppoſe the thing to be ſo, viz. that
the Curve AMD is a Semi-cycloid, genera-
ted by the Rotation of the Semicircle BSN,
(whoſe Centre is & the Centre of the Semi-
circle 4 EV) along the Arch BG N deſcribed
from the Centre O; and conceiving the ſaid
Semieirele BF N to remain in ſuch a Situation
BGN, that the deſcribing Point A falls in the
Point M, draw the right Line OX thro’ the
Centres of the generating Circles ; which, by
conſequence, will paſs thro the Point of Con-
tract &; then drawing XK SE, we may obſerve
that the Triangles OXE, OX M, are equal, 1
and ſimilar, becauſe the three Sides of the one 1
are each equal to the three Sides of the other. |
Whence it follows, 1*, That the extreme An-
gles MO K, EO R, are equal; and fo likewiſe
the Angles MO E, GO B. Whence GB: =
ME:: OB: OE. 2, That the Angles KO,
E KO, are moreover equal: And conſequently
the Arches & N, BS, meaſuring them, are e-
qual alſo. The ſame may be ſaid of their Com-
plements & B, SN, to two right Angles, be-
O 3 cauſe

A 7 reatiſe

cauſe 9 appertain to equal Circles. Now
by the Generation of the Cycloid, the Arch

GB of the moveable Circle is e
Arch GB of the immoveable one.

ual to the
hetefore

SN. ME:: OB: OE. This being premiſed,

Call the known Lines OV, i; A Vor Ke;
and the unknown one & B, u. Then will OB
I-; and becauſe of the ſimilar Se-
ctors K E A, K SN. KE ( c): KSH :: AE (x)

0 E 205 bn) E M ( 9) = N 422

be cc cu

. Whence we get K’B (v) = „ |

bevy tt

Wherefore if you aſſume K heb, a

a c
from the Centres & and O do deſcribe the Se-
micircle & N and Arch BG M; it is evident
that the Curve IMO will be half a Cycloid,
deſcribed by the Rotation of _ Semieirele
BSYN along the Arch BGM the defcribing
Point falling without, within, or on the
Circumference of the ſaid Circle,

as KY (c) is greater, leſs than, or equal to KB

(EEE),
A+XC |
tel than, or equal 0

Conor. 1.

is maniſeſt that EM(y J: AE 02

ROE (A): OBxKP be- Lerch).
Now if OB ‘be ſed to become infinite,
the right Line O E Will be ſo alſo, and parallel
do O, hecaufe it will neyer meet the we
the

— —

of FLUXTONS.
the Concentrick Arches BGM, EM will be-
come parallel right Lines, perpendicular to
OB,OE; and 2 right Line EM will be to
the Arch J ER: KB: KV. Becauſe the infi-
nite right Lines O E, OB, which differ from
each other by a finite Magnitude only, may be

Naked upon ee
Conor. II.

Bag ue the Angles MKO, EXO
| the Triangles MKG,
ASE ſhall 4 LE 0 al ay ſimilar; and ſo the

ight Lines AG, equal to each other.

199

hence v if it be Canara to draw MG from Art. 43.

a given Point M in the Curve of the Cycloid

EE to the ſame, you need only de-

the Arch ME from the Centre 0. and

from the Centre M with the Diſtance E B,
an Arch which will cut the Baſe B & N in the
Point &, thro’ which and the given Point A
you oy. draw the Derpengicular requir’d.

Gn III.

Tux Point G bein iven in the Cir-
15 * cumference of – + moveable Semi-
circle ; GNM; and it be requir’d to find. the
Point M in the Cycloid wherein the deſcribing
Point 4 falls, when the given Point & touches
the Baſe; you muſt aſſume the Arch SVG,
and drawing the Radius & & meeting the Cir-
cumference 4 E in E, deſcribe the Arch
EM from the Centre O. Then it is manifeſt
that the ſaid Arch thall cut the Cycloid in the
Point M e

O 4 LY Prop,

2
1 5

Ari. 47.

1 2 chro the given Tei M, the Arch 7G is

4 Treuti iſe A

1 4 b. III.

Fa. 1365 77 1 147 AMD%e a Semi-cycloid deſcribed

by the Rotation of the Semi-circle BGN
aloug an Arch BGN equal to it; ſo that the
contiguous Parts BG, BG, as they Ain mereafe,
are conſtantly equal to one ‘ another – And let the
deſcribing Point M be aſſumed in the Diameter
BN, either without, within, or on the Circum-

ference-of the moveable Cirtle BGN. is re-

quir’d to find the Point M in the Semi-cycloid,
whoſe Diſtance frone” “Ihe it O A 9 14 e

4 maximum.”

If che point M beſappoſes Wbe that -fought,

it is manifeſt that the Tangent in M will be

rallel to the Axis O 4; and therefore the
erpendicular MG to the Cycloid, mult like-

vwiſe be perpendicular to the Axis Which it
meets in the Point P. This being ſuppoſed,

if OK be drawn thro’ the Centres of the ge-
nerating Circles, it will paſs thro’ the Point
of Contact G; and if X L be drawn perpen-
dicular to M E, the equal Angles C KL GOB
will be formed and therefore the Arch 1E,
which is the Double of the Meaſure of the

Angle GK L, will be to the Arch GB, the
Meaſure of the Angle GO B, as the Diameter
BN is to the Radius O B. Whence the De-

termination of the Point G, in the Arch of
the Semicircle BG MN, wherein it touches the

Arch ſerving as 2 Baſe to it, when the Diſtance

of the deſcribing Point M from 40 is a
maximum, will be had by ſo dividing the Semi-
circle BEN in &, that drawing the Chord

of FLuX10Ns. 201
to the Arch BG, in the given Ratio of BV

to O. Therefore the Problem is ws to
a common geometrical one, and may be al-
ways ſolved geometrically, when’ the given
Ratio can be expreſſed in whole Numbers;
but by means of Lines repreſented by Equa-
tions of ſo many Dimenſions the mo the
Ratio is more compoundec.

If the Radius OB be ſuppoſed” to Balewe

infinite, as it will be when the Baſe BNV is
a 4 Line; then the Arch 7G ſhall be infi-
nitely ſmall with reſpe& to the Arch GB.
Therefore the Secant MG then will become
the Tangent MT, when the deſeribi Point
M falls without the moveable Circle; and it
is manifeſt that when the deſcribing Point M
falls within the Circle, there will be no maxi-
um in the Caſe aforeſaid.
When the Point M falls in N on the, Cir-
cumference, you need only divide the Semi-
circumference BOM inthe Point & in the
Given! Ratio of BN to OB. For the Point
Circle BGN touches the Ri when the de-
ſcribing Point falls in the Point W 9’s

“he Pr Bag,

| 176. 7 N. every Triangle BAC, ‘ the. nh Fo. 137.
| thereof ABC, CB, and CAD the
Complement of the. obtuſe Angle BAC 70. two

right Angles being infimtely /mall, are in the [ame
Proportion as the oppoſite Sides A C, AB, BC.

For if a Circle be deſcribed about the Tri- |
angle BAC, the Arches AC, AB, BAC be-
ing the Meaſures of the Doubles of thoſe An-

les

282 A Treatiſe \ .

1 will en . ſmall; and 10 will not

m their Chords. or Subtenſes.
Sides AC, AB, BC of the Trian-
not infinitely little, wiz. are of a
zenitude., it follows, that the circum-
le muſt * une rear; be-
baving a

of the

italy mall

ely {ſmall

5 uber at tes
£ X

Hat og $732; SY \ S111. y 5

Aw £O 51 – n , IV…

Fre. 138. .. TH] HE ſame Ch pg: it is
OT, op on e C 6
e P . M G. M it a the

Evoluze of the Cycloid. oli!

aul Conceive as Ae alt mg infi-
nitel near MG, and which by conſequence cuts
the lame in the fought Point C, draw the right
Line En, aſſume the ſmall! Arch Gg in the
Circumference of the moveable Circle, equal
to the, Arch « Eg in the moveable Citcle, and
draw the right. Lines Meg, 1g, Kg, Og. This
being done, if the ſmall ke Gg, 0 g be con-
ceived as ſtraight Lines perpendicular to the
Radii Kg, Og, it is clear that when the lit-
tile Arch Eg in the moveable Circle coincides
With the Arch Gg of the immoveable ane, the
deſcribing, Point T will coincide with N ſo
that the” Triangle & Mg will coincide with
the Triangle Gm g. Whence. it ap *
= fple un is equal to the An g=
GOg ; becauſe the ſame An 2 70 85
905 being ded. to. Nh 99 =: WAG HP
two Agar Angles. .
212 “Mow

of FLUXIONs. 203
Now calling the given Lines OG, bz XG, a;
GM or En, n; GT or Ig, n; and we ſhall |
have, * 1*, OG: KG:: Gg GOg. And At. 176.
0G ():0G+EXK or OK +0): 0 Kg:

GKg + GOg of MCm= e K g.

2% Ig: MI: Cg: Bg I. And et Bid: |
or MG(m): Ig: See Tr Ig a

iGKg: & Mg or Gmg 2K 3% The

Angle v MCmor MG mm 260 2 id.
SKY o GK): -Gm(n):GC=

n i;
Tam . Zum bn And conſequently re *

dius M C. of Evolution ſought wil be =

. 2amm + 2bmm_ .

Zam + 2bm — bu, |
If the Radius OG bb) of the 8
Circle be ſuppoſed to become infinite, the
Circumference thereof will be a right Line;
and ftrikmg out the Terms 2amm, zam, as be-
ing infinitely little with reſpect to the others
2bjmm, 2bm— n, we ſhall * M.-C =

21MM
zu — 1

Sn I.

1 Bic the Angle MG m = —

GK g, and Arches of different Circles — to
one another in a Ratio compounded of the
Radi and Angles that they meaſure : therefore

Gg:Mm::KGxCGKg: * GK g.

4 And

204

47 reatiſe

And Ab allo K OA 2 + *

gz; er (Which is the ſame thing 3

. >

Fi c. 135.

MGxGg: :OK a+3):OG(b). Which is a

conſtant or ftanding Ratio. Hence it appears,
that the Dimenſion of the Part or Portion of
the Cycloid I MD, depends on the Sum of
all the Rectangles MCxGg i in the Arch GB;
which is what Mr. Paſcal has demonſtrated 1 in
common Cycloids. ©

Mr. Farignon found out this Property *

4 manner very 4 from this here.

= S6lp Wr At.

* the deſcribing Point A falls

zy.. without, the Circumference of the
moveable Circle, there will of neceſſi jty hap-
pen one of the three following Caſes. _ For
drawing the Tangent MT, the | Point of Con-
tact & will fall (15.) in the Arch TB, as it is
ſuppoſed to do in the Figure for the Inveſti-

gation; and then MC (= Dee

zam T 2bm—bu
will be alwiys cater than MG (). 2*, 10
the Point of Contact 7; and then MC
2am + zbmm
— — 94 — ==, becauſe 1G 0 var
niſhes. 3e, In the Arch TN; _ then the
Value of @7 (x) bein now negative, we ſhall
amm I zbmm
have . 80 that MC
will be leſs than MG (n), and always poſl tive.
Therefore in all thoſe Caſes it is evident, that

the Value of (MC) the Radius of Evolution
is always politive.

Conor.

of FLUXIONS. 205

ConoLi. III.

W x the deſcribing Point M falls F : 5-136.
within the Circumference of the

moveable Circle, we have always M C =

22 7. — 15 and it may happen that vn

is greater than 2am + 1 ſo the Value of
(MC) the Radius of Evolution negative:
Whence it appears, that when it ceaſes to be

poſitive to become negative, as it happens

when * the Point M becomes a Point of In- ,,, g.
flexion, then of neceſſity we muſt have bz = |
zam + 2bm and therefore MI MG (nn mm)

— 2amm . mm

— – Now if the given Line
K MM be called e; from the Nature of the Cir-
cle we ſhall have M Ix MG (Ep EP =

BH (aa—cc);and fo the unknownQuanti-
ty MC(m)= ED. Whence if from the
given Point M as a Centre, with the Diſtance
MG —V aah — you deſcribe a Circle; this
24 +6 | :
will cut the moveable Circle in the Point G,
wherein it touches the immoveable Circle
ſerving as its Baſe, when the deſcribing Point
M falls in the Point of Inflexion F.
If MR be drawn perpendicular to B N, it

—— —

5 yaab—bcc\ ill
is eyident that the ſaid MG ( 222 ) w
be leſs than MR (Vaa cc), and that it muſt
be equal to the ſame when þ becomes infinite,
vix. when the Baſe of the Cycloid becomes a
right Line, 3
Note,

A Treatiſe –
Note, That! in order for the Circle deſcribed

with the Radius MG to interſect the moveable
Cucle, it is neceſſary for M & to exceed MN,

that is, for 2 to exceed a— -c; and

a+6

nikeſt, that in order to have a Point of
Inflexion i in the Cydoid IMD, XM mutt be

aa
— 14 KN, and greater than r

LEMMA III.

Fic. 138. 181. ET two Triangles A Bb, CDd each

ha ve one Side 10 d, and D d) infinite-
Iy ſmall with reſpect to the others: I ſay, the Tri-
2 le ABb 3s to the Triangle C Dd in a Ratio
coma; of the AngleB Ab to the Angle DCd,
and of the Square 0 2 Side A B or Ab to the
Square of the Side CD or Cd.
17 or if from the Centres . and with the
Diſtances AB, C D the Arches B E, DF be

Art. 2. deſcribed; it is manifeſt, * that che Triangles

AB, C Dd do not at all differ from the Se-
1 of the Circles ABE, CDF. Whence,

If the Sides 4 B. C.D are equal; the TH-

angles 4Bb, CDd mall be to each other as
their 48 B 46, DC d.

PRO. V.

Fic. 135. 182 THE ſame things being ſuppoſed; it is
required to ſquare the Space MG BA

n under the Perpendiculars MG, BA
20

e | “I
of FLUXTONS. 207
to the Cyrloid, the Arch GB, aud the Portion

AM of the Semi-cycloid AMD granting the
An. of the Circle.

The Angle G Mg (= CK e’s is to rhe

Angle MG m (= bens GK 2). a8 the * lit- Art. 181.

tle Triangle MGg, —_ Baſe is Gg the Arch
of the moveable Kircle, is to the little Tri-
angle or Sector — m; and therefore the

Sector & Mm = N =

Mg + _ Meg by calling M1, p;

and putting pn for m. Now * the little *4r:. 187.
Triangle or Sector X Gg is to the little Trian-
gle Mg in a Ratio compounded of the
Square of KG to the Square of MG, and of
the Angle G Xg to the Angle GMs; that is,

::4aaxXGKg:mmu%x 5 645 And therefore
the little Triangle M Gg = — — & Kg. Now

putting this Value in — £2 22M Gg forthe
Triangle Mg, and Wn wil come out the

Sector G Mm = 24 T2 Ge + = e
b aab

KGg. But becauſe of the Circle C MMI
(pm) =B Mx MN (cc-aa), which is an in-
variable Quantit Fug {o in all Siruations of
the deſcribing oint M; and conſequently
Mm g or et that is, the ſmall cy-

cloidal Space GMmg = — 22. G g +

a4+5

——— — — — — — — —

;
1
[
f-
0
ih
ly
i
b

208

Art. 96.

Frs. 139.

Art. 180.

| a TINA

» A Treatiſe .

ab xt — K Gp. 2. Therefore A unt

aab
is the F Juxion of the cycloidal Space MG BA,
and Mog the Fluxion of the circular Space
MB contained under the right Lines MG,
MB, and the Arch GB, and likewiſe ſince the
little Sector KG is the Fluxion of the Sector
KG; it follows * that the cycloidal Space

MCB 4 = —— « MG B+ (

aab

K GB.” Which was to be found. |

When the deſcribing Point M falls without
the Circumference B N of the moveable
Circle, and the Point of Contact & in the
Arch NT; it is manifeſt * that the Perpendi-
culars MG, mg interſect each other in the
Point C, and then will p-. Where-

fore the little Sector G M m = — 24—26

| b
MGg+ * 205 MGg=— 24 —=26 *
a G 8 * er Kg, by putting (as be-

aab

| fore) 22 ” Ke Gg for its Equal the little Triangle

MGs, vo therefore G Mm N Gg or m Gg,
that is, Aq Cf = W

— KGeg, by ſubſtituting c—aa for

aab

pw its Equal. Now if TA be ſuppoſed to be

the Poſition of (TM) the Tangrat to the
moveable Circle, when the Point T thereof
touches the Baſe in the Point 7; then it is evi-
dent that M Cm Cg =MGTH—mgTH,

VIZ,

of FLUXTONS. 209
diz. the Fluxion of the Space MG T, and

that MGg is the Fluxion of MGT, and like-
wiſe KGg the Fluxion of K 7. Therefore“ Art. 96.

the Space MG H- G-

ec aa
ab – 5 = x KGT. But, as we have alrea-

dy demonſtrated, the Space HT B {= =

MTB+ rz. Whence in all

Caſes we have always the Space MGA |
(MGTH+HTBA) = TEC? T

or MG e XGT TTT 5 or

aab
KGB. |

Wherefore the whole Space DN A con-‘Fic. 135;
rained under (DN, BA) two Perpendiculars

to the Cycloid, the Arch BGN, and the Se-

cycloid AMD, is — 2 + —— X
KNGB, becauſe the Sector KG B, and the

circular Space MG B, do each become the
Semicircle K NG B, when the Point of Con-

tact G falls in the Point M. |

When the deſcribing Point M falls within F : 6. 1364

the moveable Circle, we muſt put aa uc, in- |
ſtead of c- in the foregoing Expreſſions,
becauſe then B Mx MN =aa=cc. wr

If you make c==4, we ſhall have the Qua-
drature of Cycloids, whoſe generating Points
are in the Circumference of the moyeable Cir-
cle; and if & be ſuppoſed infinite, we ſhall
have the Quadrature of Cycloids whoſe Baſes
are right Lines.

P Ax o-

—— —_——
2 — —⅛ — EE CET

äN—— f — —

CE 22 —ů —
_ — — — – — a4 — » — » — —— — — —
— ——ͤ ü—4mü4.— PININY —_ EY

4 * ow = —_ *
—

— * = © 2592 — —
ang

_ *
——-—
_

Fie. 140.

A Treatiſe ©

Ne 16

bi W. rn the Radius O D deſeribe the
Arch D, and with the Diame-

ters IV, BN, the Semicircles AEY, BSN;
and draw at pleaſure from the Centre O the
Arch EM between the Semicircle A EV, and
the Semi- cycloid 4 D, as likewiſe the Or-
dinate EP. Now it is required to find the
Quadrature of the Space AE M comprehend-

—

eld under the Arches AE, EM, and the Por-

—
. be
.
£ 2%

tion AM of the Semi-cycloid AMD.

To do this, let there be another Arch em
bn fon: and infinitely near to EM, ano-
ther Ordinate ep, as likewiſe another Oe in-
terſecting the Arch ME continued out (if ne-
ceſſary) in the Point F. Now call what is

variable, vi. Oe, 23 VP, u; the Arch AE,

*; and (as before) the invariable Lines OB, 5;
KB or XN, a; KY or K A, c. Then will Fe

— Pp=z4, O P=g+hb—c+2, PE= 20
.u 15 the Arch EM*= ===; 3 and therefore the

Rectangle under the Arch E M, and ſmall
right Line Fe, viz. * the little Space E Mme

Now becauſe of the right: angled

ale Tec a 2bu; which thrown into Flu-
xions, and Ze aue big. Now putting this

Value for 2 2 in =, and the little Space.E Mme
vin be =

U
C on oo +4, =y

aa . —
R

Now,

ws X

of FLUXIONS. 211
Nov if the Semi-cycloid 4 ZT be deſcri-
bed by the Rotation of the Semi- circle 4E
along the right Line T „ Eads to VA,
and the Ordinates PE, pe, be continued out
meeting the ſame in the Points H, Y It is ma-
nifeſt * that EH Pp; that is, the little Space Art. 1721

E Hheis=xu; and ſo EA )

E Hhe (xu) :: aa tab: bc. Which is a ſtand-
ing Ratio. But becauſe this is always ſo, let
the Arch EM be where it will. Therefore
the Sum of all the little Spaces E Mme; that
is, the Space AEM is to the Sum of all the
little Spaces E he, that is, the Space AE H
:: 4a απ bbc. But we have * the Quadrature * 477. 99.
of the Space AE H by means of the Quadra-
ture of the Circle; and therefore alſo the Qua-
drature of the Space AE M ſought.

This may be demonſtrated without any
analytical Inveſtigation, as I have ſhewninthe
Atta Eruditorum tor Auguſt, in the Year 1695.

The Quadrature of the Space AE H, may
be had otherwiſe than from Art. 99. For com-
pleating the Rectangles P ©, pg, we ſhall have
or HR: Pp orRh::EP: PA or HO. Art. 18
Becauſe * the Tangent in His parallel to the
Chord AE; and therefore 1 9x 9g = EPx
Pp; that is, the ſmall Spaces H, E Ppe,
are always equal to each other. Whence it
follows, that the Space 47 contained un-
der the Perpendiculars 42, , and the
Portion AH of the Semi-cycloid AHT, is
equal to the Space 4 PE contained under the
Perpendiculars AP, PE, and the Arch AE.
Therefore the Space AE will be equal to
the Rectangle P © minus twice the circular
Space A PB; that is, to the Rectangle under

PE and KA plus or minus the Rectangle un-
: P 2 | der

212

A Treatiſe 4
der KP and the Arch AE, according as the
Point P falls below or above the Centre. And

conſequently the ſought Space AEM = 255 <

xA A+ *

Cor. I.

18 Wu n the Point P falls in K, the

Rectangle K Px AE vaniſhes, and
the Rectangle P Ex AA becomes equal to the
Square of & A. Whence it appears that the
Space AE Mis then = — 2 and conſe-

uently it may be ſquared without the Qua-
ture of the Circle.

Ce or. II.

IX the Sector AKE be added to the
| Space AE M, the Space AK E Mcon-

_ rained under the Radii 4K, KE, the Arch

EM, and the Portion 4M of the Semi-cy-
cloid AMD will be (when the Point P falls above
1 6) C 244+ 555 —

AE+ p Ex KA, and therefore if you

zaac T Zzabe- Q bec
2 V (u) DES
the Value of bec + 2aac + 2abc — 2aau—2abu
| 2bc

AE nothing,) we ſhall have the Space AKEM
4232 pBxKA. Therefore it appears {till

IS
that the Quadrature thereof is had independent
on that of the Circle,

(which makes

Hence

8 |
of FLUXIONS. 213
Hence it is plain, that among all the Spa- |

ces AEM and AK E M, there are only thoſe
two above-mentioned that can be ſquared.

Note, What has been demonſtrated of exterior
Cycloids, extends likewiſe to interior Cycloids,
viz. thoſe that are generated by the Rotation of
the moveable Circle along the concave Part of
the immoveable one; but then the Radii KB
(a), KV (c) will be negative; and ſo the Signs
of the Terms in the foregoing Forms wherein a
or c are found of odd Dimenſions, muſt be
changed. |

SCHOLIUM.

186/T*HERE are ſome Curves which ſeem
to have a Point of Inflexion, and yet
have not; which I think proper to explain b
an Example, becauſe ſome Difficulty may ariſe
about this Matter. 17 0
Let N DN be a geometrical Curve, the Fig. 141.

Nature whereof is expreſſed by a _Z—== 2xXxX— a4 (AP=x, PN = Z), wherein it is evident, 1,
That x being ], PN (z) vaniſhes. 25, That
when x exceeds a, the Value of z is poſitive
and when the ſame is leſs, the ſaid Value is ne-
gative. 3®, That when x = ia, the Value
of PN is infinite. Whence it appears, that
the Curve ND N extends itſelf on each Side
the Axis, cutting the ſame in the Point D
ſuch, that AD Sa; and that the Aſymptote
thereof is the Perpendicular B & drawn thro’
the Point B ſo, that AB Ad.

Now if another Curve E D F be deſcribed
of ſuch a Nature, that drawing the Perpendi-
cular M N at pleaſure, the Rectangle under

Fg 7

4 2
— 32
—— 2 — Z OED as Gt * _
— : — — 4 — – AB — pY — —
© II… — —

— T4

EB *. —
2 — IR

* =
— .

— ⁵— ³—˙ . — — r —

£2 ie — . .
— 2 * * = Ca 2 *
—
„
— —

Sx
„ ——— > aha; 6
2 2 8 1 —

Pr
—

ä—D— — —— —-—-—. — —

— DD eee ea ee…

—_— mu * 2 — —

. *

214

NIC:

v 474.78.

*7r7..81.

AD, be equal to the correſpon

— : „ 15
A Treatiſe *
the Ordinate PM, and the ſtandin ng Quantity
ent Space
DPN then if PM=y, it is manifeſt wah
AIDA Rm (% MP pn or NPxPp

XxX —AAx

97 z and therefore Rm (): Pp or

FN: AD. Whence it follows,
that the Curve E DF touches the Aſymptote
BG continued out in the Point E, and the
Axis AP in the Point D; and fo it ought to
have a Point of Inflexion in D. Vet the Va-
Jue of the Radius of the Evolute of it will be

found #——, being always negative, and be ·

comes e ual to — fa when the Point M falls
in D. Whence we infer*, that the whole
Curve pailing thro? all the Points M, is con-
vex next to the Axis AP, and fo has not a
Point of Inflexion in D. Now to unravel

this.
If you aſſume PM on the ſame Side as PN,

there will be formed another Curve & DH,

21.142.

which will be in all reſpects ſimilar to E DE,
and muſt be a Part thereof, ſince its Genera-
tion is the ſame. This being ſo, we muſt con-
ceive the Parts of which the whole Curve con-
fiſt, not to be ED 7, GDV, but LDH, GDF,
which touch in the Point D; for by this the
above mentioned Difficulty is ſolved. For
Example.

Let DMG be a Curve, whoſe Nature is
expreſſed by y. =x+ aaxx— (AP x, PM
=y). From this Equation it is manifeſt, that
the whole Curve has two Parts E D H, EDE,
oppoſite to each other, as the common Hy-
perbola ; Ft that the he Diſtance DD or2 4D

W —

of FLUXIONS.

will vaniſh likewiſe; and therefore the ‘vwo
Parts E DH, G DF, will touch one another
in the Point D.: So that one would think the
{aid Curye had a Point of Inflexion or Retro-

greſſion in D, according as the Parts thereof

were ſuppoſed to be EDF GDH or EDG,
HD EF. But this Deſception will eaſily ap-
pear, by finding the Radius of Evolution;
which will be poſitive always, and equal to :
4 in the Point D, as aforeſaid.

By the way, we may obſerve that the Qua- F 1c. 141;

drature of the Space DPW is dependent on
that of the Hyperbola; or (which comes to
the ſame thing) the Rectification of the Curve
of the Parabola; and the Part of the Curye
DMF, ſolves the Problem x ed by Mr.
BERNOULLI, in Tom. 2. of the Supplements
to the Acta Eruditorum, page 291.

215
If b be ipod to vaniſh, the Diſtance DD Fc. 143.

4
„
4
£

21 6

14 Treatiſe |
. , *
— = 4 { \ p | | W * | i | 75
i LY . P)/ WL þ »
7 — ar \ << k

4 \

== IT va)
ws 3:4 63664 577 9 8 |

s E Cr. X.

The V/ of Flugxions in Geometrical
“Curves after a new Manner; from
hence is deduced the Met bod of
\DyscarrEs ang HuppE.

bin. I.

‘ET * D B be a Curve ſuch, that the Pa-

rallels K MM to the Diameter 4 B there-
5 meet it in two Points M. N; and con-
ceive the intercepted Part MN or P to be-
come infinitely ſmall ; then is that called the
Fluxien of the Abſciſs AP or XM.

Co ROT. [.

nr rhe Part MN or Pꝰ becomes

g W infinitely ſmall; it is — that
the Abſciſſes AP, 4 O, will each become e-
qual to AE, and the Points M, N, will coin-
cide in D; ‘ſo that the Ordinate ED is the
greateſt or leaſt of all the Ordinates P M, Ng,
=_ it.

Con o L. II.

Av oNG all the Abſciſſes A P, it is evi-
dent that AE only has a Fluxion,

becauſe there can be none, but in that Caſe
where P © becomes infinitely ſmall.

CoRoL

of FLUXIONS.

Conor…

189.T F the indeterminate Lines AP or XM

be called x; and PM or AK, (y) be
invariable, it is evident that x will have two
different Values, viz. KM, KNor AP, 42.
Therefore the Equation expreſſing the Nature
of the Curve AD B muſt be clear’d of Surds,
that ſo the ſame unknown Quantity x expreſ-
ſing the Roots thereof (for y is looked upon as
known) may have different Values. Which
muſt be obſerved hereafter.

PARA OP. I.

7E Nature of a Geometrick Curve ADB

Being given: To determine the preateſt or
leaſt Ordinate thereof.

Tf the Equation expreſſing the Nature of

the Curve be thrown into Fluxions, with y as
a ſtanding Quantity, and x as a variable one;

ZI7

it is plain * that a new Equation will be had, » y,, 188.

one of whoſe Roots x, ſhall expreſs ſuch a
Value AE, that the Ordinate ED will be a
Maximum or Minimum.

For Example, let #-þ4=axy. This thrown into Fluxions with x as variable, and y as a ſtanding Quantity, and 3xxx==ayx z and there- fore i. Now if this Value of y be put for it in the Equation of the Curve x | j*=
ax) ʒ then will AE (x) =” 43/z be ſuch, that
E is a Maximum, as has been already ſhewn
in Art. 48. 2 |

- ————— — Oo — —
T—_— — a —
—

218 ATreatiſe
It is manifeſt that by this way we not only
can determine the Points D, when the Ordi-
nates ED are Perpendiculars or Tangents to
the Curve AD B; but likewiſe when they
are oblique to the Curve, v/z. when the Points
D are thoſe of Retrogreſſion of the firſt or ſe-
cond Kind. So that this new Manner of con-
fidering Fluxions in geometrical Curves, is
more ſimple and leſs intricate in ſome Caſes
*Szer. 3. than the firſt *. |

SCHOLIUM.

Fic. 146. J 91. Ix Curves that have Points of Retro-

greſſion, we may obſerve, that the

£P M* parallel to AA, meet them in two Points

M, O, juſt as the X M parallel to AP do in

the Points M, N. So that AP (x) continuing

the ſame, y has two Values PM, PO. There-

fore in finding the Fluxion of the Equation of

that Curve, x may be conſider’d as invariable,

and y as variable. Conſequently if x and y be

-. . – » taken as variable Quantities, in throwing the

2 aforeſaid Equation into Fluxions, all the Terms

affected with x on one Side, and all thoſe with

J on the other, muſt be equal to o. But we

muſt obſerye that * and do here denote the

Fluxions of two Ordinates iſſuing from the

ſame; Point, and not (as before, $2. 3.) the
Fluxions of two Ordinates infinitely near.

CoRroLl.

I 9 2 7¹ F after having order d the Equation ex-
preſſing the Nature of the Curve where-
in there is only the unknown Quantity x vari-
able, the ſame be thrown into Fluxions; it is
evident,

[
15
N
EY

of FLUXIONS.

evident, 1*, That in doing this, we only mul-
tiply every Term by the Exponent of the Power
of x, and by the Fluxion of x, and afterwards
divide it by x. 25 That the Diviſion by ,
as well as the Multiplication by x may be o-
mitted, ſince in every Term it is the ſame.
3% That the Exponents of the Powers of x are
in an arithmetical Progreſſion, the firſt Term

thereof being the Exponent of the greateſt

Power, and the laſt o; for the Terms that may
be wanting in an Equation, we have repre»
ſented by a Star. |

For Example, let & * – . If eve-
ry Term be multiplied by the Terms of the
arithmetical Progreſſion 3, 2, 1, o, there will
ariſe a new Equation 3x%—4ayx=0.

x) * -% + =.
35 25 I, O. 8
34 F—azx * =0,

Whence there comes out y = = the ſame
as would be found by taking the Fluxion the

common way. £44

This being premiſed, inſtead of the arith-
metical Progreſſion 3, 2, 1, o, you may take
any other at pleaſure, as m- gz, z, + 1,
mio, or n, (m being any poſitive or negative
whole Number or Fradhion). For multiply-
ing * —azz+=0 by xn, and we ſhall have
K, c. oo; every of the Terms of which
muſt be multiplied by thoſe of the Progrethon
mz, z, n 1, n; each by that anſwer-
ing thereto, to get the Fluxion thus. N

AQ ws

219

47 reatiſe

1 -a xt + x”=—=0.
mz, z, met, n
T * — ahr Tm. x”=0.

Wpenes there ariſes w. Pz Nie
m ; and dividing by x, there comes
out n+ 3x—m—+1ayx-+my=0, as was at firſt
found by only multiplying the propoſed Equa-
tion only by the Progreſſion m- gz, M 2, 1 1,
mM. |
If m – z, the Progreſſion will be o,—1—2,
—3; and the Equation ſhall be 22 * — 3) .
If n= —1, the Progreſſion will be 2, 1,0,—1,
and the Equation 2x%—y*=9.

The Signs of all the Terms of the Progreſ-
ſion may be changed, viz. inſtead of 0,—1,
—2,—3, and 2, 1, 0,—1, we may take o, 1,
2, 3, and —2,—1, o, 1; for doing this only
alters the Signs of the ‘Terms of the new E-
quation, which is to be made equal to o. That
is, inſtead of 2ayz—3 y*=0, 2x%—y=0, we
ſhall have —2ayx+ 39% — 2K ＋ . =0.

No it is manifeſt, that what we have de-
monſtrated in the aforeſaid Example, may be
apply’d after the ſame Manner to any other.
Therefore if after an Equation having two e-
qual Roots, be duly order’d, the Terms there-
of be multiplied by the Terms of any arithme-
tical Progreſhon taken at pleaſure, it is evi-
dent that a new Equation will be formed, one
of whoſe Roots ſhall be equal to one of the
Roots of the firſt Equation. By the ſame
Reaſon, if this new Equation has two equal
Roots likewiſe, and it be multiplied by an
arithmetical Progreſſion, we ſhall form a third

Equation,

of FLUX10NS.

Equation, having one Root thereof equal to
one of the equal Roots of the ſecond Equation,
and ſo on. So that if an Equation of three
equal Roots be multiplied by the Product of
two arithmetical Progreſſions; by that means
a new Equation will be formed, having one
Root equal to one of the three equal Roots of
the firſt Equation: And, in like manner, if
the Equation has four equal Roots, it ſnould
have been multiplied by the Product of three
arithmetical Progreſſions; if five, by the Pro-
duct of four, Sc.

This is what the Method of Mr. HupDpE
preciſely conſiſts in.

PRO p. II.

221

1 9370 draw a Tangent T H M from a given Fic. 147.

Point T, in the Diameter AB; or a
given Point H in AH, parallel to the Ordi-
nates.

From the Point of Contact M, draw the
Ordinate MP, and call AT, 5; AH, t; (one
or the other of which is given) and the un-
known Lines AP, x; PM, y; then becauſe
of the Similiarity of the Triangles T A HT PM,
Sto+tx Sy—sf i |

Dy ; and putting theſe Va-
lues for y or x in the — Equation expreſſin
the Nature of the Curve 4M, we ſhall
have a new Equation freed from x or y.

Now it a right Line T D be drawn cutting
the right Line AH in &, and the Curve AMD
in two Points N, D, from which are let fall
the Ordinates VO, D; it is manifeſt that
when F expreſſes 4G in the foregoing Equa-
tion,

A Treatiſe

tion, x or y will have two Values A 2, AB,
or ND; DB, which become equal to one
another, viz. to AP or PM ſought when 7
expreſſes A 1; that is, when the Secant TDN
becomes the Tangent 7M. Whence it fol-
lows, that that Equation muſt have two equal
Roots. And ſo we will multiply it by any
arithmetical Progreſſion at pleaſure; which
mult be repeated, if neceſſary, by multiplying
de Nouveau, the fame Equation by any other
arithmetical Progreſſion; that ſo comparing
the * ariſing therefrom, we may find
one of them affected with either of the un-
know Quantities x or y, and having one of the
given Quantities s or t in it. The following
Example will be ſufficient for explaining this,

E X AM pP L E.

J ET ax=yy expreſs the Nature of the
Curve AMD. If inſtead of x we

— —— z then will 755, &c. Which muſt

have two equal Roots.

ty j—asy +ast=0.
I, o, — I.

| #9) astro.
Therefore multiplying orderly (as you ſee
here) theſe Terms by thoſe of the arithmetical
Progreſſion 1,0,—1, we ſhall have a2 y=
ax; and conſequently A P (x)= 5s. Whence
taking AP A, and drawing. the Ordinate
PM, the Line 7 M will touch the Curve in
M. But if 4H (7) be given inſtead of AT (5),
we muſt multiply the — Equation 7), 5

put

of FLUX10NS. 223

by this other Progreſſion ©, 1, 2, and the
Gight Tangent PM (5) will be=27.

The ſame Conſtruction may be found by
adds * dry te. Ws
putting for y, in ax =yy. For there
ariſes ztxx, &c. the Terms whereof mulriplied
by 1, 0,—1, produce xx=5s, and conſequent-
ly A P (x) —

Conor.

1 95. N ow if you ſuppoſe the Point of Con-

tact M to be given, and the Point
Tor H, wherein the Tangent MT interſects
the Diameter A B or the Parallel AA to the
Ordinates, be ſought; you need only, (in the
latter Equation expreſſing the unknown Quan-
tity x or y with reſpect to the given one 5 or t,.
look upon this laſt as the unknown Quantity,
and x or y as known.

PAO. II.

A parallel to the Ordinates. Likewiſe call
what are unknown, viz. LAs; AN, t; AE,
*; EF y. Then becauſe of the ſimilar Tri-
angles L Ak, LEP,y Ee and,;
ſo that if theſe Values be put in the Equation
of the Curve for or x, we ſhall get a new
2 Equation

11 ͤœ e

1%. xa: — —— ——

— — oO —— — —
—— ——ÿ ͤ O— ey; Arr. ee 2

—_—

—— EF. — — — . *
. — ˖C—⅛ͥL—Ĩ j —r*T , ] 3 SE SAS
© l

— — 2

_— –
1
— —
93 6

Art. 193.

Art. 67.

90 —

—»

ET |
Equation freed from or y, as in the laſt Pro-
polition. |

Now if a right Line T D be drawn inter-
ſecting the right Line AK in I, which touches

the Curve AFD in M, and the Abſciſs in D,

from which are let fall the Ordinates MP,
DB, it is evident, 1, That when s expreſſes
AT, and :, AH; the Equation found as afore-
ſaid, muſt have two equal Roots, viz. * cach
equal to AP or PM, according as y or x be
made to vaniſh, and another Root AB, or
BD. 2, That when s expreſſes JL; and t,
AA; the Point of Contact M coincides with
the Point of Interſection D in the Point F
ſought: Becauſe * the Tangent LF muſt both
touch and cut the Curve in the Point of Infle-
xion F; and ſo 4 P, AB, the Values of x, or

PM, PD, the Values of y become equal to

one another, viz. equal to AE or EF fought.
Whence the ſaid Equation muſt have three
equal Roots. Conſequently it muſt be multi-
plied by the Product of two arithmerical Pro-
greſſions at pleaſure; which muſt be again re-
peated, if neceſſary, by multiplying it in like
manner with another Produk of any two a-
rithmetical Progreſſions; that ſo by” compa-
ring thoſe Equations reſulting therefrom, the
nown Quantities s and 7 may vaniſh.

EXAMPLE.

| 197. = T ayy=xyy+aax expreſs the Nature
of the Curve AFD. 1—.— be

put for x, there will ariſe 5z%—5?y .
.

— — —

of FLUXIONS;
H ty Taag—aast o.
— at
I, o, — I, — 2.
375 25 1, 0.

39 * aa .

Which being multiplied by 3, o, —1, o, the
Product of two arithmetical Progreflions 1, o,
—1,—2, and 3, 2, 1,0, gives us ) =; and
putting this Value in the Equation of the
Curve, the unknown Quantity AE (x) will
be = a. 5 | |

ANOTHER SOLUTION,

I 98. T HE aforeſaid Problem may be ſolved

likewiſe, in conſidering that but only
one Tangent LF or XH can be drawn from
the ſame Point L or K, ſince it outwardly
touches the concave Part AF, and inwardly
the convex Part FD; whereas from any orher
Point Tor H, taken in L or AX between
A and L, or A and AX, we can draw two Tan-
gents TM, TD or HM, HD, the one to the
concave, and the other to the convex Part:
So that the Point of Inflexion (F) may be con-
ceived as the Point of Coinciſion of the two
Points of Contact M and D. If then AT (5)
or A H (t) be ſuppoſed to be given, and you
ſeek * the Value of æ or y with reſpect to s or

; we ſhall get an Equation having two Roots

AP, AB, or PM, BD, which will be each
equal to the ſought Quantity AE or EF,
when s expreſſes. A L, and t, AX. Therefore
that Equation muſt be multiplied by any arith-
metical Progreſſion, &c. en

Q EXAMPLE

226

Fic. £49,

Art. 194;

— ¶ ́ ë—ũ—ͤ—U— — —
3 R

226

A Treatiſe –
© |

199 L.. as before, ayy =xyy+aax; then

again will 5 y\—5ty35—at yy +aasy—
aast=o0o, which being multiplied by the a-
rithmetical Progreſſion 1, 0,–1,——2, and there
ariſes % – —-2aat o, freed from 5, hav-
ing two une qual Roots, viz. PM, B D, when

t expreſſes AH, and two equal ones each to

Fic. 151.

EF ſought, when 7 expreſſes AK. Therefore
multiplying de nouveau this latter Equation by
the arithmetical Progreſſion 3, 2, 1, o, and

there will ariſe 3) – aa =; and therefore EF

(S. Which was to be found.
2! p 2,” IV.

NO Ma given Point C without a Curve
Line AMD, to draw CM perpendi-

cular to that Curve. ©

Draw MP, CR, perpendicular to the Dia-
meter AB, and with the Diſtance CM de-
{cribe a Circle from the Centre C; then it is
manifeſt that it ſhall touch the Curve AMD
in the Point AA. Now calling the known

Lines A P, x; PM, y; CM; r; and the un-

known ones AK, 5; KC, t; and we ſhall have
PKor CES =, ME=y+t, and becauſe
of the right-angled Triangle ME C, y=—t+

OOO MTS FRI IN – -i

So that putting theſe Values for y or & in the
Equation of the Curve, we ſhall get a new

Equation freed from y or x.

Now if another Circle be deſcribed from
the Centre C, cutting the Curye in two Points
“© 1 2 1. 5 D,

6 Fivxioxs. 227
N, D, from which the Perpendiculars N 9,
DB, are let fall: It is manifeſt that when #
expreſſes the Radius CN; or CD, in the E-
quation aforegoing, æ or y will have two Va-
lues 49, 4B or N, DB, whichlbecome
equal to one another (vi⁊.) to the Quantity AP
or PM ſought, when 7 expreſſes the Radius
CM. Whence that Equation muſt have two
= Roots; therefore it muſt be multiplied,

Ex AMY 1. E.
| 201.P =x ax==yy expreſs the Nature of the

Curve AMD, in which putting
rr—tt—=2i—)y for its Equal x, and then as
—y =ayrr—t—2—yy: So that ſquaring
each Side, and afterwards duly ordering the
Equation, we ſhall get 57, e. which muſt
have two equal Roots when y expreſſes P M
ſought. |

„ * — 2 , 2aaty + aass So.
aa — aarr
-—aatt

— +4 35 Ys F ty war
a7 *—qany Þ aaa *

2.44

—— 0
— 0.

Therefore multiplying it by the arithmetical
Progreſſion 4, 3; 2, 1, o, (as here you fee it
done) and there comes out 4y*—4a53424ay
+zaat=o; and the Value of y will be
PM fought. | –
If the given Point C falls in the Diameter Fic. 152
4 B, then will z=»o, and conſequently all the
Terms affected with it will go our. Whence
$45—240a = 4yy = 44x, by putting yy inftead
Q. z of

— — creo — :
*% “x — — * — Ip 88 ——
ay” * 2 —
— „
hy *

Fic. 152.

Fi6. 153.

A Treatiſe

of ax, which is equal to it. Therefore we get
x az; that is, if CP be aſſumed equal
to + the Parameter, and the Ordinate PM be
drawn perpendicular to 4B, and the right
Line C M be drawn, it will be perpendicular
to the Curve AD. 0 |

N ow if the Point M be given, and the
Point C be ſuppoſed to be that ſought;
then in the laſt Equation expreſſing the Value
of AC ) with rele to AP (x) or PM (5),
we mult eſteem thefe latter as known, and the
other as unknown. | “1. a.

COR OL.

PRO P. V.

AN Y Point M being given in a_ Curve
| AMD, whoſe Nature is alſo given :
To find MC the Radius of the Curvature in
that Point: Or to find the Point C, being the
Centre of a Circle of equal Curvature with
the Curve in the Point M. 1

Draw MP, CK, perpendicular to the Axis;
denote the Lines by the ſame Letters as in the
Problem aforegoing. Then we ſhall get the
ſame Equation as there, wherein we muſt ob-
ſerve that the Letter x or y does here denote a
given Magnitude, tho! it did there an unknown
one; and on the. contrary. 5, 7, eſteemed as
known ones there, are in Reality here un-
known as well as 7. „„ 8 Mga 7’s

Nov it is manifeſt, 41, That C the Centre
of the Circle of equal Curyature with the
Curve in the Point M, muſt be in M per-

8 | pendicular

of FLUXIONS. 229.

ndicular to the Curve. 2% That a Circle
can always be deſcribed which will touch the

Curve in M, and at leaſt cut it in two Points

(the neareſt of. which is D, from which the
Perpendicular D is let fall); becauſe we can
always find a Circle that will interſe& any
Curve Line, except a Circle, in four Points
at leaſt, and the Point of Contact M is equi-
valent but to two Interſections. 38g: That the
nearer the Centre G thereof is to C the ex-
treme of the Radius ſought, the nearer will
the Point of Interſection O be to the Point of
Contact M. So that when the Point & falls
in C, the Point D falls in 7; becauſe * the An. 76.
Circle deſcribed with the Radius C M ſought,
muſt both touch and cut the Curve in the
ſame Point M. Whence it is manifeſt, that

s exprefling A F, and t, FG, the Equation
muſt have two equal Roots, (viz. * each equal . 4, 200.
to APor PM according as y or x are made to
vaniſh) and another AB or B D, which be-
comes likewiſe equal to /P or PM, when s
and t expreſs AK, KC, ſought; and fo the.
ſaid Equation muſt have three equal Roots.

EXAMPL E.

L* T ax — yy expreſs the Nature of the

Curve AMD. We ſhall get *), Sc. » 4,2. 201.
which being multiplied by 8, 3, 0,1, o, the

Product of two arithmetical he ms 45 FL

2, I, o, and 2, 1, 0, —1—2, and there ariſes
8)*=2aaty. See the Operation.

3 g*

* — 2a zaat + aass = 0.

aa — AArr
| | > aatt
% 11, O.

23 Is Og N I, = 2

$833 * Za * s.
| . 48 |
Whence KC or PE (t) ==.

If it be required to find an Equation expreſ-
ſing the Nature of the Curve paſſing thro” all
the Points (C), we muſt ſtill multiply 94, Cc.
by o, 3, 4 3 o, the Product of two anthme-
tical Progreſſions 4, 3, 2, 1,0, and o, 1, 2, 3343
and then will 8a5y— 4aay =6aat: Whence
(making, for Brevity’s ſake, 5 — 4 =) there

| zat | a7
comes out y , and % = ge 4413
and therefore 164z=274tt, Conſequently the
Curve paſſing thro” all the Points (C) is a ſe-
cond cubick Parabola, the Parameter to the

Axis bcing = _ and the Vertex is diſtant
from the Vertex of the given Parabola by + a,

nce un- 4.
When the Poſition of the Parts of the Curve
adjoining to the given Point M, is alike on
each ſide that Point, as it happens when the
Curvature there is a Maximum or Minimum.
Then one @f the Interſections of the touching
Circle cannot coincide with the Point of Con-
tact, except the other does ſo likewiſe; ſo that
the Equation muſt have ſour equal Roots.
Now if , Sc. be mulWlicd by 24, 6, o, o, o,
the Product of three arithmetical Progreſſions
4 37 2, 1, o, and. 3, 2, 1, 0,—1, and 2, 1, o,
a; we ſhall get 20 . Therefore
Ne

of FLUXIONS. 231
the Point M muſt fall in A the Vertex of the

Parabola, in order for the Parts of the Curve
adjoining to it on each Side to be ſimilar or

alike.”
ANOTHER SOLUTION.

207.JF you call to mind what has been de- Fi. 154-
monſtrared- in Art. 76, viz. that but

one Line C M can be drawn from the Point C

ſought perpendicular to the Curve AMD;
whereas from any ſother Point & in the ſaid
Perpendicular, there can be drawn two Per-
pendiculars MG, G D, to the Curve. From

this Conſideration we can ſolve the Problem

thus: Suppoſe the Point & to be given, and

ſeek * the Value of x or y with regard to s and i. 200.
t which are given: Then it is plain that the
Equation muſt have two unequal Roots, viz.

AP, AB, or PM, BD, which become equal

when the Point & coincides with the Point C

ſought. Wherefore multiply that Equation

by any arithmetical Progreſſion, &c.

EXAMPLE E.

T Er, as before, ax==yy, then will * 4%, Art. 101.
(9c. See the Operation. |

4) * — 4a + 2aat=0.

244
2, Is Og —
dy * — 2aat=0.

Which being multiplied by the arithmetical

Progreſſion 2, 1, 0,— 1, and there comes our,
| ä

28 * before, =. i

tore, 2 Art. 204s

8 Coro.

1 | | Gr 2 ? * # * * ;
=— 25s Nie
BE – ee + $15
Fre. 153, 207. I is manifeſt that the Point wherein
7 the Radius of the Curvature touches
rt. 203, the Curve, may be conſider’d * as the Place
where the Point wherein the Circle of the
ſame Curvature with the Curve touches it,
coĩneides with the Point of Interſection of the
Art. 205. ſaid Circle; or elſe as * the Point wherein
two Points of Contact of different concen-
trick Circles coincide. Juſt as a Point of
Art. 196. Inflection is looked upon *, as that wherein
a2 Points of Contact coincides with the In-
Art. 198. terſection of the ſame right Line; or * as the
Coinciſion of two Points of Contact of two
right Lines iſſuing from the ſame Point.

Por. VI.

Fic. 155. 208.J”O find an Equation expreſſing the Na-
ture of the Cauſtick AF G K, gene-
rated in the Quadrant CAMNB, by the re-
fleted Rays MH, NL, Cc. the incident Rays
PM, QN, Sc. being all parallel to CB.

We may obſerve, 15, That if the roflected
Rays MF, NG, touching the Cauſtick in F,
G, be continued out to meet the Radius C B
in the Points H, L; then will HH be=C H,
and N L=CL. For the Angle CHH=CMP
MCH; and in like manner the Angle CNL
CN NCL. |
20, That from a given Point Fin the Cau-
flick A FX, there can be drawn but one right
Line MH equal to CF; whereas from a given
Point D between the Quadrant AMB, and

2 | the

of FUxIO&s.
the Cauſtick 4 F A, two Lines MH, NL,

may be drawn ſo, that MHC, and VI.
=CL. For but one Tangent MF can be

233

drawn from the Point F; but from D two

Tangents MH, NIL, can be drawn. This
being well underſtood. re it]

It is required to draw the right Line MH
from a given Point D, in ſuch manner, that

it be equal to the Part CH determined thereb
in the Radius CB. is
Draw MP, DO, parallel to C B, and MS
parallel to CA; call what is given, viz. CO
or R, u; OD,z; AC or C, a; and the
unknown Quantities CP or MS, x; PM or
CS, y; CHor MH; r. Then becauſe of the
right-angled Triangle SH, rr=rr—27y+

YAxx; and ſo c N. Moreo-

ver, becauſe of the ſimilar Triangles MR D,
MSH, MR (x-): M&S ():: RD (z—9)

😀 —— . | And therefore C & + Hor

Xx 1

| 2-1 44 Ear xs
= 7 ce

for xx+yy. Whence (multiplying crofswiſe)
and there ariſes aax—a2au=22zx3z—249yy; and
putting aa xx for yy, there comes out 22
= 4ax + aau—2uxx. Then ſquaring both
Sides, to get rid of the Surds, and again ſub-
ſtituting aa – for 0, and at length we
have 4uuxt—4aaux— 4aauuxx + 24% fr aus
42 Aa

+ 24
Now when « expreſſes CO, and 2 O D, it is
manifeſt that this Equation muſt have two

9 unequal

— — — —

. ——ͤ— — — A Ou 2

bs fr We STD ani ee —— .. * — — —̃ —⅜ — C7»

— 4 — — — —o_
r

— 2 4 rn,

— — —

— ——5

—— — —
— — — 22

— 7 Ln.

— —
— 6

—

—
— 222

+74

A Treatiſe ©
unequal Roots, viz.,,C P, C Q; and on the
71 7, when # expreſſes C Ez and 2, EE;

will become equal to C; fo that then it

: 91 have two equal Roots. Therefore if the

In. thereof be multiplied by the Terms of
© Arithmetical Progreflions, hs 35 2, i, 0.
and o, I, 2, 3, 4, there will be two new

uations formed, from whence there will a-

_ riſe this Equation, after the : yaknoWn Quan-

tity x x is We out.
7 NF. in — 244. 4 60
þ > #þT zun Ga- I fa*un
IA! 48a
464
n the Netten between the Abſcik
GE. (u) and the Ordinate EF (2). Which

Was to be found.

The touching Point F may be determined
by what is explained in the eighth Section.
For if another incident Ray pm be conceived
infinitely near P 2, it is plain that the refle&-
ed Ray n h ſhall cut MH in the Point F ſought;
from which having drawn FE parallel to PM,
and making CE Su, EF 2, CP =.
PM ), CM=a; you will find as before

| 40 DE

—=22. Now it is manifeſt,
that GAL CE, E F continue the fame while

CP and PM vary. Therefore with 4, u, z, as

ſtanding Quantities, and x and y variable ones,
throw that Equation into Fluxions, which
will be 22yxxx-+aauyx—aaxx)—aauxy+ zur

=o. Wherein ſubſtitute — 25 2 for its Equal

v, (becauſe they are ſuch by hieewling yy =
e Ny * luxions) and then aa — xx for 9,

and

of PFLUX1ONS. 235
and at length there comes out CE (4) =
a | WO

——

aa 8.

If the Curve AMB be not a Quadrant,

but any other Curve, whereof the right Line

MC is the Radius of Evolution in the Point

; it is manifeſt, * that the Portion Mm Ari. 76.
thereof may be eſteemed as an Arch of @ Cir-

cle deſcribed about the Centre C. Therefore if

from that Centre the right Line C P be drawn
perpendicular to = incident Ray PM, and

aſſuming CE = — Pes CMSa), you

draw EV bnd to PM; it ſhall cut the re-
flected Ray MH in the Point F, wherein it
touches the Cauſtick AFK.

If thro” all the Points M, m in any Curve
4 M be drawn ſtraight Lines M C, M to a
given Point C in the Axis AC of it, and other
right Lines MH, mh terminated by C per-
pendicular to the Axis; in ſuch manner that
the Angle CMH == M H, and Cab —mCh;

and it be requir’d to find che Point F, in eve
Line M Wherein it touches the Curve AF
formed by the continual Interſection of the
ſaid right Lines MA, mb, we ſhall find as before

CH 2 — From whence we
If © NW.
4 , —
ger 12, Which thrown

into Fluxions (with # and æ as ſtanding Quan-
ities, and x and y as variable ones) will be
2.x? * — UXXYX — UxxYx — xy + ux*y + xxyyp+
uxyy) —uyx—0; and therefore CE (4) =

8 e N
-. Now the Nature of

Xa +F) — 0 l |

the

: ( —— PET —
-n Ä 2 4
= by _
“- 2 a * * —

—— 3

236

Fig. oo

4 Treatiſe
the Line 4 MB being given, we ſhall have a
Value of ) in x, which being ſubſtituted in

the Expreſſion of CE, and the ſame will be
whe . F luxions. |

PRO p. VII.

209 5 5 T AO be an indefinite right 3

the Beginning whereof is the ſtable
Point A z and let there be an infinite Number of
Parabola’s BFD, CDG, having the right Line
AO as a common Avis, 70 which the right Lines
AB, AC intercepted between the ſtable Point A
and their Pertices B and C, are the Parameters.
I is requir’d to find the Nature of * Line AFG
touching all thoſe Parabola’s.

We may obſerve, 15 „That an any. two of theſe
Parabola’s BFD, C D G interſect one another
in the Point D, ſituate between the Line AFG
and the Axis 4O; and that when AC 4B,
the Point of Interſection D coincides with the
Point of Contact F. This being well under-
ſtood, _ |

It is required to draw a Parabola thro’ the

iven Point D having the denoted Property.
raw the Ordinate 30. and call the given
Quantities 40, u; O D, z; and the unknown
one AB, &; then by the Nature of the Para-

bola given IBN BO (ur- xx) — DO (z2);
and ordering the Equation xx — ux 2 D.

75 Now when # expreſſes 4O, and z OD; the

fad Equation will have two unequal Roots,
viz. AB, CA: and. on the contrary when «
expreſſes AE; and z, EF; AC becomes equal
to. AB, that is, the Equation then has two
equal Roots. Therefore it muſt be multiply’d

of FLUX10ONS.

by. the Arithmetical Progreſſion 1, o, 1:
and ſo x —2z, and ſubſtituting z for x, there
ariſes 4— 22, which muſt expreſs the Nature
of the Line AFG, Whence it follows, that
AFG is a right Line making the Angle FAO
with 40, being ſuch that AE is the Double
of EF. | |

If the Problem is requir’d to be ſolved ge-
nerally, viz. let the Parabola’s BFD, CDG
be of what nature you pleaſe. Recourſe
is to be had to the Method explain’d in
the Eighth Section, and the ſame muſt be uſed
thus. Call AE, u; EF, z; AB, x; then will
u * x =2″+” expreſs generally the Na-
ture of the Parabola BF. This == 12-091
thrown into Fluxions (making x and z in-

variable, and x variable) and we have
—ů — nant, —

— m XU—X XXX”+ aux XXU—x 0;

ma. 1 1

and dividing by x „ „ there

2
comes out x == –

m + n $ le

Now ſubſtituting theſe Values for

u; and therefore 4—x =

mM
mn |
4 — x, and x in the general Equation; and
. . 8 7 3 8 N

making (for brevity’s fake) nar at *
=q, mr, and then will z be ={/p”9″.
Whence it appears that AFC is always a right
Line, be the Parabola’s of what Nature ſo-

ever, the Ratio of AE to E Fonly varying.

From what has been explained in this Section,
it evidently appears how Deſcartes and Hudde’s
Method muſt be uſed in the Solution of Problems
of theſe Kinds when the Curves are geometrical.

237

But

238 A Treatiſe, &c.

But it is not comparable to the Method of Huxi-
ons, which furniſhes us with general Solutions,
and extends to all Kinds of Curves, without any
Neceſſity of clearing Equations of Surds. Where-
. as, by the former Method, we only get particular
Solutions, and are neceſſitated to throw out the
Surds: which very often cannot be done.

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Being an

APPENDIX,

Containing the

Inverſe Method of Firuxrions,

Application thereof in the Inveſtigation

of the Areas of Superficies, Lengths of
Curve Lines, Contents of Solids, and the
Determination of their Centres of Gravity
and Percuſſion.

Wherein are Examples of Solutions, accord-
ing to the excellent compendious Way of
the late Learned Mr. Ro GER Cor Es, by
the Meaſures of Ratios and Angles, or Ta-
bles of Logarithms, and natural Sines and
Tangents.

APPENDIX.
OF TAE
Inverſe Method of Ftuxioxs.

— —

SECTION I.
Of the Reduction of Fractional Ex-

preſſions and Surd Quantities to in-
finite Series. N

HE Fluents or flowing Quanti-
ties of Fluxions expreſs d fraction-
wiſe, or of ſuch wherein there
are Surds or radical Quantities,
in general cannot be found till
the ſaid raftional Expreſſions are freed from
their compound Denominator, and brought to
ſimple ones, and the radical Quantities from
their Surds, by throwing ſuch Expreſſions in-
to infinite Series. Which may be done by
the two following Problems.

R PRO B.

. —
— – th
= — — = — —

— – ——— = = ; —
— — : —— — –

— —

—
—

— 2 d >
— — —

APPENDIX,
P R O B. I. |
“* T O throw — (a and b being ſtanding,
and x a variable Quantity) into an infinite Se-
ries, whereby it will be freed from its binomial
Denominator. 4 5 9 “a
Divide the Numerator & by the Denomina-
tor a+ x, after the very ſame manner as you
do decimal Fractions, by adding o to the Re-
mainder, and repeating the Operation till you
have gotten 4, 5 or 6 Terms in the Quotient;
after which, in many Caſes, you may find as
many Terms as you pleaſe, by conſidering the
Law of the Pon of thoſe Terms alrea-
dy found. And an infinite Number or Series
of Terms ſo found, will be the exact Quotient
of the Diviſion; bur uſually a few of the firſt
Terms are ſufficiently near the Truth for any
Purpoſe. 55
EX AMP L E I.

ITS OO 0 Ee

APPENDIX.
For dividing 6 by a, the Quotient is 2. The
Product of Aby a+x is = = or +5 |
Which ſubſtracted from the Dividend &, and
there remains —E Again, if © —Zbe di-

vided by a, the Quotient will be — – . There-

fore the Product of a+ x into —— that 1s,

7
abx bx* —- be bw -:
C0

the Dividend — leaves © + 4 W hence

the Law of the Continuance of the Diviſion

is evident. Now the Quotient conſiſts of an
infinite Series of Terms, whoſe Numerators
are the Powers of x, leſs by 1 than the Num-
ber of the Order multiply’d by 5, and deno-
minates the Powers of a, whoſe Exponents
are equal to the Number of the Order of the
Terms. For Example: In the third Term,
the Exponent of the Power of x in the Nu-
merator is 2, and of à in the Denominator
is 3.

Tn like manner, if you put – the firſt Let-
ter in the Diviſor, and then divide h by x + a,

as before, the Quotient will be – — + =
x 2

, &c. $0 that it’s plain there will be as

many Quotients or. infinite Series gotten by
this Diviſion, as there are Terms in the Divi-
ſor; and thoſe Terms of the Diviſor which are
reateſt muſt ſtand firſt, as well as thoſe in the
ividend, in order to have a true Series.

R 2 For

„ 1
4
.

ApPEND IX.

For Example: Let þ=1, x 1, and a =:.

| Then if the Diviſion be perform’d with à as
the firſt Len * the Diviſor, you will find

1 | Rela += = * Sc. which

W iS

know to Ri true * other Princi-

ples. But if be the firſt Letter, then will

— — be =1 —-2+4—8 +16, Sc.

15 2
which is falſe. For this Series diverges, and
differs ſo much the more from the true Quo-

tient, as the Number of Terms is greater. For

Example: One Term 1 exceeds 4 by 3; two
Terms are deficient by 4; three Terms exceeds
it by 55 four Terms are deficient by 55 and fo
on.

E * 4 M y.L &- I.

5G 3 e &c.

44x? ar-
UN * —

APPENDIS.
If 3 be put firſt, then will the Quotient be
aa 44 aab* ‘aah*

„ 200 SL x XC.
Again; the Quotient of 7 — will be

found 1 — x + x* — * + , &c. or (ma-
king xx the firſt Term in the Diviſor)
XA rmx . Fr &c.
2* —x? 2
Moreover,. freed from its com-

C L+ x ——3x. as
pound Denominator by Diviſion will be brought

4 U
to2x —2x+7x—13x + 3 4×5, &c.
And laſtly, This Fraftion _
IIA -H x* + Tra x* — ie, &c.
I = r = , —Trb x —rhebx, &Cc.
having both Numerator and Denominator in-
finite Series may be freed from its compound
Denominator, or brought into an infinite Se-
ries by dividing the Numerator by the Deno-
minator, as before: the Operation being alike
to that whereby one interminate decimal Fra-
ction is divided by another.
This Quotient or infinite Series will be
ITU THAN NH , &c.
ria Trab“ rab E abs
— 1 ra — ra
Tr TT

rtr.

P R O B. II.

T2 free a compound Expreſſion fram Sards,

by throwing it into ap infinite Series, Sup-

poſe Vaa , xx.

Extract

r —_— 8
k – *

APPENDIX.

Extract the Root thereof, as you do the
Root of a decimal Fraction by the Addition
of Cyphers, and bringing out ſo many Terms,
till you may diſcover the Law of the Progreſ-
ſion; that from thoſe already found you may
continue on the Terms at pleaſure, and the
thing 1s done.

ExAMPLE I.

| | af . . c
a 16a 12897 uy

OTXX
**
— 22
93 nt
x4 as *

- *
  O- wh
  T 5. 644®
* 8 470 ***
Da 164 54 2564″?

For the ſquare Root of aa is +a, for the
firſt Letter or Term of the Root. Which
{quar’d and ſubſtracted from aa+xx, there
remains xv. Which being divided by 2 4 (as
in the Extraction of the ſquare Root) and the

Quotient will be __ – the ſecond Term cf the
Root. Which added to 24, and the whole

multiplied by – will be 2 This ta-

ken

APPENDIX. 7

ken from xx, and there remains TRL <2 Which
divided by 24 — the Double of the two

firſt Terms, and the Quotient —85 will be
2
the third Term of the Root. This added to

2 —4 „and the whole multiplied by — —
5 „ * * a
will be Fer 8 ＋ Gan” Which taken
—8 . 7 |
from — , and there remains o- 8a. Tar

And after this manner you may find any Num-
ber of Terms. And ſo continue the Series on at

Pleaſure.
If xx be the firſt Term of the Surd, then

will the Root be x + Z — © + A

Fa * 08 * 16x –
1 28x” 2 FM

And here, as well as in the Diviſion afore-
oing, the greateſt ‘T’erm of the Surd muſt
firſt, otherwiſe the Root or Series will

not be true.

_ EXAMP L EL.
xxx (» —i$x* —+ x* <Iz* E &c.
—

1 0 ; Y
g—x”

_ x” + 42

irn
O—y Xxt—z x
erer
O = —#7 ο e
— rx tix ATN r

—
— ——— —
— 1 » © os

—

—— —
P ³¹ Ä — —— —

— {cue

— —— 2
6— — — — ror oo on . chat.

—
— —

<< – =

———— 3

—

1 1
2 — —_ .

— —
1 l

Af –
—— bb –
.

—
—
TI pre
r

—
—

Ar PE N Dt X.
Ex AM L E III.

erge- pay N

eee Ke.

„„
oK 5m, &c.
2 2x + „„ MS 1h

1.9. an. ty Ke.

941
We
Fa

2 F

After the fame way you ma extract the
Cube, Biquadrate, c. Root a Surd, even
if it be an infinite Serics.
But theſe Extractions, as well as the Divi-
ſions aforegoing, will be very much ſhorten’d
by a Theorem invented for that Purpoſe by

Sir Ian Newton, which i is this: P 4: P **
=p* ＋ e kes

1 T | DEK. g Sh BE tae BY. 2 3 0; 10.

“= PP 9 1 is “the aht) Abele
Root or any Dimenſion, or Root of the Di-
— is to be found; that is, expreſs’d by
Aan infinite Series. P is the firſt Term of that
Quantity: ** reſt of the Terms divided by

6＋„—özẽHw «4

firſt; and — — is the numeral Index. . of the” Di-

menſion of P+P9. Moreover,-4, B, C,
D, &c. are uſed for Thi Terms end in the

— —

Quotient, vix. A 2 the Gol Term . B for

= the ſecond— 2 485 and fo on.

eter k — a , -..J A few

T A * – ry oc — _ 8 p

W N PLE
_— 1 aw TE
$ n *

APPENDIX. 9

A few Examples will ſhew this wonderful
Theorem’s Uſe.

EXAMPLE I.

„% Fix or r 18 2442 eb

24 8a
. for this Cal P1
16% Ia C. IoOr in t IS Cc, n
Xx
= — = =2 4 = P%= 4a; )=58.
= m=1,n » 4 (=P aa” )J=4
mm XX “x
=- A) =—. C (= B
e
at La.)
— — VC.

erer II.

Ye 4 af Fo 1 ==
— Or 4 X—X

6
. a + — — +: ante — + &E. for
; Fat 25

here m=1,n=5, Pa and 2

PR Oo ad

WS ca –

——DM@eGR—OQ OO EDI. OS Oren. op _ —— – ” PRY

_ 3 — AT —— —— * ——
ab 4
4 2 \

by *
97 — 9 * js . 2 – wet Sa a – * oh ©. Fs 7 I
. * 1 oY * a+ 4 – * % „ a P „ 632
. =

Av? EIS. .

7
14 „

E Xx AMPLE IVV.
: | 1 3 – TY | 7
e or N LA Pe lay
2 —
=£&+ Kc. for P=a. ==. 1 =

; 8 145 ay f W 2

2 23. oor d, ce.

Ex AM YT “Ea
—

2 P or 2 is ener He.
＋ 10 ** 4 N For P a. =

—— —

| m=5, and . Z Er). 08
f 249) = Feix. and ſo C o. D =

ro E = 5a. = X*, and EC (=
2 #2): .

Exanyix VI.

— — Pans —

+ —

| — a2 20 | 7 —

1 491 9

— ann A for in this Cafe P = =

ArPEN DIN.

Do Sc. So that the ſaid wonderful Theorem
7

m frees Fractions from their Denomina-
tors, as well as extracts Roots.

4
—

)

) ExaneLe VII.

;

a nn — 3 x x

a x 1 4
Lo **

9 & 1 2

Ex AMG LE VIII.

OO . —
And F=—==b x a+x is = K
6 VNA ; 8 | a
Arn 1 ©;
—— + — 7 Kc.
347 94 814

EXA NT L Y IX.

I 2xx Ca

“T5 &c.
Zya 12yay”?

In the Philoſophical Tranſactions, No. 2 30.
Mr. De Moivre has given us the following

Theorem for pn, infinite Series to a given |

Power u, or extracting the Root thereof, viz.

— 1 ; jg, — 7
azo+bzy+c2+dz*-e3f+f25, &c. Will be

1 =a>2>

—
to
7
to

1 * — . – — –
= — — ———— : A nee *
‘Y

1
:

APPE ND IX.

1 111 3 +
nn 2 n —3 326
1 2 * ==”
„„ „
“Oy F. 1 |

— Mow
. I 2
5 / al 1
N
1 R N 73; M4 an—s ys ?
F
5 — 1 „ —2 MWm—
=: x EI == 3 gn—433
“2 2 3 I
; FP
21 92
7 —1
» 45% “fi R 1
I T7 |
E
1 I
© Y |
— MT gn—2;
i * Y $1
“hs b 2
Ds N * roy “If

APPEND IX. 13

M 15 Mm 3M A Ca 6 56

— ——Uti¹ de *

EPS i $1443
+. te Et : 1 * + 42 — 59
F

1 2
| N M—
1 — X „ — 7 —32—.5
N WP mm

* . an —3 Pe
2 Wh — 2 |
= xXx Can — bd he
—+ * 1 A |
I I
I 2
2 x .

For underſtanding of which, it is only ne-
ceſſary to conſider all the Terms by which the
ſame Power of z is multiply’d: In order to
which two things in each of theſe Terms muſt
be conſider’ d, 1%, The Product of certain
Powers of the given Quantities or Coefficients
a, b, c, d, &c. And, 2% The Unciæ or Pro-

ducts of * , c. prefixed to them.
1 2 |

Now to find all the Products belonging to
the fame Power of z. For Example; to find

that
J

APPENDIX,
that Product whoſe Index is r (r bein
any whole Number) the faid Produc malt
be diſtinguiſhed into ſeveral Claſſes. Thoſe
which immediately after ſome certain Power
of a (by which all theſe Products begin) are
Products of the firit Claſs: as an — %% is a
Product of the firſt Claſs, becauſe h immedi-
ately follows am-. Thoſe which immedi-
ately after ſome Power of a have c, are Pro-
ducts of the ſecond Claſs. So an d is a
Product of the ſecond Claſs: Thoſe which
immediately after ſome Power of à have d, are
Products of the third Claſs, and ſo of the reſt.
This being underſtood, 1®, Multiply all the
Products belonging to 2 – (which im-
mediately precedes 2+”) by , and divide
them all by a. 29, Multiply by c, and divide
by a, all the Products belonging to 2 .
except thoſe of the firſt Claſs. FE Multiply
by a, and divide by à all the Products belong-
ing to zu – except thoſe of the firſt and
ſecond Claſs. 4*, Multiply by e, and divide
by a, all the Terms belonging to 21

except thole of the Grit, ſecond, and third

Claſs; and ſo on, till you meet twice with the
ſame Term. Laſtly, Add the Product of an —-
into the Letter whoſe Exponent is 7» I toall
theſe Terms. |

Note, The Exponent of a, Letter is the
Number expreſſing what Place that Letter has
in the Alphabet, as 3 is the Exponent of the
Eetter’c. Ar?
By this Rule it is manifeſt that it is eaſy to
find all the Products belonging to the ſeveral
Powers of , if you have but the Product be-

longing to 2, Viz. a“. ö

| Now
4

APPENDIX.
| Now to find the Unciz prefix’d to Every
Product, you muſt conſider the Sum of the
Units contained in the Exponents of the Let-
ters that compoſe it (the Index of à except-
ed); then I write as many Terms of the Series
mMXEmM—1 XM—LXM—}, Cc. as there are U-
nits in the Sum of theſe Indexes; this Series
js to be the Numerator of a Fraction, whoſe
Denominzktor is the Product of the ſeveral Se-
ries I 2K 3 4X5, Cc. IX 2X XAN Y, r.
I X2X 3 4XF X6, (96. the firſt of which con-
| uv as many Terms as there are Units in the
ndex ob ] the ſecond as many as there are
Units in the Index of c; the third as many as
there are Units in the Index of d, &c.
The Demonſtration of this ſee in the above
cited Tranſaction. | 1
Here follows an Example or two of the Uſe
of this Theorem. |

EXAMPLE. I.

C
1 | o raiſe this infinite Series – LY
, | S ,X” Þ*
Se. to the ſecond Power, or to {quare it.
In this Caſe in the Theorem #2 = 2, z=x,

bus. I — 2. — 3
g = n, ee K..
xx? ot? * 5 * | *

——

2
b

— 3 7 7 1 ̃ | 3 * * * _ C1
—+ —+ IA, r. will be- = + — +
4 * a

ae” x. 2 5
27.4, G forthe firſt Term a”z*(=Zxax))

is . The ſecond Term d b

| 3
2

22 K x 1 is 2.

The

16 APPENDIX.

The Fourth Term 0 4 — —Lim—2;

a0
| 6
O

I l
2 K —X—XXX%X 0
2

1
2 NI NI XI XI XI —
e

E x A M L E II.

Pe. ſquare this infinite geriet 1— * 2 4

, &c.

i In this Take in the Theorem = 2, 2=2.

þ=1.c=—1.4=1, &c. and

Fes

fo 1 — -+ x* ee Nc. will be =1—2
„ ＋ 4 4 e Ke. for am 2m

—— b.) =I—2x xx. Siam —

4 1 RA t r 3 i

a” =

APPENDIX. 17

n
1 2 Zm E 2 —
2 an —1 6
1

2—2
n XI X xt

| =—x* + 2. x*, &c.

SI X— I X x
y.- 2

ETC ANT WI. HL

T* raiſe I —x+ – + x”, &c. to the
third Power, or to cube it.
Here z. T & . 4 == 7 — I, b *_

c=1, d=0o. and fo the third Power will be
I —3xX+3 +25 —6 xt, Cc. for am 23

— 3
1
( word ** =1—3x + 3x x gf,

7 ;
r _ (gat — I X OXx* }=O,

— — 1 am— 2 52
: Zm + * =
2 am—;
1
—C____

I |
R XO*X x9 So
— |

I
zx ——IXIXX =3x* —Cxtx3 x8

3 — 6x* + 3×7.

And becauſe , and alſo 4, therefore the

next Term of the general Theorem will be o.

And thus you may proceed on.
‘E

— —

——

Ex Au-

:

{
ry
x
!
q
‘
1
1
4
(i
1
|
TY
4 :
|
|

|
|

I8 APPENDIX.

r. ne IV.

T extract the Root of an infinite Series;
that is, if z be Sa T C + dx+

ex*, Ge. to find the Value of x in an infi-
nite Series of Terms affected with z, and free
from x.

Firſt, Let us ſuppoſe x =fz + bhz* + k 23
+124 + m25 +125, Sc. Then by the Theo-
rem x 2 afbe+ ZH +K* 25

2fkz+ 2f12 he
＋ 2fmz*
| CC.
x f + 3f bz + 3fÞ 2! +h* 25, &c.
3/*k2’+3f .

＋ 6f bkz*

“+4f bz* H b, &c.

*. . Af
wes. rs 6 &c.

5 Now ſubſtitute theſe Values in the Equation

O – Tax TC +dxÞex, Cc. and
then will — z = — z.
bSax=bafz+ahbz* + akz* + alz* + ams Tanz, &c.
The = * bf* T +2bfbz* TI +26f1z* bk” ,
S2bbkz4 – *
N –2bbkz* +2bfmz*
hat ebe fat,,
FAE Cc.
| 301 H

| —+ 6fhkzs
dS #* e T+ifz* +adf’2’Taf’b’2

| * o+44f* bz.
8 295 Þ r * 2 ef hz

Nov if the Sum of the Coefficients of eve-
ry Term in this Equation be made equal to
no-

APPENDIX. I9

nothing, we may get the Values of the Cocf-
ficients f, %, K, 1, m,n, thus. The Sum of the

Coefficients of the firſt Term 77 > 2 will be
af—1. Whence if af=-1 o, then will f
be =. Tn like manner the Sum of the

a
Coefficients of the ſecond Term — * will

be ah + bf. Whence if 2h + bf So, there

will ariſe == == In like manner

the Sum of the Coefficients of the third Term
made equal to o will be ak+ 2b ef So.
3 3
Whence þ = e — Again,
a a
al+bb+2bfk+3f *h4 df . Whence =
zA -Z __ —b a ＋ 2

: a — 139g a”—4b* a*+36b6
+= = [aL – — “go bkewidimin
4 2 * 1
— 145 + 64bd 21abc 2 1

—4255 4-84.45) c=284″ bi==2Ba* %. -Z ic. beabf
Il —_ | — 9

| 4
W hence at length ſubſtituting theſe Values
of the Coefficients f, %, K, I, in, u, in the aſſumed
Equation x=fz . E + 1z-|-mz + nz*,

c. and the Root ſought will be x = A

I 20 —2c _, , Fabi==fb’ —a’d „

+ 7 2 +
4 2 3 – : 8 3
145*-F62 bd . +34 —2 2275 Ge. |
Note, If there are any Terms wanting inthe
propoſed Equation, it is plain that they will
; T 2 .

likes

APPENDIX.
likewiſe be wanting in the Root. For Exam-
ple: If z be as Ken, Te, &c. then will

3 2 + 2 21, Cc. Ih
a 4a a?

like manner, if z be = ax+bx* ex + 4×7 +
ex?, oY then on the contrary will x be =
2 * + Ex 4 8abe—and— 1243
. 7 28 4

25 Lea 8 — 6 2,

42
Sc.

Sc HO LIV M.

Tus BE two Expreſſions for the Root x
being thus found, will now ſerve as
Canons for finding the Root of a propoſed
infinite Equation by Subſtitution.
For Example: It you would extract the

| 3 yt
Root of this Equation z . + —=— —

— 7 17
+=, Se. Then ſubſtituting i in the firſt Ex-
preſſion 1 for a, — f for 5, 5 for c, — 5 for 4

and + for e ; there will come out x 2 +2 2

+32? + #27; Ce. ens the Root of

f 7 . 22 N *
this Equation a „ n on 7
ec. will be x = 75 22* J. 1727

= N * 7 75 n zien
pn Sc. by 2 4 in the Expreſſion 1

for a, ** b A for 2 St for a,

APPENDIX.

S EG] TON HE.

Of finding the Fluents or flowing Quan-
tities of given fluxionary Expreſſions.

DEFINITION.

HE fluent or flowing 3 of a
given fluxionary Expreſſion, is that
Quantity whereof the given fluxionary Ex-
preſſion is the Fluxion. As the Fluent of & is
x, the Fluent of x 4-5 is x + 9, the Fluent of
x3 +) is xy, the Fluent of mx=»—*# is xm,
| an mM +
the Fluent of ax? + is. —- Mer
m +1 |

CO ROL. I.

4 H= e if the Ordinate PM 9) of a Curve Pfd. 1.

(or ſtrait Line) A Mat right Angles to
the Abſciſs AP (x) drawn into Pp G) repre-
ſents any given fluxionary Expreſſion; then
the Area of the Space APM will be the fluent
or flowing Quantity of the given fluxionary
Expreſſion; and, vice versd, the Rectangle un-
der the Ordinate PM, and # the Fluxion of
the Abſciſs AP, will be the Fluxion of the
Area or Space APM. For this Rectangle
may be taken for the Trapezium PM p, which

„

ud —
— »

APPENDIX,

is the real Fluxion of that Area: becauſe their

Difference is only the ſmall Triangle Mm n,
infinitely leſs than PMmp. And ſo it may be
rejected. (Axiom 1. Part 1.) |

II.

Hexen likewiſe it appears that the in-

COR O1.

verſe Method of Fluxions is a kind of

a general Way of ſumming up of Series,
SCHOLIUM.

H* RE we may obſerve, that a Fluent
| can have but one Fluxion; but on the

contrary, a Fluxion may have an infinite Num-

ber of Fluents. For Example: The Fluent
ax will have but only this Fluxion ax. And
if b, c, d, f, g, &c. be conſtant Quantities;
then will ax >b, ax Kc, ax d, ax ax dg.
or ax, an infinite Variety of other conſtant
Quantities, be each the Fluent of ax. W hence
accurately ſpeaking, the Fluent or flowing
Quantity of ax is not ax, but ax+p. p be-
ing any given Quantity, which may be equal
to any other Expreſſion whatſoever, conſiſt-
ing of conſtant Quantities. The ſame may be
underſtood of the Fluents of other fluxionary
Expreſſions.

As it is eaſy to raiſe a given Quantity to any
given Power; but on the contrary, any Root
thereof cannot be had in finite Terms; ſo like-
wile in the Buſineſs of Fluxions, it is eaſy to
find the Fluxion of any variable Quantity, or
variable and conſtant Quantities any how com-

pounded together. But on the contrary, the
Fluent of any given Fluxion cannot be had in

finite

am o– A # . A £m v©O£

APPENDIX.
finite Terms. For as in Algebra we have re-
courſe to Approximations in the Extraction of
Surd Roots, where they cannot be exactly
expreſs’d; fo in the Inverſe Method of Flux1-

ons we make uſe of infinite Series, where Flu-
ents cannot be had exactly.

P RA O B. I.

7 78 find the Fluent of a given fluxionary
Expreſſion.

Caſe I. 1. When fluxionary Expreſſions
conſiſt of no Powers of flowing or variable
Quantities, but Products of flowing Quanti-
ties multiply’d by Fluxions, aud alfo the Flu-
xion of every flowing Quantity that is in the
Expreſſion, as yx + xy, or xyz ZU ZN.
The Fluents are had by this Rule, which is
the Reverſe of the direct Operation, viz.

Iuſtead of each Fluxion ſubſtitute its reſpective
variable Quantity; and adding all the Terms to-
gether, divide that Sum by the Number of Terms.

So the Fluent of x + xy will be x5, and
of xyz + ZU +2z5x will be xyz.

When ſimple fluxionary Expreſſions, in-
volving ſome Power of the variable Quantity,
occur, viz. multiplied into ſome ſtanding
Quantity, as 2x, or 3xxx, or mx” x, or

—

1 . 3
75 x, or ax* , Which is the moſt gene-

ral Expreſſion of all of this Nature: The
Fluent will be had by the Reverſe of the di-

rect Operation. For as in the direct Operati-

on, any of the foregoing Expreſſions is found
by leſſening the Index of the Power of the
variable Quantity by 1, putting in the fluxio-

nary
2

N
*. 4

APPENDIX.
nary Letter , and multiplying the whole by
the Index of the Power of the variable Quan-

tity; ſo we come back again to the Fluent by
adding 1 to the Index of the Power of the

variable Quantity, ſtriking out the fluxionary

Letter x, and dividing all by the Exponent
thus increas’d by 1 : therefore in all ſuch Caſes
this is the Rule.

Strike out the fluxionary Letter, add Unity to

the Exponent of the variable Quantity in the
Expreſſion, and divide it by that Exponent thus
increas d by Unity.
Hence the Fluent of 2x#*, or 2 ** is *
For ftriking out x, and adding 1 to 1 (the
Exponent of x in the given Expreſſion) there
will be had 2x*; which divided by 2 (= the
Exponent increaſed by 1) the Quotient & will
be the Fluent of 2 xx.

In like manner the Fluent of 3 x* x will be

1 4 .

; of m will be a=; of = x = Xx

Bf ut = ——] , . — n

will be , of — :x!“ & will be x or
£5 OS NYE omen.

; of ax x. will be + . For in

x nn

this latter Caſe by adding 1 to the Index of

the Power of the variable Quantity, and ſtri-

king out the fluxionary Letter x, we hayc
| ＋ 2 | |

4 + which divided by IL (thenew In-
| mM

| * „
dex) and the Quotient is x „ vis.

che Fluent of ax 4.

APPEND IX.

Otherwiſe :

Becauſe this latter Fluxion and its Fluent
are the moſt general of any of thoſe of the
above-named Condition: ‘They may ſerve as
a Canon for finding of Fluents of ſuch ſimple
fluxionary Expreſſions as aboveſaid, by bring-
ing them under the ſame Form, and afterwards
ſubſtituting. For Example: To find the Flu-
ent of * Now this brought to the ſame

Form with the Fluxion ax” will be 1 x* x;
ſo that 4 is , „t, and u . Whence
putting 1 for a, and 1 for », and 2 for in

the Fluent =T; * and then we ſhall have

for the Fluent of 1 * x. 5
In ike manner the Fluent of 4y/x x (Ar)
will be 1* ( i).
The Fluent of / (ali) will be 4:7

The Fluent of , (I= 5) will be

— * T=)—x * For here a = 1,

n l, and m=—2,
The Fluent of 2, (Er 5) will b
e * of = will be

83 ) 1

Laſtly, The Fllent of 25 =1x d |
U will

we ſhall have the affirmative Values, vz.

thoſe aforegoing, they muſt firſt be reduc’d to

APPENDIX.

will be «“ =3x*=E&x1 = = Infinity. For
here a , m==—1, and =I.

Cafe II. If a fluxionary Expreſſion conſiſts
of any Number of ſimple Terms, ſuch as
“thoſe in N. 2. Cafe I. the Fluent thereof will
likewiſe conſiſt of the Fluents of the ſeveral
Terms of the Expreſſion connected together
with the Signs + and —. For Example:

The Fluent of * or * + 5 N

will be 3 +27. For (by N. 2. Caſe I.)

the Fluent of ** will be + , and that of
wg BS |

x will be «. And fo the Sum of theſe

Fluents will be the Fluent of the Sum of thofe

Fluxions.

So likewiſe the Fluent of „4 2 will
be 1 —2xf, 1
And that of 33az%—2*x+ Y will

be r* – x + 3ixt—x5,

Moreover, that of x * x 4 or that

3 | 4
of k —-x *x, will be- —2x 5
— 1
7

or —— ＋ 2K ; and by changing the Sign
ES: +4 x „ or x”—2x *.
That of *r + x* x, will be 5 3*— x.

Caſe III. When the Term or Terms of
a Fluxion are more compounded than any of

ſimple Terms like ſome ofathoſe of the fore-
going Caſes, by throwing the Expreſſion, or

ſome

APPENDIX. 2.7
ſome Part of it, into an infinite Series, ae-
cording to the Rules of the firſt Section: then
the Fluent of the Expreſſion, thus brought in-

to an infinite Series of ſimple Terms, may be
had by the Rules before laid down. For Ex-

b
ample: Let _ x be a given Fluxion.
Now this thrown into “1 NO Series (by
E. I. Hect. 10 will be —_—— =;
a* a
—= x, &c. the Fluent 4 which by Caſe 2.
b
will be 2 — =; of bg , &c.

In like manner the Fluent of xX=*» At. 1.

1 — . +34″ Ai &c. will be * – Art. 7.

The Fluent of —. 8 — 1 Art. 1.
1 +x*—3x

47x; —1 3 + 345125 &c. will be * * Art. 7:
— * 591 * &c. ;
Alſo the Fluent of Va +xxXx=*ax + * Art. 2.

II 8 *

22 . 1 . . ll 8

2 a af Fx? |
4 165 ri 240 Ts. 1524” nn
The Fluent of LL = will by

ah 1-4 5

a By

f a

AP PEN DIxX.

By the ſame Rules the Fluent of -F x
may be found: Where , e, F are any given
Quantities, and m, u, p the Indices of the Pow-
ers of the Quantities to whom they are affixed.

1m +1

For make = +7 =% =>

7
xe+f ‘=9, and n -n t then will

c 1 as B
luent be ond 21 amd PL
hos 2 S —1 fo” Th fx

r. g
$—} 38 “Fe? Se. The Let
ters A, B, C, D, &c. expreſſing the neareſt pre-

ceding Terms, viz. A the Term —B theTerm

—— X 755 &c. This Series when 7 is a
Fraction or negative Number will not termi-
nate, that is, the Fluent will conſiſt of an in-
finite Series of Terms. But when r is an af-
firmative whole Number, the Fluent will con-
fiſt of a finite Number of Terms, viz. ſo ma-

ny as there are Units in 7.
Otherwiſe.

This laſt Fluxion and its Fluent will ſerve
as a Canon for finding the Fluents of fluxiona-
ry Expreſſions any how compounded, not ex-

ceoeeding Binomials, by bringing them to the

ſame Form with the fluxtonary Expreſſion, and
afterwards by Subſtitution, as you may ſee in
the following Examples.

EX AMB L E I.

To find the Fluentof /2xx5. Thisbrought
to the ſame Form with the fluxionary Ex-
I preſſion

APPENDIX.
L
preſſion above, and then 1 x*x 0 + dun
xe+ Fx i Whence d=1, m==0, e=0, a,

1 =I = == Xa, o, r = 1,
Ig. And fo by ſubſtituting theſe Va-
lues for their Equals in the general Fluent a-

I 2 —

bove, and we ſhall have —Xax” x Mar

the Fluent ſought. And generally the Fluent

SR 13
of c will be .

ExXAmMmPyerL es II.

a*x . b
—_ rough t
e ght to the Form of the

general Binomial fluxionary Expreſſion will be

— — – — — 2

ad Xxx -& , AX X- ISK rr =
P N

dne -f. In the firſt Caſe, d=a*, nl,

e cc, f=—1, 2, p=—2, Whence
4⁴
1 =I, S=—I, D = r x that is,

„ t=0. And the Fluent will be
2-2

: „
8 „ ic: 20 200
L* * eint i, Tec xx
cond Caſe, d is a., m===3,e=—1, f cc,
aq

PE —2, p=—=2, =I, = —1, D=— —_—

_ I 4
| ———y . a”xx
X — I cx „ that — ———

And the Fluent will be 9 x — 75 that is,

alxx
S C

E x A M-

vill be 118

APPENDIX.

ExAMPLE III.

| Ex brought to the general Form

will be r T x , or 41 . x

In the firſt Caſe d will be =-,
1 t, eb, f=1, p. And ſo
12, Oc. Now ſince 7 is negative, I try

che other Caſe. Here d is 0 m 2 —
e=, f=b, un=—1, p= And fo 1=3,
eee a ax +4b

3 295 9 1 +bx 1 Or ” bxx
vxx+-bx, and f = — ber ex _ Fluent

. 122
ry + 17 1 2778,

3 2 Jacks Sd gs
chat is, __2obb 1 IC A

10 h .
Y aa bx.

ExaMPLE IV.
1

ah x, brought ta
$/C— ZAcexF zaacæx; 7

the Form as above, will be bx X (ma 7 X x.
And ſo db, Mm, ec, f= —a, n, p —

25 n
122, S, 2 — 22 * c—axF*, f=7. Then

the Fluent will be A,, Nhat is
n

3 . 124 D — 7 5

APPENDIX.

ScuoLliUu M I.

G*® NERAL Forms may be found for the
Fluents of Trinomial, or other more
compounded fluxionary Expreſſions, after the
ſame manner as the general Form above; and
they uſed for finding the Fluents of fluxionary
Expreſſions propoſed, which are leſs compound-
ed, or can be brought to the ſame Form with
them. But the Work this way being moſt
commonly extremely tedious, it will be beſt to
get the Fluents by (the firſt way of Caſe 3. a-
foregoing) bringing thoſe compound Terms to
infinite Series of ſingle Terms.
But compound Expreſſions muſt not be
thrown into infinite Series before we have tri-

ed to reduce them by augmenting, leſſening,

multiplying, dividing, c. the variable Quan-
tities: For by this means they may often be
brought down to ſuch ſimple Forms as come
under Caſe 2. aforegoing, and the Fluents of
them be had in finite Terms.

Here it will be of Uſe to obſerve likewiſe,
that if in radical fluxionary Expreſſions, the
rational Part of the Expreſſion, or that with-
out the Vinculum, multiplied into the Fluxion

of the variable Quantity, be the Fluxion of

the Part under the Vinculum, or in ſome given
Ratio to itz the Fluent will always be had in
finite Terms, by Caſe 2. aforegoing, and by
Subſtitution. SD

EIn. TFT.

A
1

HE Fluent of ax . or ax a N,

where 4x is the Fluxion of 4/ax— or
x 3

ax — aa will (by Caſe 2.) be j3x—aa* =”

34X — da Var 24. EXA AM-
\

i
o

3²

APPEND IX.

Ex AMY IA II.
NE Fluent of 2* K a- Or r
X 2 Fs where 2 ** is the Fluxion of

ax Fas”, will be jr Fas? = 2:xx + 2.4

3
xx r aa
Ex AML III.

T* Fluent of a+x xx will be

— =
ge :

E x AM P I. E IV.
HE Fluent of e e. Where the
Fluxion x x without the Vinculum is to
the Fluxion of the Quantity under it, vi.

2 * x as 1 to 2, will be zxx+Faxyzx Ta.

For make / Jas =2: then 22zz=2xx;

and {o VTi E. Whence the

Fluent (by Cafe 2.) i by Subſtitution
to the Aeta Fluent. After this manner

might the Fluent of the ſecond Example have

been found by putting xx ＋ aa = 27*,

Tux Flyent of x + a „ . Where
the Fluxion * , without the Yincu-

um is to the Fluxion of the Quantity con-

tained under it, Viz. * as 1 to , will

be T . For * r

mcm

then

APP E ND IX.

then will x” + 29 = 2″, and x” + a ==

u—

zuck. Allo zue = um N NN 4
=nmx”—xx2* . And dividing by 22 t,
there comes out & x, or & x=
\ . a,

„ Whence a”—”x x FA ==; and

I e I
the 1 21 7 = 22
—ä— ö•
x” +49 »

Here it may not be amiſs to give the fol-
lowing Table of ſimple Fluxions, (from Sir
Isaac NEwToN’s Quadrature of Curves)
whole Fluents ſtanding againſt them are ex-
pr in finite Terms. By which Means the

luent of a Fluxion coming under any of the

Forms of the Fluxions in the ſaid Table may

be had by Inſpection, or elle a very eaſy Sub-
ſtitution. Here 2 is the variable Quantity,
aud 4,e,f,g,h, 1 are invariable or given Quan-
tities.

X Forms

13 Forms of : |
k Flux1ons. [ Fluents. — 2212
1 20 35 D 2
; Flu. = ©. — 83882

: 3582
II A⁊Zu n — HE 82 Q
— — 7 Janet PL or — , 2 8 & – 0
98 -.; — = 3 808 8 8
. 2 — L fz* Flu. = 24 ; 2 — — — a . 8 2
= | yed-fz u 7 „R being = V5. 2 735

42 21 1— 2 * —4e4-0 * 2 = & = 2
6a [28 | eafZ” = —_ d RS. * 5 5 2 2 5
Tre e
“Fi; | 105, R3. o 353888
— 4. | 4&3 e. . eznio/ ni gp — 82 * 2 2
<C | =o TEEN L 9 a 75 2 8
| Flu. = 22 MN 2 5 8
—— 8 3
| 42 mn | Oo Q
| ————_— | Ae 272 at 028 2
IV þ e | Flu. =- — d R. | 3 $85
„„ ————= | 22
| | v7 Flu, = 182 —89z”4+6/*z2″ , — EA
„ aA W. . ITY. 8 A ;
| ER r .

— R.
l — ͤ— 1057 |
ts | — ma a—a_a___—w4

APPENDIX.

gles written by the late Mr. Cotes, Profeſſor of
Aſtronomy and Experimental Philoſophy in
the Univerſity of Cambridge; and publiſh’d
after his Death by his Succeſſor Dr. Smith,
under the Title of Harmonia Menſurarum.

Here the Labour of throwing Quantities
into infinite Series, which in many Caſes is
very troubleſome, and on account of their
converging ſometimes too ſlowly are not fit
for Uſe, is entirely avoided, and elegant Con-
ſtructions of the Fluents of Fluxions are had
geometrically with the Aſſiſtance of ample
Tables of Logarithms of Briggs’s Form, for
finding the Meaſures of Ratio’s, and of large

| Tables of natural Sines and Tangents for find-

ing the Meaſures of Angles. And from hence
may be deduced wonderful neat and comune
dious Solutions of all difficult Problems; ſuch
as the Quadrature of Curve- lind Spaces, Re-
ctification of Curves, Cubation of Solids, Cc.
wherein the Fluents of given Fluxions are con-
cerned. Several Examples of which I ſhall
give hereafter.

In the Treatiſe before us, you have two Se-
ries of Tables of ſeveral Forms of Fluxions at
the Head of each Page, with their Fluents
underneath them expreſſed in the Meaſures of
Ratio’s or Angles; the one Series compoſed
by Mr. Cotes himſelf, and the other by Dr.
Smith. In Mr. Cotes’s Tables, which I ſhall
confine myſelf to, they being ſufficient for
Purpoſes that uſually occur, z is the variable
Quantity, , e, f ſtanding. Quantities, » any
Index of the Power of 2, 9 any affirmative or
negative Number; and the Quantities R, S,
2, always being the three Sides of a right-
angled Triangle, whoſe Values are fer down

21 at

APPEN DI X.

at the Bottoms of each Page, expreſs the Ra-
tio or Angle, by the Meaſure of which the
Fluent of the given Fluxion is had: and if R
be the ſquare Root of an affirmative Quantity,
they expreſs a Ratio; always being that of
R+Tto S. But if & be the ſquare Root of
a negative Quantity they denote an Angle,
which Angle will be always that whoſe Tan-
gent and Secant are to the Radius as Tand & to
R, where the Sign of that negative Quantity
is changed into an affirmative one. The Fi-
or in the Column of each Page of the Ta-

les, with 0 at the top, are ſome of the afhr-
mative and negative Values of 0, againſt which
the Fluents of the fluxionary Forms a top

— 9 1

ſtand. As in the ſecond Form Bis Pe
| e+f 2″ x

When 9 is 2, the Fluent will be = 55 —

„ 8 i
5 THE * 5 2 When d is
o, the Fluent will be —d R RET And

when 9 is — 1, the Fluent will be —2-
Ear EL. And foof others, “**
Bur theſe F luents cannot be ſaid to be en-
tively known till the Quantity EI NN be

found: Which is called the Meaſure of the
Ratio of RET and & to the Module R, when

R is affirmative, or till the Quantity 55

rer ee,
= _ KEN or 5 8

APPENDIX.
in the ſeveral Fluents be found; the firſt of
which is the Meaſure of the Ratio of R+T

and S to the Module 775 R; the ſecond the

Meaſure of the Ratio of R-+T, and & to the
Module = Rz and the third of R-4-T and 5

to the Module —d &.: Or when N is nega-

tive, the Quantity R N 5 8 is = Meaſure
of an Angle, whoſe Radius, Tangent and Se-
cant are the reſpective Values of R, T and &,
to the Radius R as a Module. And the Way
to find this Meaſure of a Ratio or Angle to a
given Module I ſhall ſhew preſently ; but firſt
take the following Definitions, or Deſcriptions
of Terms, in order to clear this fo far, as that
a Perſon may in ſome meaſure underſtand the
Uſe of theſe excellent Tables, without being

at the pains of reading the Propoſitions in the

firſt Part of the Harmonia Menſurarum : which
are too generally handled to be perceived by a
moderate Capacity without much Application.

LH & wy. I.

T* Meaſure of a Ratio is any Quantity

proportional to that Ratio; that is, if
M be the Meaſure of the Ratio of A to B, or
of A and n the Meaſure of the Ratio of 4

to þ, or

75 then will M Fun: There-

fore equal Ratio’s have the fame Meaſure, If
one Ratio be the Double of the other, the
*Gre

—_— CD —— ͥͤ yew rn.

APPENDIX.
Meaſure of the former will be the Double of
the Meaſure of the latter: if the former be
the Triple of the latter, the Meaſure of the
former will be the Triple of the Meaſure of
the latter: if the Half, the Half, c. So
that if the Ratio be neyer ſo much increaſed
or leſſen’d by Compoſition or Reſolution, the
Meaſure thereof will be likewiſe increaſed or
leſſen’d proportionably.

Mor:over the Meaſure of a Ratio of Equa-
lity is o; and if the Meaſure of the Ratio of
a greater Quantity to a leſs be ſuppoſed poſi-
tive, then the Meaſure of the Ratio of a leſ-
ſer Quantity to a greater will be negative,

DE xy 1 n. II.

HE Numerical Meaſure of a Ratio is the

Exceſs of the Logarithm of a Number
expreſſing the Antecedentabove the Logarithm
of the Number expreſſing the Conſequent ;
that is, the Logarithm of the Quotient of the
Diviſion of the Antecedent by the Conſequent,
is the numerical Meaſure of a numerical Ra-
tio.

D ETI N. III.

T HE Trigonometrical Meaſure of an Angle
is the Quantity of Degrees, Minutes,
Seconds, &c. contained in that Angle.

D = r 1x. IV.

Tu E Module of Brigge’s, Flague’s, c.
| on is o, 434294481903, (Fc,
by -which if you divide 1, the Quotient
2,302585092994, Cc. will be the reciprocal

Module

APPENDIX. 39
Module of the ſaid Logarithms ; that is, the
Quotient of the Diviſion of any Quantity by

the firſt Module, is equal to the Product of
that Quantity multiplied by the reciprocal Mo-
dule. |

This Definition, or rather Account of the
Quantity of the Module of the Logarithms, is
ſufficient for my Purpoſe. Thott who are
not fſatisfy’d with it may conſult Prop. 1. and
its Corollaries, and Scholia in Part 1. Harmo-
nia Menſurarum. The fame may be ſaid like-
wiſe of the following Definition, which is a
Conſequence of the Prop. in the Notes of the
ingenious Dr. Smith contained in p. 94. at the
latter End of Harmonia Menſurarum.

D E- 1 N. V.

Tur Module of the Trigonometrical Ca-
non, or the Number of Degrees contain-
ed in the Arch of a Circle equal to the Radi-
us, which is to 180 Degrees, as the Radius of
a Circle to ⁊ the Circumterence, is $7* 17′ 44”,
or 77, 29777951 zo; and dividing 1 by it, the
reciprocal Module of the ſaid Canon will be
0,01 745 32925»

Pos:

PO find the Meaſure of a given Ratio to a
given Module, or to find the Quantity of
the Expreſſion R —_ when R is the
ſquare Root of an affirmative Quantity, and R,
T, 8, the three Sides of a right-angled Triangle.

Rule. As the Module of the Logarithms

0,434294481 903, Cc. is to the Module R –
the

the Ratio RT to S; ſo is the Logarithm of.
this Ratio to the Meaſure of it, having Ras a Mo-
dule, which will be equal to R or mul-

8
tiply the Product of the Logarithm of the pro-

poſed Ratio of R-+T to & into the Quantity
R, as a Module, by the reciprocal Module
2.302#85092994, Ce. and this ſecond Pro-
duct will be the Meaſure of the Ratio of
R+T to & with R, being the Module there-
of; and. it is equal to the Value of the Quan-
tity R — ſought.

Take this Numerical Example. Let R=8,
T=6, Slo.
Module of the Logarithm o, 43429448 1903

A FE Logarithm (of +) 0,1461280 : 2,69
16777 = Meaſure of the Ratio of 14, and 10

to the Module 8, equal to 8 8 ff

Or ſhorter thus: 1: recip. Mod. ILogarithm
2.3025850929904. : : Logarithm (of =
0,1461280×8 the given Module : 2,6916777
the Meaſure of the Ratio as before. a

This Problem may be ſolved without Com-
putation by means of the Sector of an Hyper-
bola, after the following manner. Ler 4G
be an Hyperbola, C A the Semi-tranſverſe Di-
ameter, and C the Semi-conjugate, and CE

an Aſymptote: and draw A Y parallel to C.

Then make R: T:: 40: AD. And if CA
x C3 be R, the Sector CAM will be =
R — The Triangle CA being equal
to 7, When 7 is leſs chan N; and the Trian-

1 gle

APPENDIX.
ple CBE equal to 7, when 7 is greater than
R.

; „

ro. H- N E if n be the ſtanding Module of

the Logarithms, and / the Logarithm
—— then will * =

of the Ratio

Xx:
mM

Coror. II.

th RE
Her xcs — = | 2. m being any
m \s S

4¹

whole Number, and another. This follows

from the Nature of the Logarithms.

P R Oo B. II.

I 2.70 find the Meaſure of an Angle, whoſe
Radius is as R, Tangent as I, and Se-

cant as 8, to the Quantity R as a Module : Or to

find the Value of this Expreſſion R [<<

M ben R is the ſquare Root of a negative Quan-
tity, and ſo impoſſible; all of them being given
Quantities.

Rule. Firſt ſay, As the Value of & to the

Value of 7, or as the Value of R to the Value

of $; ſo is the Radius of the Tables of the
Tangents and Secants to the Tabular Tangent
or Secant of the Angle to be meaſur’d. Againſt
which ſtands the Granitty of that angle in

trical

Degrees, Minutes, & * being the Trigonome-

. ett. ii —

trical Meaſure of it. Then as the Module of
the Trigonometrical Canon 57. 17”. 44”. or
$74295779F130, is to the Trigonometrical
Meaſure of the Angle juſt now found, fo is the

Module of that Angle, viz. R, to the Meaſure
of the ſaid Angle, having R as a Module; which

is equal to the Quantity & Z. or mul-

tiply the Product of the Trigonometrical
Meaſure of the Angle into the Module R by
the reciprocal Module of the Canon o,o 1747

And this ſecond Product will be the
2 of the Angle to the given Module R.
Here follows a numerical Example. Let
R be =16, T=12, and S 20. |
As 16: 12 :: Radius 10000000 : 7500000 =
Tangent. |
And as 16: 20:: Radius 10000000: 12500000

‘=Secant.

Againſt both which in the Tables you have

1. 6”. for the ‘Trigonometrical Meaſure
  of the Angle.

Again: Module Trigonometrical Can. 57.
2957795130: Trigonometrical Meafure 36,
00033341 : : Module 16: 10.4330985 = Mea-
ſure of the Angle, whoſe Radius is as 16,

Tangent as 12, and Secant as 20, to the Mo-
dule 16; or equal to 16

16 ＋ 12
8

Note, By uſing the reciprocal Module,
0174532925 wall be ſhorter than by uſing the
direct Module: And the Operations will be {till
very much abbreviated, by uſing the Loga-
rithms to find the fourth Terms of the Pro-
— in the aforegoing Problems as well as

Tot

66
APPENDIX.
This Problem may be ſolv’d likewiſe by

means of the Sector My Circle or Ellipfis,

after the og manner. Let CA, CB, pc. 3.
the Semiaxes of the Quadrant of a Circle or
Ellipfis 4B, drawn into each other, be equal

to 2 R. Draw 4 © parallel to C B, and BG
parallel to CA; then make &: T: CB: AD,

and draw the right Line CMDG; then the

Sector CAM will be equal to R a —, Ds.

when R is greater than 7 =Triangle CAD,

and the Sector CB M = R nk . When R

is leſs than 7 = Triangle CB G.
Note, When & happens to be the Square
Root of a negative on, and ſo impoſſi-

ble, you may change the Sign, and every
ching will ſucceed right. me

COR O.

I’ be the Angle, whoſe Radius, Tan-
gent and 37 — are as N, , C, and m be
the Ending Module of the Canon; 5 then

RAFT will be = .
8 M

SCHOLIU u.

P=cavsr R, T, S are always the three
Sides of aright-angled Triangle; there-

RT; $
fore RE EN FL ſuppoſing T to

be the Hypothenuſe. For from the Nature of

a right-angled Triangle T N NN: and
ſo 37 N And

„„ K

1 o
p 7

7:3
z 1
3
i
s
: F
BJ
KL
N

Ar PEN PDITX.
of; oF

is R D — –
( — n And mul

tiplying croſswiſe, you will find that
Fs TERxX/ARTISS— —Ris Ss.
The Sum of two Numbers multiplied by their

Difference being equal to the Difference of

their Squares; therefore EI | „

n Conſequently R | — Rr x
This evidently appears alſo in Fig. 83. For
let the Hypothenuſe 45 =T, the Perpendi-
cular CB R, and the Baſe AC=S. Con-
tinue out AB till BD=BC=R, and make
BE=BC; then AD=R+T, and AE =
T—R. If a Circle be deſcribed about B with
the Radius BC, it will paſs through E and D,
and AC will touch it in C; therefore (by |

| Prop. 36. lib. 3. Eud.) FT (88) is = AD

x A E== Tr, and ſo . =
S

SECTION III.

U/e of the inverſe Method of Fluxions

in the Quadrature of Curve- lined
Spaces.

PRO B.

Te T’* /quare à given Curve-lined Space.

Having gotten the Equation expreſſing the F 16. 1.

Relation of an Abſciſs AP (x) to its correſpon-
dent Ordinate PM )) at right Angles to each
other, find the Value of y, which multiply by
x, and the Fluent of the Expreſſion thus ari-
ſing will expreſs the Quadrature of the inde-
terminate mixed-line Space contained under
the Abſciſs A, Ordinate PM, and Curve
AM. And if the Abſciſs A4 be determinate,
vi. = to a given Quantity a; and according-
ly the Curve itſelf ſo likewiſe: Then by ſub-

Goring a for x in the Fluent aforeſaid, there
will come out an Expreſſion for the Quadra-
ture of the determinate mixed-lined Space.

This will be plainer by the following Exam-
ples. |

But if the Area CDEF contained under F 1 c. 4

two Curves or right Lines DE, CF, the right
Line C D, and the Part EF of any right Line
AE drawn from a given Point 4 in the right
Line DC be ſought, draw fe infinitely near
AFE, and from the Centre A — ar |
mc

Fic. 5.

APPENDIX,

ſmall Arches Fp, £9: then from the Nature
of the Curves find the Area of the Quadrila-
teral Figure FEgpz which is equal to the
Difference of the little Sectors A Fp, A Eg, or
equal to; AE x Eq—rAFxFp: and this
will be = Space FEef, being the Fluxion of
the Area COE; the Fluent of which will
be = ſaid Area. Examples of this will be
ſhewn hereafter.

i EX AMP LE I.

2 find the Area of a Triangle ABC,

Draw A D perpendicular to one of the Sides,
as BC; in which aſſume any where between

A and D the Point P. Thro’ which draw

the Line IN perpendicular to 4D; and let
n be infinitely near and parallel to MN; and
draw Mp, Nq perpendicular to MN. Now
the Rectangle M N 9p, viz. under Mq, or Ng,
or Pp, and the Ordinate MN will be the Fluxi-
on of the indeterminate Area AMV. Which
muſt firſt be found thus:

Call the variable Quantities A P, x; PH, y;

and the given Quantities 4 D, a; CB, b. Then

becauſe MN is parallel to AB. AD (a): CB
(b):: A P(x): MN()). Whence y =. But

Pp(=Mp=Ngq) =x. Therefore the Flu-
xion of the indefinite Arca AM will be
bx.

** The Fluent of which (by Caſe 2.) will

berg Area AMN. Now if for AP (

you ſubſtitute AD (a), there will ariſe _

(Aa TAD. CB. And this will be the
| | Area

—

APPENDIX. 47
Area of the whole Triangle 4C B; which we
know likewiſe to be true from the Elements of
Geometry.

Here it may not be amiſs likewiſe to ſhew F 1s. 6.
the way of finding the Area of a Trapezium,
GPCB, having two Sides PC, G parallel,
and the Angles B, C right ones, by the Inverſe
Method of Fluxions; tho” the thing is much
ſhorter found out by the Elements of common
Geometry. But it is delightful to perceive
the ſame Truth ariſe from ſuch very different
Principles.

In order to this, continue out CB and PG
to meet one another in A. From 4draw Am p
infinitely near 46 P, and with the Diſtances
Am, Ap deſcribe from A the ſmall Arches
wr, pu. This being done, let AB =a, BG
=b, BG =x, A)). Now the Triangles
ACP, pPn are fimilar; becauſe the Angles at

and C are right ones, and the Angle P is

common to both. Whence 4G (y): 4B (a)

: Gm (&%):my= * This multiplied by + Ar

2 — k – o .. WS, mn i > nee eb mn I 3 —— — —-—-——¼ —
— — wx — =

” SS. BY. & « £ i, Bo. ali.d

— |

(= * AG = 2), and the Product — is e-
6 35

qual to the Area of the little Triangle Ar m or

An: theſe differing only by the Triangle

it Eur being infinitely leſs than either of them.
1 Again; Becauſe the Triangles AB E, AC P

„

of are ſimilar; therefore 4B (a): AG(y):: BG
11 | (b):GP =22 . Conſequently AP = L+ [7

i And ſince Am, AG, and 4 P, Ap differ from |
x) BE one another by infinitely ſmall Quantities on- 1
„Hh: therefore Am may be taken for 4G and 1.
— 4595 for AP. This being granted, and the |
4 WE Triangles Amr, Ap n, taken as ſimilar, we 1
he ſhall 4
ca f

48 APPENDIX.
| J. 4
ſhall have 4C(9y):rm =: 42(L+)):

b + ; which multiplied by; A P
. bbx 1h; L145
(Y,, andthe Product 5 +6bx + tax

ol | is equal to the Area of the Triangle Apn or
1 APp. From which if you ſubſtract the Area

of the Triangle A & (Z) before found, the

‘Þ Remainder b. + 24 will be the Area of the
\ 1 Trapezium mpnr, which may be taken for the
|} Area of the Trapezium , being the Fluxion

| | of the Trapezium BCPG. But the Fluent of this
| 14 bbx __2atx + bbx a xb
‘} r 4

1 X 2 Bu ELL is = PC. Since from the

&-. a
Similarity of the Triangles 4B G, ACP, AB

I} eG ::AC (a+0): CP n τ⏑ .

. Therefore C FPCxiBC = Area of the

Wh | Trapezium GCB. Which is a known Truth

“nn from the Elements of common Geometry,
EX AMG L E II.

7 find the Area of (or to ſquare) the com-

| mon parabolical Space ABD.

Fic. 1. Call the given Quantities J D, a; B D, h:
and the iavariable Abſciſs AP, x; and the cor-
reſpondent Ordinate PM. y: and let Pp be =
x. Now the Equation expreſſing the Relati-
on between AP (x) and PM () is px =.

Whence y = N ng And ſo the ſmall
Rectangle PM np, equal to the Fluxion of the
indefinite Space AMP will Ee V xx; that

APPENDIX.

15,9 X =p x. The Fluent of which (by

2 7 2
Caſe 2. Fett. 2.) will be = (DD =
21/x572)2xy, equal to the indeterminate Space

AMP; and by ſubſtituting à for x, and & for

yi in this Fluent; ; and then we ſhall have 345

=z3APxPM,; that is, the parabolick Space
is ko the Rectangle under the Semi- ordinate
and Abſciſs, as 3x. to xy, or as 2 to 3.

ExXAMP:Li’E III.

Fas ſquare Parabole’s of all Kinds.

If AP (x) be the Abſcils, and PA Y) the Fo. 1.

correſpondent Ordinate, then the Relation be-
tween the Ordinate and Abſciſs of a Parabola
of any kind will be expreſſed by this general

Equation pm x® _ Whence p#x* hy; and
o the Fluxion of the Area will be 9 * =

15 . The Fluent of which (by Caſe 2.

4 *
Se. 2.) will be P
) ES N

becauſe þ Pa 2 5. And ſo any Paraboloid is
to the Rectangle under the Ordinate and Ab-

ſciſs, as t f
ls, 5 1 9 xy, or to u

EX AMP E’s IV.

7 10 ſquare the Segment of the Parabolic

Space PM N Q contained under the Or-
dinates PM, NQ; the Part PQ of the 46-

ſeiſs, and the Part M N of the Curve of the

Parabola.
Z | Here

F 1c.

Which is 1 — == 12. + pxX y/pa x

APPENDIX.

Here let 4 P a be invariable, and let P
be the Beginning of the variable Quantity
PO x. Alſo let V be = y, and the Pa-
rameter =p; then A will be = a + x,
draw 29 parallel and infinitely near to NO.

Now the Nature of the Curve is AP,

xp=N®O ; that is, pa T D , and
pi +px=y. Therefore 2 Nx A =I

y/ps +px=1x*Xpa+px’x is the Fluxion

of the Area, the Fluent of which may be
found (by the latter way of C/ 2. Seck. 2.)
or more eaſily thus: Let Vpa +px =2z; then
will pa + px =2z, and px=222, and x =

£3 22 2
: therefore y 3 is = —_ : the Fluent of

7 14 — *
x y/pa Nr =3A Q NO.

Bur ſince in the Point Px is So, the Space
PMN will vaniſh; and ſo making x = 0

in the Fluent juſt now found, and the Terms

x and px will vaniſh; fo that the Fluent will
be now zaypa. Which ſhews what is to be
added to the Fluent, that ſo the Space MPN9
may become o in P; and conſequently the
Quadrature of the ſame had. In this Caſe
2ay/p4 is to be taken from it: therefore the

Area of MPN Q will be = TV + px

jay A. N- AP P.

Otherwiſe.

Draw np infinitely near to MP, and let
Da be invariable. Let the Beginning of
x be in ©, and let O be = x, PM=y;
then will A be Sa -x.

Now

APP E ND IX.

Now from the Nature of the Curve

AD — R pP =PM that is, pa — px =
. Whence /p —px=y: and ſo Mp Pp
the Fluxion of the Area will be y x = X4/pa—px.
The Fluent of which may be found as before
thus: Make pa —px =zz. Then will —px

be 222, and conſequently x = —

a
Therefore y x = — —— The Fluent of which
will be — = — — =2x—a
„p — p.

Now to find what is to be added to the Flu-
ent to give us the Quadrature of the Space
PM NV, making as before x =o in the Flu-
ent, and we have — 3 % pa. Whence it is
manifeſt, that if ＋ 2 2 added to the Flu-
ent, the Space PM N will be =j}aypa +
EX — @XPa——Px.

Conor: LL.

Tux Space PMN9 = 4N’9 — AMP,
. But in the firſt way 41 48

x 0 N= 3a +xxpa+ px; and AMP 45
x PM==34ypa: therefore PMN © ==4 AD
XN; APM. Moreover in in the latter
way : AND = A LON YB. O, and
AMP ==; AP x PM=j a—x x pa —p IX 2
therefore S NAP is TAN V-14 5
„PM here alſo.

Z 2

Co ROL

APP E N DI X.

onen. II.

Ir the Curve be not deſcribed, and the

Equation expreſſing the Nature of it
be only given, and fo it is uncertain where x
is to begin; it is evident from the Solution a-
bove, that o mult be ſubſtituted for x in the

Fluent; and ſtriking out all the Terms affected
with x, what is left muſt be added to the Flu-

ent with the Sign changed, and the whole
will be the Quadrature fought. ©

rn .

To ſquare a Curve expreſſed by this #2
tion x TAN +@& A + af =

; x+
Since = — Z- + — — Ta; th
Jy S+S+Z ++ z the

Fluxion of the Area will be ) x =
10 3 and the Fluent will

St 2 +E, ＋ 2 Ken.

be—

EN Amr VI.

O ſquare any Curve, whoſe Nature is

= expreſſed by this general Equation y =

õX 71 +a. 2

Becauſe y is =x* x + a 2, therefore the
Fluxion of the Space ſought will be y x =

re” the Fluent of which will be eaſi-

2
m

V found by making a T4 2. For then

—

The Fluent of which will be 6 3

APP END IX.
„ +a=2″; and both Sides of this laſt Equa-
tion thrown into Fluxions will be x = , x.
Whence * man, the Fluent of which
„ — — will be 5 . r a.
Now in order to know whether this be the

true Fluent, ſuppoſe x = 0; then this laſt Ex-
mW

3 a. Which muſt

preſſion will become

[
therefore be taken from the Fluent, (by Cor. 2.
Example 3.) ſo that the true Fluent or Qua-
drature of the Curve will be _—

WPI T 6

—

EXAMPLE VII.

12 ſquare Hyperbola’s between the Aſym-
_ ptotes of all Kinds; or, which is the

ſame, to find the Area of the interminate $ Pace

HMPAS, or h MPs. The former contained
under the Abſciſs AP, Ordinate PM, Ahm-

ptote A 8, and the Part MH of the Curve of

the Hyperbola; and the latter under the Ordinate
PM, the remaining Part Ps of the Aſymptote,
and the remaining Part Mh of the Curve of the

. Hyperbola.

In theſe Curves the Relation between AP(x)
and PM ()) is expreſſed by the following E-

quation aun = m x.
= n
Whence a+ Y] = ym, anda & 7
—

N
Conſequently PMxXPp=yx is = 7 *
m-

// at (7

F 10. 8.

54 APPEND:
wenn Eu PTE pg np

xy. If m be greater than , then we

m—u
have always the Quadrature of the Intermi-
nate Space HMPAS: the fame being =

— APxPM: If m be leſs than », then

HM—NR

7M
M—N |
Quadrature of the Interminate Space þ MP s
lying on the other Side of the Ordinate PM.
But when n, neither of theſe Spaces can
be ſquared, they being in this Caſe both infi-
nite. For if „ , then m 2, 11;
and ſlo HM PAIN (y/xfy) = xy. If x % ga, then m =4, ; and ſo AMPAS Sgr. If My Sa, then , z; and ſo — xy will be the Quadrature of the Space b MPs. If x3 a,, then will » =1, =,
and fo — 3x), that is, * = MSS. But

will x3 be negative; and ſo we have the

when n, then is : therefore the

nn
Numerator is infinite in reſpect of the Deno-
minator.

Aer VIII.

TO ſquare the common Hyperbola between
its Aſymptotes;, or, which is the ſame
thing, to find the Area of the Space C MP
F c. 8. contained under the Ordinates Cc, PM, the Part
CP of the Aſymptote, and the Part cM of the
Curve of the Hyperbola.

Let the given Quantity AC be , and
let the Beginning of x be at C. N
; Then

APP EN DI X.

Then the Equation expreſſing the Relation
of AP (b+x) to PM (y) will be &=by+xy.

And fo © = yz and xy the Fluxion of the

Ares will be .
b+x

F luent of this (by Call 2. Seft.1.)
. 1 A 2 ax ‘ AM * of
will be * * * * J Se. == to

the Area Cc MP. And if you ſuppoſe a =

=1, then will x —3x* + ix* — 3×4, Sc. be

= Area aforeſaid.
Otherwiſe by the Meaſure of a Ratio or Angle.

. « +a

The Fluxion x 2 2

1ſt Form in Mr. Cotes’s Tables. For making z==x,
=1, i, d=, eb, f=1; then will

may be referred to the

2 Me And the Fluent ſtandin
e bs –
againſt 8==1, * A= 27 = 4C
N

. to the Meaſure of the Ratio of 4A
and AC to AC for a Module, which you
may find by Art. 9. = Area of the Space
CSM; and if the Aſymptote AS be not

perpendicular to As, and ſo the Ordinate PM
parallel to it, not perpendicular to the Abſciſs
AP; the Meaſure of the Ratio of AP to
PM, with the Parallelogram AJ Cc as a Mo-
dule, will be {till the Area of the Space CcMP.

This is demonſtrated ſynthetically Page 12.
Part 1. Harmonia Menſurarum, |

Exams

56 APPENDIX.
EXAMPLE IX.

Fic. 9. 26. 78 find the Area of the Space AC MP

contained under the Part AP (x) of an
10 infinite right Line perpendicular to the Axis of
1 the Hyperbola, the Line AC (a) being the Conti-
| nuation of the ſame Axis, and any Line or Ordi-
nate PM (y) parallel to AC.

Here aa + xx =1y; and fo y=y/aa + xx.

Conſequently ) y/aa + xx. The Fluent
of which (by Caſe 3. Sect. 1.) will be Sa

64 — 1122” ‘ 11524”

AJC ſought. Which gives the Quadra-
ture of the Segment D CM of the Hyperbo-
la, by ſubſtracting the ſame frem the Rectangle
ADMP (x9); and if @ be = 1, then will the

3 5 7 9
. .
T7 20 Tim 11724 &

Otherwiſe by the Meaſure of a Ratio or Angle.

The Fluxion x V Þ xx comes under the
fourth Form of the Tables of Cotes. For if

, l, 125, I, a, cl, then will

42 /; + be V + xx. The

Fluent of which is 4 P + 4,4 R =
n y

and making P (= J= yaa + xx,

The ſaid Fluent will be-

come

APPENDIX,

come = V xx += . =z AP
«PC++ TT +7 C = Area ACPM;

that is, APxPC plus the Meaſure of the
Ratio of 4 pc and A C to 1 AC, as a
Module will be the Aid Area.

The Quadrature of the Hyperbolick Space
AMP may be had thus, as laid down by
Mr. Cotes in Harmonia Meuſurarum.

Make the Semi- conjugate Diameter CH, Fo. 10.

the Semi-tranſverſe C Aa, C P=x, PMA;
then from the Nature of the Curve we have

= =y; and ſo * „aa aa = Fluxi-
on of the Space AMP.

Now making 4 ==, Z==x, o, ij==2, e
—aa, i; the Fluxion of the 4th Form in
the Tables of Mr. Cotes, viz. dar- K 40 —1
e will become= 5 . | And the

Fluent of it againſt $=>, is = 4254. 7
— And making P ( e) —

as, R(=f)=1, E —

Ve — da, & (= VS )=< ; the ſame will be-

„ e ah 2
xy 450
If. 2 by ſubſtituting y for its Equal

b AA

5 8 APPENDIX.
= Daa. And Art. 11. where the Ratio
of x +y/xx —aa = to a, is ſhewn to be equal

to the Ratio of a to „E .

4 is the Fluent of the
9
5

Fluxion 2 rad. Which may be thus

conſtructed. :
AfſumeCF:CA(a)::PM(y):CB(b); that

is; make FN. And aſſume CG: CA (a)

::CB (b) : PM ()); that is, make c
Then if PA be taken equal to the Meaſure of
the Ratio between CA (a) and FP (=x — 20

to the Module 75 that is, if PH be taken

= 9 M
y ab

„& —

And you draw the right Line

y a
MA; the right-lined Triangle HM is =

Space AMP. For it is ==X x —-

[== 2 Ü And to find the
LT I @

Quadrature of the external Hyperbolick Space

CAMP after the ſame manner, we may pro-
ceed thus.

. Let AC, CB be Semi- conjugate Diameters,

—

— — ECT

— — —

APpPENDIX. 59

ada is =y, from the Nature of the Curve.

T his Equation thrown into Fluxions, and there
bbax . _ blax _ bux

will ariſe — Y — — — 4
au han aby ** – 4 ay —
But ) x X is = Fluxion ofthe Hy-

ay Xx 4a

perbolick Space CAMP.

Now making d=2, Z=x, uz, =I, e=
5 : : 201 1
— aa, f=1 ; the Fluxion -— == ofthe
ſixth Form in the Tables will become

S*
And the Fluent of it againſt f,

ay e

will be Z 5d P— 4 4 RE — Z. Whence
275

thu! P 195 —)= 1 RI

= I, EF * en, M’s.

the ſaid Fluent will become 2 _ = v/x aa

ee 4

—
qua
| _

ſubſtituting * for Wt and Art. 11.) =

bets
Fluent of the Furien ======. The Cov-
ay NA 4d

ſtruction of which may be thus.
Aſſume CF: C Ft (b)::PM(x): AC * > 5

is make CF=E, and CG: C ()::
wer : 7

Fic.

APPENDIX.
O PM (ﬆ); that is, make CS L. Then

if CH be taken equal to the Meaſure of the
Ratio between BC (3) and PF 29.

(W hich is equal to the Ratio of a to ͤ 5 )
tO the Module CG (=) b the ſaid c H will
be = ba

And you draw the right Line

MH, the right- lined Triangle is equal to the
Space CAMP. For it is = 7 * —— 22

—— 2 , the Fluent of the

given Fluxion.

IF; = ſquare the Circle, or, which is the
ſame thing, to find the Area of any He-
mi- ſegment AP M thereof.

Make AB=1, IP x, PM= =. Then
from the Nature of the Circle APx PB =

PM 1 that is, 533 5 and ſo = xx.
Conſequently x y =/x—xx = PMx Pp is the
Fluxion of the Area AMP. The F luent of

which (by Caſe 3; Seck. 1 will be T* — ke

— A* — „c. = Area AMP, or 45
into 3 x —x ö +.

E Others

APPENDIX. 61
Otherwiſe.

Tf the Radius C M be = a, and CP be ſup-
poſed = x, PM); then from the Nature
of the Circle CM = CP +PM ; that is,
aa =xx + yy, yy = aa — xx, and y=y/aa—xx.
Therefore xy =xy/aa — xx is the Fluxion of
the Indeterminate Space PMDC, and the

: a * x5 *
Fluent of this, viz. a x „

i- Sc. = Area PMDC. Now if
a be = 1 = x, then this Series will become
I —7 —#. rein, Cc. = equal to the
Area of the Quadrant 40 C. And four times
this will be the Quadrature of the whole Cir-
cle ADB A; or if the Diameter be 1, the

ſaid Series will expreſs the Area of the whole
Circle.

Otherwiſe ;

Let AE the Tangent of + the Arch AM F 1s. 13.
be = x, the Radius C =. Let AB be
the Tangent of the Arch 4M. Draw the
Secants CE, CB, and the Sine MP of the
Arch AM. Let pm be infinitely near PM,
and from the Centre C draw the Secant Ch
thro’ the Point ; alſo from the Point draw
Mt perpendicular to pm, and from the Point
B draw Bs perpendicular to C6.

Now we propoſe here to find the Area of
the indefinite Sector ACM; the Fluxion of
which, being the little Sector MC, mult
firſt be found thus:

Firſt,

APPENDIX,
Firſt, the Tangent AB of the Arch BN

will be = _—=-. Now becauſe the Angle

1—

AC is biſected by the right Line CE, there-
fore AE (H: AC (i) :: ER (== ) CB=

I —XX

1% Allo becauſe of the ſimilar Triangles

I—x*

ACB,PCM,CB (LES, ): AB(LE.,)

2X 1 + x*
::AC1): PM= rs 5 and C 8
:AC(1):
a the Fluxion of which will be .
IT D

Pp or Mit.

Again, the little Triangle Mt right-
angled at 7, will be ſimilar to the right- angled
Triangle CMP, the Angle :- 1m being =
Angle PMC, and the Angle tm = Angle

PCM, as is ‘eaſy to prove. e MP

(=>): 40 0::M (== =): Wk

2.x

And ſo + MC(1)x Mm = “I
= Area of the little Sector MC m, being the
Fluxion of the Sector AMC. The Fluent
of which will be x —Fx* +5%* — 3x” +5),
Sc. Area of the Indeterminate Sector AMC.
And when the Tangent AE (x) of half the
Arch 4M becomes =1 Radius, then the
Sector ACM will become a Quadrant; and
the Series aforeſaid, expreſſing the Area, of the
_ will be 1 =3+F—5+5—# +75,

APPENDIX. 63

c. and if the Diameter of the Circle be==1,
the whole Area will be expreſſed by that Se-
ries.

The ſaid Series may be found ſhorter thus:
If AB be =x, then will BY , and CB
y/1 + xx. Now the little Triangle BS right-
angled at &, is ſimilar to the Triangle ABC,
the Angle 4 BC differing from the Angle 5
only an infinitely ſmall Quantity, and ſo they
may be taken for Equals; therefore CB VI + xx)

= :4C(1)::Bb(%):Bs = 8 Moreo-
‘Y VII xx

ver, ſince Bs is infinitely ſmall, CB and Cs
differ from one another only by an infinitely
ſmall Quantity: therefore C (VI +xx): Bs

SS . — —
C cs =) :: MC (1): Mm | A W hence
the little Sector MC = —=— = Fluxion
2+ 2.xx

of the Area of the Sector AMC. The Flu-
ent of which will be 3 x -& + 77x —7 x7
rr, Sc. = Area of the ſaid Sector: ſo
that when the Sector ACM is an eighth Part
of the Circle, viz. when the Tangent AB (x)
= Radius AC is = 1; then the aforeſaid Se-
ries will become; — ; + 75 — 37 + #7, &c.
which doubled will be 1 — 3 ++, Oc. = Area
of the Quadrant as before.

EXAMPLE XI.

Te ſquare the Elliptic Space : or, which is
the ſame thing, to find the Area of any

indeterminate Elliptic Segment ACM P contain’d F 1 6. 14;
under the Semi- conjugate Diameter A C, the Or-
dinate

ä 2 hd

—U— : ˙n—— — 2 ———̃ ⁵⅛ ͤ⅛w ,

FI. 15.

APPENDIX.
dinate PM, Part of the Abſciſs AP, and CM
Part of the Curve of the Ellipſis.

Call 4 C, a, AB or A D, l, AP,x, PA.;
then from the Nature of the Ellipſis MH =

7 | a4 a

and ſo y =” aan. Conſequently 5 *

Vaa – xx will be the Fluxion of the Space
AJC PM, and the Fluent of this will be þ x —
7

. — 0 Sc. = Area AC PM.
64″ 404* 1124

Now if you put à for x in this Series, it
will become a4b—#ab—#F-ab—+>;4b, &c.
= Area of the Quadrant 4 CD of the Ellipſis.
And if a be = Axis B D, then this laſt Series
will expreſs the Area of the whole Ellipſis.
And if /a be = 1, then the Area of the El-
lipſis will be 1—#— #—++r — 7755, (9c. and
therefore an Ellipſis is equal to a Circle, whoſe
Diameter is a mean Proportional between the
Tranſverſe and conjugate Diameters of the El-
lipſis. And fo an Ellipſis is to a Circle of the
ſame Diameter with the tranſverſe Axis, as ab
to 4*, or þ to a, viz. as the conjugate Axis is
to the tranſverſe Axis.

Otherwiſe.

N
To find the Area of any Sector CMA of the
Ellipfs. A

Let CB be the Semi-conjugate, and

C 4 the Semi-tranſverſe Diameter, MP a Se-
mi- Ordinate. Now draw np infinitely near
3

APPENDIX. 65
MP, join the Points C and M, and mz by the
right Lines C M, Cm; and from m draw the
ſhort right Line m H perpendicular to MP,
cutting MP in J, and CM in ; as alſo the

little Line A perpendicular to CM continued
out.

This done, let AC a, BC=1, AP x,
PM ()), CM = u, and C PZ. Now the
firſt thing to be found muſt be the Area of the
ſmall Triangle C Mm. Thus: Becauſe the
Triangles CPM, H 1M are ſimilar; therefore

PMO): CP (Z) :: M0) :1 [Iu W hence
(ſince Pp = Im=x=—2) Hn=2—z. A-
gain; becauſe of the ſimilar Triangles CMP,

HKM. 0 PM (y):: Hm (2=z):

mK = 21 2. Which — into half the

Baſe CN 00 z and then the Area * the fluxio-
nary Nn C Mm will be = — 1-25; and

throwing the Equation of the Curve ——

3 44
=yy.into Fluxions we have — _ =.

Which ſubſtitute i in for y, and there

. — 8 — — 4 8
ariſes— 2 _ 2. 8 In which

6
— —

zaay 2 24a)
ſubſtituting aa — 22 for its Equal aayy, and

— IIS 2a a
it will be —:

————

zaa) 2ay za

ſince x is =— z. And again ſubſtituting
v/ 24x==xx for ay; and the Triangle Cm will
B b be

————— — —— —ͤ —— — — — — ñ— ———— —’

2
LD Bp ů — . ATE Yo ner wen. — LU LE ene <>

—
—
— ͤ— —— — a

66 APPENDIX.

4 2

Do. 16 be= =PFluxion of the Sector 4CM
ZV LAX XX 2
of me Ellipſis. And if you make DT — xx i

Zaun

1 75 then will x be found = “Ia and x = 4

and by due Subſtitution — =

I un 2y/ 24X — x
an an
— The Fluent of which is a2 —
I un 3
an; an

_ 7 7 „Sc. = Sector of the Ellipſis

And after the fame manner you will find
the Sector of the Hyperbola (Fig. 16.) to be

ab e +, Sc. and the Sector of

5
the Circle will be =» — 1 += —— 15 Se.
and this becomes 13+ — 55 Ge. by ma-
king #=1.
This may be done Gmething ſhorter, thus:
Lc Draw the Tangent 4 D, and continue out
is, © CMrtocut the fame in D. “Draw Cd infinite-
| by near CD, and from C deſcribe the ſmall
Arches u, De. Make C BY, C A==a, AD
=x, CD=y, CE P=z; becauſe the Triangles
CAD, Ded are ſimilar, the Angle A differing
from the Angle D-only by an infinitely ſmall
Angle, which may be rejected, and the Angles
at e, and A right ones. Therefore CD (3):

AC(a)::Dd (8): Der 77 And becauſe of
the ſimilar Friangles C =) CAD, AC (a):

CD(3)::CP(2):CM 2 . Likewiſe ſince
the Sectors C De, C Mn are ſimilar, CD 63 5

Ap PE ND IX.
Del YOM Y) ef. Now

J
WOE 7 WO 2K. +
2 2 2.4

rea of the Triangle C Mm, being the Fluxion
of the Sector CAM. Again, from the Na-
ture of the Curve, and the Similarity of the
Triangles CPM, CAD, in the Ellipſis we

have PM E ) P

AC (a) and in the Hyperb. PM E y a)

:CP(z)::4D(s):AC(a). And multiplying
the Means and Extremes, we have 2 x =
by/aa+zz in the Ellipſis, and zx=by/—a2a+2zz

a PS. 3
in the Hyperbola. N 2
in the Ellipſis, and zz =—-—— in the Hy-
perbola. Which being ſubſtituted in =, we
agbbx abbx
have age the Ellipſis, and r

in the Hyperbola, = Fluxion of the Area of
the Sector CAM, the Fluent of which will

3 © ax* axs av a n

be 712 70 105 Sc. in the Ellip-
A. A ax* ax! 3

ſis, and 3 Se. in the
Hyperbola = Area of the Sector CAM and
making CB ( =1, the ſaid Fluent will be
tax -g – Ea -A , Sc. in the El-
lipſis, and va x -g – – α , Oc.
in the Hyperbola. And in the Ellipfis, if &
be = b, the Area of the Sector CAM will be

4b—tab+%ab—746, 9c. and when «x

Bb 2 Da,

APPENDIX,

==4, the Area of the Sector CB M will be al-
ſo ab- gab +x-ab=-774ab, c. the Sum
of which, viz. ab—3jab+Fab—534 b, Sc.
will be the Area of the Quadrant of the Ellip-
fis ABC.

Gan 0 L:;

II ENCE if a b, viz. when the Ellipfis is
a Circle; the Area of the Quadrant of a
Circle, whoſe Radius is a, will be 44 —344

Faa—5aa, Sc. and if a be = 1, the A-
rea of the Quadrant will be 1—3 + 5—7,
Sc. .

The Fluent of —%*_ , the Fluxion of the

2bb=2xx

Sector of the Hyperbola, may be found in the
Meaſure of a Ratio; for it may be referr’d to
the ſecond Form in the Tables of Mr. Cotes;
ſince making 2 x, d abb, , n=2, e=
2bb, f=—2, the Fluxion in the ſecond Form

SET as |
3 2 — and the Fluent of it
R＋ 7

inſt 0 20, is 2 =
againſt 8 =o, s —d R rag Now R (

—

= rdf,

=/bb—xx. Whence – d R EXT — rab

b+x , CB+ 4D |
=rACxCB|_==== ; that is
0b — xx VCB —AD

the Arca of the Hyperbolical Sector CAM is
equal to the Meaſure of the Ratio of CBA

to „ 45, the Triangle 40 B being
the Module. | 5 ;

EX AM-

APPENDIX, 69

ENA In NE

TY [quare the Space A Mp, called the Fi-
gure of the Tangents.

The Nature of this Figure is, that any Ab- PiS. x9;
ſciſs AP is equal to any Arch (AP) of a Cir- 20. |
cle, and the correſpondent Ordinate PM at
right Angles to it, is equal to the correſpon-
dent Tangent AT of the Arch.

Draw the Sccant CT, and the Secant Cz in-
finitely near to it; and from C with the Di-
ſtance CT deſcribe the ſmall Arch Te. Make
the Radius AC Da, the Arch 4 P(= Ablciſs
AP) x, and the correſpondent Tangenr A
(= correſpondent Ordinate pM) =y. Now
becauſe the Triangles ATC, Tic, are ſimi-
lar, the Angles at 7, 7, differing from each
other only by an infinitely ſmall Angle, and
the Angles at and C right Angles; therefore
TC (yaa+)3)) : AC(8)::Ti0):Tic=
ay
Vaa + Jy 3
ctors C Pp, C Te, therefore CT (yaa +33): Te

ay 3
(= NO. Pp (=x). And mul-

tiplying the Means and Extremes, we get *
— — aa)
Va TI) =

Again, becauſe of the ſimilar Se-

| yaa + Jy
there will be had x =

and dividing by /2aa-kxx,

aay
as TI
the Abſciſs Ap; and multiplying by y, there

any)
aa Jy
1 AMps

= Fluxion of

will ariſe x y =

Fluxion of the Space

120 APPENDIX.
AM pz and x Fluent of it will be )

2 + 25 975 +, Cc. = Area ſought.

— oe tn Mwu V!Cðvb ˙ — — —

Otherwiſe by the Meaſure of a Ratio.

l This Fluxion . may be compared
is Fluxion — I y P

with the firſt Form in Mr. Cotes’s Tables.
For if TA), 25 g=1, d=aa, eaa, f=1 3

bo — vil be = . —

_ 22 OO. OE at « Any “I ogy :)) Ws ..
— *

And againſt 8 (=1), the Fluent * be

F e which

2107
AC

is equal to AC For putting 2 / for the

—

2 and mn for the
C

Logarithm of the Ratio =

Module of the Logarithms; then will; AC”
Fo 1418

1
— * 1. — —
Att. 11. TT be*#=17C IIC x —. But from

the Nature of the Logarithms I is equal to the

Logarithm of the Ratio £ 5 therefore Ic

1
f
&a
9d
o b]
}

Art. 11. ** = Ac 12 werbe

*r Area Ap Mis ei to the Meaſure of the
Ratio between the Secant CT of the Arch
AP = Ap, and the Radius AC, having the
Square of the Radius for a Module.

E x AM-

APPENDIX,
E x AM P L R XIII.

O ſquare the Space A Bp M, called the
Figure of the Secants.

Here the Abſciſs Ap is equal to the Arch Fc. 21.

AP, as in the laſt Example; but the corre-
ſpondent Ordinate pM is = Secant CT, and
the Ordinate 4 B = Radius AC, every thing
elſe being as in the laſt Example; only let CT
be = y. Then from the Similarity of the

Triangles ACT, Tee, we have AT (% aa)
:: AC(a):te O): Te

aa
of the ſimilar Sectors C Te, C Pp, therefore
CIO): TT eO: Pr);

and becauſe

4a
and multiplying the Means and Extremes, there
„

= Fluxion of the Space

„ — aa
AB Mp, and the Fluent thereof will be very
eaſily – found in the Meaſure of a Ratio, it

coming under the fixth Form of Mr. Cotes’s

Tables. For making z=y, y=2, o, A aa,

9 E ©
—=—. Al in this Form R (Af) r,

( 2 *
e N
„
( Au 556) =

e aa, fr, we ſhall have

3 Whence

y —aa+ Jy

Ari. *

APP EN DI X.
the Fluent (8 being = o) 2 TRE 7 will

be aa ER =AC CA.
| a N
Conſequently the Area of the Space ABMp
will be equal to the Meaſure of the Ratio be-
tween the Sum of the Secant and Tangent of
the ſame Arch, and the Radius to the Square
of the Radius for a Module.

0K © 1.

ENCE. the Meridional Parts in Mercator’s
Chart may be computed for any given
Latitude AP. It is well known *, that the
Meridional Parts of any given Latitude 4 P,
are to the Length of the Arch A, as the Sum
of the Secants in them to-the Sum of as man
Semidiameters; that is, as the Area of the
Curve-lin’d Space AB Mp to the Rectangle

BAxpA;’or as AC [ro o AC*x AP;

or as AC 5 1 775 to AP. Whence the
| Meridional Parts are equal to AC 2
“AC

or * AC 8 71 7

K X AMP L E XIV.

TO find the Area of the Space CPO

contained under the Parts of the Con-
choids CPD,cqd of Nicomedes, the Part Cc
of the right Line AC drawn from the Pole A
perpendicular to the common Aj/ymptote BG, and

the
See Philoſophical Tran/ No. 176,

APPENDIX.
the Part QP of any right Line drawn from the

ſaid Pole A, and intercepted between the Curves.

Let Ac 2a, cB=CB=b,cC=26, AB PIC. 22.

Ac, AG =), BG=x. Draw Ap infinitely
near AP; and from A, as a Centre, deſcribe
the little Concentrick Arches 4% Mr, pn.

Now the Nature of theſe Curves is, that BC
or Be is = GP (pm) or G (ung). Becauſe
the Triangles 4 B, r Gm are ſimilar, the An-
gle at & being common, and the Angles at 7
and B right ones (the Arch rm being taken

for a ſtraight Line) therefore 4 (y): AB (c

::GM(*):rm _ Again, ſince the Se-
ctors Ar m, A59q are ſimilar, therefore Am =
AG (Yin (=) : A= AG AN
cy bx |

(3b): 54 = And again, from the

Similarity of the Sectors Ar m, Anp; there-
fore AG (: mr =): AP (3 +0): pu =
2 —.— —— pn, is = —*

And ſo PT FP = _ is = 3
the little Space s P pg Space 2 Ppg = Flu-

2Cx

ion of the Space CPY9c. And ſubſtituting

2ch&

Vcc Hu
Which comes under the ſixth Form in the Ta-
bles of Mr. Cotes. For making d ch, ax,

o, „z, e cc, f=1, the {aid fixth Form

Vet for y, the Fluxion will be

09 . 2—1 1
il become ===, Ther lu-
Ve NEU Ver

Wc ent

— A ⁰ ⁰˙

o, n=2, e aa,

APPENDIX.
cnt of which ſtanding againſt 0 So, is 2 IR
Di And making R (VF) =1, T (=
AE) =2 VF _—

= )== NY, and S = )=
— the ſaid Fluent will become 1

Fluent of the given Fluxion = Area of the

BG+ AC
Space CP9c, =Bex AB| —— That
is, the Area of the ſaid Space is equal to a Re-

ctangle under Br, and the Meaſure of the Ra-

tio between BG + AG, and AB to the Mo-
dule AB.

But to find the Area of the Space CP GB
contained under the Part PC of only one of the

Conchoides of Nicomedes, BE being the Aſym-
ptote, A the Pole, and ABC at right Angles to
BE.

Call AB, a, BC, B, AC, y, AB, x. Draw
Ap infinitely near to AP, and from Adeſcribe
the ſmall Arches mr, pn. Then by reaſoning

as before, we ſhall have — + —
244-4 2.Xx Hau
Area of the little Space C Ppm = Fluxion of
the Space CPG B. The firſt Part of which
Fluxion comes under the ſecond Form in the
Tables of Mr. Cotes : for making d abb, x x,
=2, the Fluxion of the
1

MR . .
ſecond ; “hes vill become
ſecond Form, viz 272 WI
abbæ

„ 42
. : I 4:
2 Os Flaent of which 1s <-

APPENDIX.
2 In which ſubſtituting for R (= V+ )

y/—aa, for 7 (= 2 x, and for & (=/* = =)

yaa + xx, and we ſhall have 25

for the Fluent of the firſt Part

24a + 2:xx

of the Flu-

Again, the latter Part

yaa + xx
xion may be compared with the ſixth Form
in the Tables of Mr. Cotes by making d ab,
2=x, o, n=2, e=aa, =I; and the Flu-

ent thereof againſt o, viz. 170 N | a —

by making R= „T= Vaa + Xxx» and S 25
x + aa + **
a

will be 45

Therefore the Sum

E t os <4

yaa t xx
. 1 will be the
Fluxion. Conſiſting likewiſe of two Parts,
the former of which is the Meaſure of an An-

gle, becauſe R is Laa; and the latter of
a Ratio, becauſe R is = y/1 =1.

of theſe Fluents + 535

Fluent of the whole

The Conſtruction of which Fluent may be pC. 24.

thus: Draw CF perpendicular to AC, aſſume
B H equal to the Meaſure of the Ratio be-

tween BGA, and AB to AB taken as a
Module, that is, make B H=3 —

Ce: More-

APPENDIX.

Moreover, aſſume BI equal to the Meaſure
of the Angle GA, to the ſame Module AB,

= The Radius, Tan-

/44 E

that is, 24

gent, and 5 and Secant of which being as a, x,

and yaa + x xx, or AB, BC, AG. This done,
draw ATE and HD parallel to the ſame; then
will the Trapezium BZ C be equal to the
whole Fluent; and conſequently equal to the

Space CPGB. For ſince the Triangles A BN,

ACD are ſimilar, therefore A B (a): BI
(- ==) ::ACa+0D:CE=

e
122 yaa + xx
«BC=5 T. 5 ax =
aa i xx 2 aa PN

Trapezium BIE C. Again, HI (=B H —

BI e | —. a + x i
a aa + xx

which multiplied Itiplied by BC (Y), and the Product
ITV _ ;| a –X
oy | “OH 1 F ＋ Xx
logram DEI) added to the Trapezium BIEC
ab CAT 4 * bbj ax ab b

will be 2 — — –

yaa -j- xx yaa x

8 bbl a+x

a EN |
= + a

V ντ wx
2.2 . y | – | L
U — „ Fluent to be conſtruct-

= (= Paralle-

ed, = Trapezium BHC.

or wy —_—

APPENDIX.
If the Conchoid C PD be of ſuch a Na-
cure, that ABx BC (ab) be AGG (9b);

by reaſoning as above, we ſhall have 69 7

3 9 aa
4 = for the Area of the little Space
2 — a8

G Ppm =Fluxion of the Space GPCB,; and
the Fluents of each Part of this Fluxion may
be had from thoſe of the Fluxion of the fifth
Form in the Tables of Mr. Cotes. For making
d=aab, X , o, n=2, e= —aa, f==1, the

0n—1
BY :
Fluxion of the Form, viz. x will

bovine Shad the fb Part of the
aa 5

Fluxion. And making d D I,

1=2, e aa, f=1, the {aid Fluxion in the
| abby 0
Form will become

” , the latter Part
2y/JY—aa

of the Fluxion; and the F mel} = the former
Caſe correſponding to , is — ll ix =;

_y in the latter, the Fluent nt j=—1,
—zd P ＋ 22 25 4 R E Conſequently

for p ed pn e for R for R
(Sve) /=an, for T (CV F ae,
and for $ (= „EY y; the former Fluent will

become a 4 nn * 7 and the latter b b
— 479
2 2 2 ; and the Sum of theſe

Fluents

F 10. 23.

F 1 6. 24.

Tluents VI — 5 424425 3-Y þ + 1

is = Fluent of the Fluxion above. Wich
may be conſtructed thus:

Make AG ()): T1 AB a):: TBC (19: BN

= And AG(3):GB a):: BN

F418. .

— =): BM= = /jj—aa; thenif to BM

you ns MO = to the Meaſure of the Angle
B AG (whoſe Radius, Tangent and Secant are

as AB(a), BG A, and 4G (y) to the
Module AB 4.5 _ (a + =) the Rectangle
CB x BO will be cqual to the Fluent

= iy —aa — e . Area
of the Space G PCB.

Err .

L 7 DPEBQD be half the Lune of Hip-

pocrates, A being the Centre of the 2
DQ, Nin C the Centre of DPE; it is requi-

red to find the Area of the Space QP EB, con-
tained under B E, Part of a 2 CEB Join-
ing the Centres A, C; the Parts QB, PE of the

Arches forming the Semi-Lune; and the Part.
QP ef any right Line AQP drawn from the

Centre A of the larger Arch of — I ying
between the ſaid Arches.

Draw Ag gp infinitely near A4 & O, and from
the Centre A deſcribe the ſmall Arches Gr, ps,
and join the Points P and E.

2 This

Let DCF be at right Angles to ACE.

: “hr

APPENDIX.

This done, let AC a, CGS x, AG y.
Becauſe the Triangles 46 C, rg & are ſimilar;
therefore 4 (3): AC (a):: Gg (x): Gr =

And ſince the Triangles AGC, AE P,

are likewiſe ſimilar, the Angle 4 being com-
mon to both, and the Angle 4 PE in a Semi-
circle equal to the right Angle AJC; there-
fore AG (5): AC (a):: AE (2 a): AP =

7 Moreover, from the Similarity of the

Sector Aer, Asp, we have 46 (): Gr
=): AP (== 7 N * Which

e into 2 A P (= y 5 : te Product

= will be = Area of the little Sector Ap,

= little Triangle 4p P. For theſe differ only
by the Triangle p Ps, which is infinitely lets
than Aps. Again, becauſe the Sectors Ar,

Asp are ſimilar, we have A0 (0: Gr (=
:: 49 (4%): e Which drawn

into x 4 (< 2 2), and the Product 5 will
be equal to the Area of the little Sector or

245 x
Triangle 49 9; which taken from 2 1
the Area of the Triangle 4p P, and the Re-

2 — — will be = Area of the Tra-

2 ”
pezium © Ppg. Which is the Fluxion of the
lunar Space EP B.

mainder —<

Again,

APPENDIX.
Again, becauſe y3j=2a-þ wx; therefore xx=
4$y—az: and throwing this Equatiort into Flu-

: PE FE, yy
X10 eren and i == —

8 Hang A 7

by putting for x its equal Va; then if
this laſt Value be put for x in the Fluxion of
the ſought Lunar Space, the ſame will become

4 “oy +: be Hp e.
25% e

e ee e eee e
Both Parts of which come under the fiſth
Form in the Tables of Mr. Cotes. For in the
firſt Part, making dz, z, = — 1; n=2,
eg aa, t, we have the Fluxion of the

2 On —1 1 —,
Form 1 – S | In like manner,

Len „el
in the ſecond Part, making d= -a, z=0,
o, „2, e- 4a, f=1, we haye

e f
_ Tze Fluent againſt

Se =
EA AR N
Nez nee S 9

x33} —aa Þ+ aa * 5 by ſubſtituting
x V1 —aa for P Web 2), a for R (Je),
Vene, for 7 (39a), dy e
| . 1 2. KR * *
And that againſt oo, is R | MER

My — by ſubſtituting the ſame
Values. Therefore the Sum of theſe Fluents
— jj —aa +aa 2 X. – n ——

eee 0
— *. is ==,/ yy —- ad ſince the
Jy
Meaſures

APP END IX. 81

Meaſures of the ſame Angle to the Module aa
are, the one affirmative, and the other nega-
tive, and conſequently deſtroy each other.) =
Fluent of the Fluxion of the Lunar Space
EP9B, = EP9B.

Now this Fluent may be eaſily conſtructed;
for if a right Line be drawn from the Centre
C to P, the Iſcoſceles right-lined Triangle

——

CPE will be = Fluent — — OS AEM
JY VI b /

(ſince x = 4/yy—aa) = Lunar Space EP9B,
as may be ſhewn thus:

Draw PH parallel to GC; then the Tri-
angles AG C, A PH are ſimilar: therefore AG

(9) : AP (=) SO: PH= 222%,

Jy 1
And ſo 1 PH (= xCE(a)is = ZZ =
J) JJ

= Fluent to be conſtructed = Area of the Tri-
angle CPE. Hence the Area of the Semi-

Lune is = CExXzDC=75a4. And thus you

have the Quadrature of the aforeſaid Space by

the Method of Fluxions; tho’ indeed it may

be ſhewn much ſhorter by common Geome-
try.

| EXAMPLE XVI.

ſquare the Cycloidal Space, or to find
the Area of any Segment AMG of it.

Let AP B be the generating Circle. Let pc. 26.

AP be any Ordinate, PM the correſpondent
Abſciſs; let 2p be infinitely near M; let
eM touch the generating Circle in P, and
M the Cycloid in M. Then from the Na-
ture of the Cycloid, the Subtangent PT =
PPM = Arch AP. Draw AG perpendicular

D d 2x”

ae ie omen Avenues

— wy “Ie >
= — ” — —
—— EE I . —

— — — —
Gn ———————r— — *
— —ũUä

8
— *
— — — 1 _

APP END IX.
to AB, and from the Points M, n, draw MG,

mg per pendicular to AG.

Now let I gx, 4B —=1. Becauſe TP
— PM, the Angle MT P = PM; and there-

ford the Angle TPO =2T MP. But the

Meaſure of the Angle 4, is + the Arch
AP, which is alſo the Meaſure of the Angle

TPA4: therefore APD =T MP = Mms.

Conſequently the Triangles APD, Mm & are
ſimilar, therefore © (x) DN : DP (xxx) : :

MS (&): 1582 == DE 7. Bur + W —.

thrown into an infinite Series, will be x
Z3 2

– R N x—7T7x* , &c. = the Flu-
xion of the Ordinate © © Mrto the Axis 4B of
the Gycloid, and the” F luent of this, </z,

2x —ix* eee Um. — n 2 will be the ſaid
Graner M9. Wine 9 Mxx= the Flu-

xion of the ä Space AMY is =

2 A * * — ox Maha Ar xa 2 the Flu-
3
7

ert — which will be + x Ret TN. — Xo

IHE „Sc. = Area of the indefinite Cycloi-
dal Space AN 2

Now if 12 = be drawn

into GM=49=x; then will the Fluxion
GA SG of the Area 4 M be equal to

xx. Therefore fince this is the ſame as
the Fluxion of the Segment of the Circle
APD, the Space AM G will be equal to the
Segment APY of the Circle, and fo the Area
ADBC of the whole Cycloidal Space A D
is equal 1 to the Arca of the Semicirele APB.
22 Co RO.

APPENDLIS:

+Y” WY

B ECAUSE DB is equal to ? the Circumfe-
rence of the Circle; if you call the ſame
p, and AB, a; then the Rectangle AB DE
ap = Area Semicircle 4APB. Whence the
external Cycloidal Space 4E D M A= A ap.
Therefore the Area of the Semicycloid 40
—Zap AMDPA=zap. Conlequently the
Area of the Cycloid is the triple of the gene-
rating Circle.

EXAMPLE XVII.

3.7 ſquare the Ciſſoid of Diocles, or to F ie. 27.

find the Area of any Segment APM of it.

Let ADB be the generating Circle, BH

the Aſymptore to the Curve (47) of the Ciſ-
ſoid, at right Angles to the Diameter 4 B let
the Diameter 4 Bbe= 71, the Abſciſs A P=x,
the correſpondent Ordinate PM to the Ciſſoid

— —
— —

Now the Equation expreſſing the Nature
of this Curve will be PBx N =AP ; that

. 3
is, „Y- =x*, and fo = — . Whence
I—x

2 XJ) xX 2 ;
11 e I *; and x
I—X I— |

—— —

— XI— x „ Fluxionofthe Area APM.

The Fluent of which will be = + = +

3 * ä
11 1 8
4.9 ＋ 46.11 T F

APPENDIX,

to = 575 x5 ＋*＋ — + SEE. 3122 5 19s. by
7 T 75.15 4.6.8.13
Sc. = 8 A PM And vom „ =I, then

will this Series become 2 += 42 lt, 2 141

F749 748.17
+7 * 5 z Sc. = Area of the infinite Space
4B HL

| Otherwiſe:

Let AB a, and PVE. Now becauſe
from the Nature of the Curve ay -) =x; this thrown into Fluxions, and there will ariſe 2% — 295 —x=3x*x, and 24 — 2K *) =

C == But fince from another Property

of the Curve x* =2y; therefore ==2. Now

make a—x = PB=2; then we ſhall have 223
—Yx=32x; and fo the Fluent of the one
will be equal to the Fluent of the other. But
⁊ x is the Fluxion PN 7p of the Segment AN
of the Circle; and becauſe «== PB—=O M and
z=mR—=Ooc, mMOo, will be the Fluxion
of the Area AMO B, and yx the Fluxion of
the Area AMP. Now when the Fluent of
45 expreſſes the Area of the whole Cycloidal
Space AB HA, the Fluent of x will be the
fame Area; and ſo the Fluent of 22 —yx=
Fluent 25. Therefore ſince in the ſame Caſe
the Fluent of z is the Area of the Semicircle
AN B; and becauſe the Fluent of 2 =

3 2X, the whole Ciſſoidal Space AB HTA

will be the triple of the generating Semicircle
ANB.

E xAMs

APPENDIX.

ET KML AVAlk

TO ſquare any Interminate Space HMI p, 5. 28.

contained under the Aſymptore PH, Or-
dinate MP, and Part MSI of the Logarith-
mick Curve MI.

Call the Subtangent P 7, (becauſe from
the Nature of this Curve it is a ſtanding Quan-
tity) and the Ordinate PM.). ‘Draw the in-
finitely near Ordinate pm, and from M draw
Mꝗ perpendicular to 29. Now becauſe of
the ſimilar Triangles MT P, mMq, MP ()

4 (#)::mg (3): My==>: Whence MP

x Mq =} 9 ah, is the Fluxion of the

Area 1M PH, and the Fluent thereof will be
ya; that is, the Area of the infinitely extended
Space MPH is =M PXPT.

CoRo L.

the Ordinate O & be =2z; then the Inter-

minate Space 18 © 7 = az, and conſe-
quently & MPa -a N -; that
is, the Space contained under any two Semi-
ordinates M P, $ ©, the Part of the Abſciſs
P 9, and the Curve MS, is equal to TPM
— z; and ſo the Space BAPM is to the
Space BMS, as the Difference of the Or-
dinates A B, PM to the Difference of the Or-
dinates P M 0 O ,

— OO On Ore
—— — *

DO YR II — —

D : = – _ — — m—_
-———— ͥ ——

3 ˙ꝛ—• —ͥ On ——
— —— rr — — >
—— 2 a 3 aa rt
— oy 2 2 — — — — *
I Rs S – – A \

—ͤ ——-—-

Ftc. 29.

APPENDIX,

EXAMPLE XIX.
37; 78 ſquare Spiral Spaces.

Let the Semi-diameter of the Circle, viz.
AC=a, the Periphery , any Arch AB
x, as an Abſciſs, and the correſpondent Or-
dinate CM=y. Conceive the Radius C6 in-
finitely near C B, and draw the ſmall Arch ME.

Now the Nature of Archimedes’s Spiral is
ACx A B=Periphery b drawn into CM; that
is, ax==by. This being granted, the ſmall

Arch ME=!z, ſince CB:Bb::CM: ME.

Therefore i C M x ME — Area of the little
Sector MCE = to the little trilincal Space
CMm, which is the Fluxion of the Spiral

Space 2, but from the Nature of the

2 2

A x* X ©

nary Expreſſion, and 22 it will be = 9

the Fluent of which will be Z 7 the Area of

the Segment of the Spiral RM and if for

be put 5, the whole Circumference, the
whole Spiral Space will be =;ab.

Again; The Nature of all Kinds of circu-
lar Spirals will be expreſſed by am x = n m,

therefore _ Sym, and — L of Conſe-
_

quently ** = 7 The Fluent of which
24 . 8

Dk will

will be —

x, and the whole Spiral Spaces will be

mah

APPENDIX.

2n Lm

max ®

2n*

4u+ zm

4 n

Moreover, i

yy, and & =

5-4

4ap—y*

Much after the ſame manner you may find
the Area of the Space contained under the
Arch AB, and the Spiral AM; whoſe Flu-

xion is the Trapezium BMUmb=Bb+Mmximb.

Now let the Curve be a Parabolical Spiral;
ſubſtitute 7 = for its Equal x: then will

f the Arch 4B be to BM as
the Abſciſs to the Ordinate in any Algebraical
Curve, the Spiral Space may be ſquared after
the ſame manner as above.

For Example: Let 4B be to BM as the
Abſciſs of a Parabola to the Ordinate; then
aſſuming p for the Parameter px = a* — 2az+

Therefore putting þ for

, the Fluent of which will be

But Bb , Mm ==, mb a-); there-
fore BA, (A EN D E.
re DO 14 * 2574 Y — 24

29 1-47 1 — 0
—.— — be the Fluxion of the

ap
Space 4B M, and the Fluent thereof

88 APPD END IX.

—— y+ wall be equal to the ſaid Arca.

44455

. rd

— 8 „ @

F16. 30. 38. 7 ſquare any Quadratical Space ACM
contained under the Baſe AC, the Ab-
ſciſs AP, a correſpondent Ordinate P M, and

Part C M of be Quadratrix.

Let Cc be a Quadrant of the generating
Circle, A being the Centre. Draw Amn M,
and from n and M let falt the Perpendiculars
we, ME. Now let AC=1, FP =x, PM
==y. Then the Nature of this Curve is, that
AB (x) = EM is = Arch Cm of the Qua-
drant; therefore (by Schol. 1. Example 4. of
the next Section) the Sine e will be =1 +

– He, &c. and the Sine Comple-
ment Ae =1 a Y Tr -, &c. But
becauſe the Triangles Ame, 4 PM are ſimilar,
therefore me (IT -N + ix, &c.) : Ae
(i- NU x, &C.) : 4 P(s): PM
(9) =1I —Faxt=—75 xt —377 x0, XC.) W hence
-n -r, -, &c. = Flu-
xion of the Area AC MP, and the Fluent will

be =x— 3a} — rr v, &c. = Area
ACPM. od

SECT.

SCF .
U/e of Fluxions in the Rectification of

Curves.

P R OB.

12 refify any given Curve, or find a ſtrait
Line equal to the fame.

N 3 88 5 1 2 n

Since a Curve 1s conceived to be made up Fe. 31.
of an infinite Number of ſmall right Lines;
one of them being found by the direct Me-
thod of Fluxions, that is, the Fluxion of the
Curve; and the Sum of them, or, which is all
one, the Fluent of that Fluxion, will be the
Length of a right Line equal to the Curve.
Now let PM ()) be the Ordinate at right An-
gles to the Abſciſs AP (x) of any Curve AM,
and let TM touch the Curve in M. Draw
Pm parallel to PM and infinitely near PM,
and from M draw Mm perpendicular to pm.
Then Pp(=Mn) =x, and m =), and Mm
will be the Fluxion of the Curve AM; which
muſt firſt be found, either by getting the Va-

lue of MT» (, or #3 (53) in the Equation of
the Curve thrown- into Fluxions, becauſe

+y* = Mm; or by making the Subtan-
gent T P, or Ordinate PM (y) to the Tangent
TM, as the F * the Abſcils Mn (x)
S —

5
1
— |
—
.

|
”
5
25

EQ. 33.

APPENDIX.

or m (3) to a fourth Proportional, which will
be = Mz the Fluxion of the Curve 4 M.
For the little right-angled Triangle Mn is
ſimilar to the right-angled Triangle TMP;
and the Flucnt of that Fluxion will be the
Arch fought. Some Examples will make this
evident.

PM ():: Mn (x), or am (53) : Mm the Flu-
xion of the Curve AM.

+4.

TO find the Length of any Arch AM of

the Curve of the common Parabola.

2 i TE l.

Here A Pxa PII; that is, a x =yy; and
both Parts thrown into Fluxions is ax =2 97.

Whence aa , and & =_ . And
adding * to this laſt Expreſſion, we ſhalt have
— . So / |
„5 +S55-=5 varF ay = Mm
the Fluxion of the Curve AM. The Fluent
of which will be y + 7 +; *
Sc. = Arch AM. | +,
Otherwiſe by the Meaſure of a Ratio. ;
The {aid Fluxion of the Curve of the Pa-
rabola = yaa + 4yy may. be referred: to the

fourth Form in the Tables of Mr. Cores. 2
2 i

APPENDIX.
if you make z=7, y=2, $==o, d==, e aa,
ſ=4- Then will Hh 41 e be =
Zi Fay. And the Fluent of this Form
(againſt 8=9) is 4P Þ+ i T; d R —— And

e 2 Aer K

(SVH Dr, Sue == Joh /aa PA,
S — V A) =* . The ſaid Fluent will be —

yaa + 439 + 4 2 Z. Which may

be thus conſtructed: SE the Vertex A draw
AB biſecting the Ordinate PM (7) in B; then

will AB (= SAP + PB) = 25 Vaa + 45.

And multiplying the Numerator and Denomi-

nator of che Ratio 22341121 by = the
| a 4

3 um 1
fame will become 22 V t 497 * 42. That is,
75
re y 7
a] A + gvy +29 is = EVA +499 *
4 4 4 8 ‘

But 25 Vaa -, 2 += s AB N＋ BP, and

—

FI. 32.

7 is = PB; alfo.5.is = focal Diſtance from

the Nature of the Parabola. Therefore
e 2 a

APPENDIX.

a |/aa . ü ABEL AT +
4 PB

the Meaſure of the Ratio between AB + AP,
and PB to the focal Diſtance A, as a Mo-
dule. This added to AB, and the whole will

Lyn Fan +5 a] /20 + I +23 = py.

ent of the Ns L a 1-439 = Length

of the Arch AM of the Parabola. Hence
the Rule for finding the Length of the Curve
of the common Parabola is this.

Let A be the Vertex, F the Focus, 4 P
the Axis, and PM an Ordinate to the ſame.
Draw AB biſecting the Ordinate PM in the
Point B; to which continued out, add BC
the Meaſure of the Ratio between AB EAP,
and PB to the Module AF; and then 40
will be the Length of the Arch 4M of the
Parabola. This is the Conſtruction given by
Mr. Cotes in Harmonia Menſurarum, p. 12.

In this Curve the Subtangent TP =2 4P

=2x, Whence TM =y4xx ax. There-
fore T p (2 Y): TM(y/axx + ax):: Mu): Mm

= LK „ x = Fluxion of the Curve.

Y
Or ſince ax =yy; therefore x = 75 and 2. and 2x

14
= , and 4xx = — =: Conſequently —— — 4) r /

_ Now PM (9) : U ty)

X. 00
um (: Mm = as =j LETS

——

APPENDIX.

as before; and the Fluent of x B12 T — 5

2 *
ſuppoſing a=1, will be x* +7 Vn, 4 U,
+F/x”— , &c.
Co R O.

1 AC, DC be the Semi- conjugate Axes of F 1 6. 32.

an Equilateral Hyperbola, and AC (=DC)
be ſuppoled = a, the Latus Rectum of the
Parabola z and the Ordinate PM be =2 y, and
the Abſciſs O x: then will A P be =x—4a.

Conſequently becauſe PCx AP = PM 3 that
is xX—4a==4), and ſo xx=43y aa. There-
fore x = 4/4yy aa. And if 9m be drawn
infinitely near n, then =; and ſo the
Fluxion Om M of the Area of the Hyperbo-
lick Space COMA will be =) 4/aa +4 yy
= Fluxion of the Curve of the Parabola.
71 herefore the Quadrature of the Curve of the
Parabola depends upon the Quadrature of the
laid Hyperbolick Spare. And fo the Rectifi-
cation of any Curve may be brought to the
Quadrature of a Curve, by ſuppoſing the Flu-
xion of the Curve to be rectify’d (found as a-
bove) as an Ordinate, and the variable Quan-
tity in that Fluxion as an Abſciſs to that Or-
dinate. Conſequently the Buſineſs of rectify-
ing Curves ſometimes may be ſhorten’d, from
a Pre-knowledge of the Quadrature of that
Curye it may be reduced too.

APPENDIX.

rern I

1 10 reftify a Parabola of the ſecond Kind ;

where ax i, or makihg a = I, *

== 4

| Becauſe * ; therefore 2 x x = 33* y, and

Y) __ o9ty*

4 =909%z*, Whence 4 = =—
4 Gf 4x 49

cby ſubſtituting 47 for ax*) ; therefore

Art. 19.

v/*-13 =y/235* +5* (by adding) to each

Side of the Equation x=23 y) L +ag
=7Ijy/9y +4, = Fluxion of the Curve, and
the Fluent thereof willbe 7x 9y + 4*y/93 .
But in order to find whether any thing is to
be added or taken from this, make S o, and
then the Remainder will be 4 /4 =; there-
fore the Length of the Curve will be * =

X93 + 4X9 -.

Con 0©0 i.

E T the Parameter of the common Parabo-
la be 1, 4 P=1,P9=2y; then will 49
=2y-|-1, and becauſe the Parameter is 1,

ON (23 +1) == Conſequently ON

of a Parabola of the ſecond Kind expreſſed by

ax =y* depends on, or is the Quadrature of
the common Parabola, and may be had in fi-
nite Terms.

EXA M-

APPENDIX,

EXASCEE Ss RL

T* rectify Parabola’s of all Kinds.

Let the Parameter be i; then the Nature
of infinite Numbers of Parabola’s of different
Kinds will be expreſſed by this Equation ex.
The Fluxion of which will be n 3 = 3
and ſquaring both Sides #* * ”—? 52 . And
if for Brevity’s lake you 3 zu — 2,
then will mw y5 =x; and adding y to each
Side, and afterwards S the ſquare
Root, „ n Jo =p +7
Fluxion of the Arch of a Parabola of any
Kind ſoever. The Fluent: of which, viz.

wy may 1 I 2 6437 ＋
5 5 Fl I 2.4.27 +1 2.46. 37 +I

— 2 Fw) „Kc. will be the Length of
T7 ==
the/Curve . Now ſubſtituting 22.2
for 7, and then the lame Je will be =

| ho — —
7 | 2. 71 | 2.4: 47th; 2 we” om=x
= 1.3. ems ys 8

=” 2.4.6.8 Jun

‘Fre. 33.

APPENDIX.

ExXAMPLE IV.

Te find the Length of any Part BC of
the Equiangular or Logarithmical Spiral.

Let the Radius AB=a, AC=5: Let BD
(e) touch the Spiral in B, and 4 D be perpen-
dicular to B D. Call BD, p. Alſo let PE
touch the Spiral in P, and let AE be perpen-
dicular to it. Moregyer let Ap be infinitely

near AP; and from 4 deſcribe the ſmall Arch

Pm. Call 4? — AB F, y.

Now ſince from the Nature of this Curve
any Radius AB cuts it in the ſame Angle, or
makes the ſame Angle AB D with the Tan-
gent B D; therefore any Triangle, as APE,

Will be fimilar to the given Triangle ABD.

But the ſmall Triangle m is alſo ſimilar to
the T 8 APE, the Angle at being a
right Angle E, and the Angle at p common
to both Triangles : whence the fluxionary Tri-
angle mpP is ſimilar to the Triangle AB D

and ſo MAI (= =
Fluxion of the Part BP of the Curve; and
the Fluent — will be che Length of the
Part BP; and putting 6—4a for 5, we ſhall
have — = Length of BC; that is, as BD
(0): A (a):: AC-AB: Length BC.

Ex AM-

APP E N DI X.
EXAMPL E V.

TPO find the Length of any Arch AP of

the Spiral AP C of Archimedes.

Let CAM be the generating Circle. Call F. 34.

AC, a, the Circumference b, the Arch CM, x,
the Ordinate AP of the Spiral y, and let PT
touch the Spiral in P; interſecting the Line
AT drawn from A the Centre of the Spiral
perpendicular to 4 P, in the Point 7. More-
over, let Mm==x, draw Am, and with the
Diſtance Inu, from A deſcribe the little Arch
2p. Now ax=by expreſſes the Nature of this
Curve. Becauſe of the ſimilar Sectors A Mm,
Apn, AM(a): Mm (x) :: Ap AP (3): Pn

== . And fince the Triangles Ppn, PAT
are ſimilar alſo; therefore p P (y):p7 (=): –
AP (3): AT the Subtangent 5 But
ax =by; whence 1 1. Which ſubſtituted
for x in and we 1 AT = 2 Agai
ay ? aa Balm,
ſince TAP is a right Angle, AT (= —

2
TP M FB = xx, Da
AP (yy) =TP = — and ſo TP=

;; and AP (3): rÞp(2 VN)

1 P 😛 .I — J ya*+bbyy, =Fluxion of

the Part AP of the Spiral. Which comes
under the fourth Form of Mr. Cotes’s Tables.

F f For

APPENDIX.

. | I
For making Z=), 25 g=0, 42 pv

7 F Abb, we have 4 in verſe” = |

— SY Va#Fbbyy. The Fluent of which is –

45 + % And writing for P, R,

of
25 , or SF vt, , their Va-

lues 7 Y, 5, 5 P55, J, chere
aa [Va TIE 4-o6

ariſes = Y * 2 =

aa
Fade of the Fluxion aforeſaid.

This Fluent may be conſtructed thus Bi-
ſe&t AP (y) in the Point L, and draw DL pa-
rallel to the Tangent PT cutting the Subtan-

gent in D; then will DL be 22 a* + bby),

ſince it is 3 PT = „55 Again, the
Module — is had by making AD (2);

AL (2): AL(2 IE AF ==. And

the Ratio — 22 +5 is =

725 Vai +2- –

8
viding the Numerator and Denominator of

as you will find by di-

this laſt by =; therefore that Ratio is =

N. Conſequently if to D L you add

L9,

a ed kk. to

SY =] ll

weak

APPENDIS.

L, which is the Meaſure of the Ratio of

LD+ AD to AL, the Line DF being the

Module; the whole Line D will be equal

to the Length AP of the Spiral; and thus

you may get the Length of the whole Spiral

oy making y=a, the Radius of the generating
ircle.

41
7 the reciprocal Spiral APC.

About the Centre A, with the Diſtance
AC, deſcribe the Quadrantal Arch BC MH,
and continue out AP to M; then the Nature
of this Curve is, that any Radius AC of it is
reciprocally as the Angle BAC it makes with
the firſt Radius AB, or as the Arch BC;
that is, AP: AC:: BC: BM, and fo AP
xXB M=ACxBC.

Now make 4B or AC==a, the Arch BC
==b, the Arch B M=x, and AP); then
will aß be =xy. Draw P to touch the
Curve in P, and at right Angles to AP draw
AG. Let Ap be infinitely near 4 P, and con-
tinue it out to interſe& the Arch in , and
with the Diſtance Ap from A deſcribe the
ſmall Arch py. Becauſe the Sectors An p,

EXAMPLE VI.

A Mm are ſimilar, AM(a): Mm ():: AP (9)

2 = = And becauſe of the ſimilar Tri-
angles OG, nP Pb therefore Pn (): n 2
::AP(y): AG _ And throwing the E-
quation of the Cur ve a4b=xy into Fluxions,
We have x = which put for x in =, and

Ff 2 we

O find the Length of any Arch CP of Fi. 35.

APPENDIX.

we have 46 =; that is, the Line drawn
from the Centre A perpendicular to any Radi-
us AP of the Spin, to interſect the Tangent

to the Spiral at the Extremity of that Radius,

will be a ſtanding Quantity, viz. = Arch BC,
which Arch will become a ftraight Line, per-
pendicular to the firſt Radius AB, when the
Point C is at an infinite Diſtance, and the An-
gle BAC infinitely ſmall; that is, it will be
equal to AI D the Diſtance of the Aſymptote
DE from the Centre. Hence AP (3): GP

GNP (3): Pp=5 N Fu-
xion of the Curve AP.

This may be referred to the third Form in
the Tables of Mr. Cotes. For making z==y,

, _ l, e=bb, f=1, we have
3 e = CES + yy. And the

Fluent of this 2 – 4 P — —d R A 75 4 „ by Wri-

ting for P (= VIE ) . for R (e)
b, for T (=y/eFfz”)) ybb+97, and for &

y
(=yf)1, vilbe Na.
And if 4 be ſuppoſed invariable = _ then
the Fluent bb zz — 12 = will be
= Arch APC. And the Difference of theſe

Fluents, viz. bb ZZ — VD +6
e IVV

r —b|— 5 — [0 FG N＋
e. e = Len ength

of the Part PC of the Curve; that is, if 8
| ; the

„ r WY 4&4 as

the Curve; or becauſe AF

APPD END IX.

the Difference CF — P of the Tangents,
you add the Difference of the Meaſures of the
Ratio between AF PG and 4 to the Mo-
dule AF, and of the Ratio between AF
CF and AC to the ſame Module 4 F, the
whole will be the Length of the Part PC of
AF+ PG. _
AL £4
| B AF+CF
to AF PU and AF 2 equal
A — — 7 (by Schol. Art. 14.) there-
fore if LM be the Meaſure of the Ratio be-
tween AC and CF— AF, to the Module AF,
and in like manner Im the Meaſure of the Ra-
tio between AP and PG— AF to the Mo-
dule AF; then the Aggregate of the Diffe-
rence of the Tangents CF -P, and the
Difference of the Meafures % — LM will be
= Length of the Part PC of the Spiral.

ET AMP LE VII.

46.70 find the Length of the Arch CP of Fic. 36.

the Logarithmick Curve CPG.

Let AC, a P be Ordinates perpendicular to
the Aſymptote Af, and let C F, Pf be Tan-
; ay Alſo let » be infinitely near P, and

raw np parallel to af.

Now call A4 C, z, a P, y, and the invariable
Subtangent AF or 4 b. Then becauſe of the
ſimilar Triangles a Pf, » Pp, we have a P ())

FP (bb ):: P): PS == VI =
Fluxion of the infinite Part P G of the Curve.
2 | W hich

101

102

Art. 14.

” Art. 14.

APPENDIX.

Which compared with the third Form of
Mr. Cotes’s Tables, and the Fluent (as in the

laſt Problem) will be y — .

In like manner the Fluxion of the infinite Part

CPE will be Z yI7Fzz; and the Fluent
thereof will be /#5 L — 5 | a= LT – =

and the Difference of theſe Fluents, viz.

JETE—EF +6 — —

D 7 F+4Pf
J

=CF— 1 —_

2 or * =CF— Pf AE
3

be * . toCF—AF, and al equal to
Pf – A, and LM, Im be drawn arallel to
Af; then from the Nature of the thine LM
(= Abſciſs Az) will be the Logarithm of the
Ratio of the Ordinate AC to the Ordinate
Mn; that is, of AC to AL: and confequent-

; ; AC :
I ; ;
y ZM is the Logarithm of Th So like
wiſe Im is the Logarithm of .

Pf—-AF
ſuppoſing the invariable Subtangent AF, or
af () equal to the Module of the Logarithms,

the Line LM will be = 4p [4 3x_CF _

„ n
3 221

2 .

APPENDIL
4F|; oP _; therefore if to Cf P/, the

Difference of the Tangents, you add In LM,
the Difference of the Parallels; the whole
will be equal to the Length CP of the Curve.

ETA Ii Va

HE Sine PM of any Arch PM of a
Circle being given, together with the Ra-
dius : to find the Length of that Arch.

Let AP be = x, the Radius AC=1, and p, ,,
the Sine or Ordinate PM==y. Then from

the Nature of the Curve, APx PB = PM,
that is, 2#—xx=zy. And throwing this E-
quation into Fluxions, we have 2x—2xx =

245, and x = —. . Which ſquared will be

becauſe from the E-
I —2x+xx 1—95˙

quation of the Curve 2x —xx=yy; therefore
—__
: 2 — 2 — = —— —

942 4
Sms * 2 =I %; which will be

—

the Fluxion of the Arch AM. And the Flu-
ent thereof will be 4 „

| 1 % 3470 *
r 1

2 3 + 2 &c. Length of the
Arch AM.

Nov if the firſt Term be called A, the ſe-
cond B, the third C, the fourth D, &c. and
the ſecond Term be multiplied by x, the third
by 3, the fourth by 5, Cc. the Series 1

wil

—— — —

— — — — — < =
p = —

— 22 2 .
— . – — 2 «- Y nn – * — —
PE — – —— = a 2 r — — L— — — — —X—k..- ( —
8
* — — – _ A A — 8 — 22 . 2 — So. — — DEA oe

— 6

:
by EEE
. — —

104

APPENDIX.
ill b 1 hi F L 44 = B a 7,”
will beco 9 i * +

1 2 7 2
85105 2 5 557 47 „Ke.

Otherwiſe :

Draw the Radius MC, compleat the infi-
nitely ſmall Rectangle PM p, make AB=1 ;
and as before, A P=x, PM=y; then x—x x
SD), and becauſe the ſmall right-angled- Tri-
angle Mm n is ſimilar to the Triangle PMC.

Therefore PM (Y/x—xx): MCC: Pp or My
— 7 = YZ x5 =Fluxi-

2Y/ XxX 2X—2XXx |
on of the Arch A M; and the Fluent thereof

. A 3. 4 2 2
will be x* TX ++ ] rx fL Cc.

=x into 1 + 3x+ #x* Tr ＋ 345 x4&c.

If the Coſine P C be made x, and the Radi-
us AC be i; then the Fluxion Mm of the
Arch CD, being the Complement of AM to

. I * 21 – l
a Quadrant, will be + or VE for
| V – X—xX ..

ſince the Nature of the Circle is 1—yy =xx,
this thrown into Fluxions is — 2 xx, Viz.

3 . . L. #- 4 * x? x* 4 **
1 ; Ty 5 I=ax
„ *

. 2 2
and thence /x”x-p5 N = + ** =x

. Fluent of which will be x

1—

, 7 3 x. EN, 697, = Length of the Arch

Scho-

APPENDIX. 105

Scholl iD M .

r the Arch AM (ſuppoſe z) be given in the 1
Series found above, and you want the ver- – 1
ſed Sine or Baſe AP (x), or the Sine MP;
then you muſt extract the Root of this Equa-

WH +: +# rx, Sc. where PM=x.

S’cHuoLivM II.

x F Ro M this Example we have the Inveſtiga- pro. 3y.
, © tion of the Theorem of Hugen’s for find-
ing the Length of any Arch AM of a Circle
by having the Chord AM thereof, and the
Chord Am of half the ſame given. The Theo-

rem is L- = Arch AM nearly.

Call the Radius AC, a, the Arch AM, 2,
the Chord AM thereof A, and the Chord Am
of half the ſame B. Then will 4 (mp =
twice the Sine of Am =+2) be* = z — A3.

y A 25 a ;
—, Cc. and B =; 2 —
4.644 TIT 208″ * 8 ?
Ay 25

ä — —, Oc. Now mul-
CET r 4

tiply B by any ſuppoſed Number x, and from the
Product ſubſtract A, and that the ſecond Term

; 123 25
O — = \
f theRemainder r Tong ny vaniſh,
make it. Then there comes out z; and ſo

$ Bom A=32 * LE +, Se. that 18,

Scho-

[ 106 APPEND IX.

SchoLIUM III.

y the Sagitta AP of any Arch M Am be
| en I AS out, and it 4 requir’d to find
[if the Point F in the ſame, from which if the
{i} right Lines FME and Fme be drawn, they
ll may include the Part Ee of the Tangent to |
the Circle in Avery nearly equal to the Length
of the Arch Mm. |
Let C be the Centre, and AG a the Dia-
meter, and the Saggita AP x. Then will

3 4
2

PM ( -=) be -= —
oat 8 4?

2 | 12
= s. and AE (= Arch AM) =a* x
164* |

1 I 2
n EZ + — ＋ Oc. But becauſe
6a? 40a 11 2.47
the Triangles FMP, FE A are ſimilar, there-

| 4 E 2
j fore AE — PM Eos)
l 44 123 64⁴³
L 2 4 I 2
AP :: LE (+24 Ren
6a* 40a 1124

&c.) AF=ja—jx— PX or- Ge.

Now let us ſuppoſe I F Yα -g. Then if
AH be taken =; AP (x), and & F be taken
equal to HC, a right Line drawn from Frthro’
M, and another thro’ zz, will cut off Ee near-
ly equal to the Length of the Arch M Am;

the Error being only == e or — &c.

z ra
oy Scho-

APPEND IX.

SCHOLIUM IV:

107

Ax if the Area of any Segment MAm of p, d. 39.
a Circle be wanted, nearly true, reduce

the ſame to an infinite Series, viz. let the Seg-

34 a Fe

ff 148

oat yu
, Oc. to which + A M-+ PH APwill
364 K* 3

be nearly equal, the Error being only og ar-
Sec. too little.

EAM n

78 reftify the Ellipfis, or find the Length
f any Arch AM thereof.

Let the Semi-tranſverſe Axis be 4 C, a, and F 1 6. 40.

the Semi-conjugate Cc ; and let AP =x,

PM=y. Now the Nature of the Curve will
APXPaxCc ——

be expreſſed thus = —=PM that

2 25 —xx=yy, And throwing both

Sides of this Equation into Fluxions we ſhall

have 2aX—2xx => X JJ, or c-. 5 9

, * .

and fo # =497, and ſquaring both Sides & =
| —*

e :
Canin a And ſince from the Na
| Gg 2 ture

108

APPENDIX.

ture of the Curve bb x 24x —xx =aayy, and

b4 X 24x — xx =aabo)y if aabbyy be put for
its Equal in the Denominator, we ſhall have

“— INS a yy 3% ayy*
W — * and adding

j* toboth Sides; 4 ＋π = = =; — 5

and conſequently Vx +5 = =Y 5 EEE a” 75 yy

= Fluxion Mm of the 1 A 5 of the
Ellipſis. And ſuppoſing 5 , and 4 — 56
(=4a*—1) =c; — will Vm the Fluxion of

the Arch be yy =; and the Fluent of the

mom 22. 2
ger ag 3. 7241 1

1 dy?
7 Z. 1 OE”
But to find the Length of the Arch Me of
the Ellipſis, you may proceed thus: Here let

PC be x; then will e —PM

_ F – ia
that is, aa * – =7); and ſo aa—xx N bb

=aayy. And chiowing both Parts of the E-
quation into Fluxions, we ſhall haye 2x

= = 240), or —bbxx =aay), and a * £2
bbs

and ſquaring both Sides ; x EET wh Now
from the Equation of the Curve we ſhall get
Ye p and lo bx” =aabi—}**, and

ſo

„Oc. ſuppoſing u=1 8

APPENDIX;
ſo ſubſtituting aab1—4y for 5x in the De-

; | : 5 451; Ls +.
ominator, and —
J a aab—aby e

and then adding) to both Sides, there will be
had x* ＋ 3 — ( 2 5 FIT TE 25 —— L

bi
r

xy”) = — N 2 25 ”’; therefore x57?
— a —tfyy 5 *
== Fo 25 = Fluxion of the cs
Mc. Which is the fame as the Fluxion of
the Arch AM, when AP was cqual to x;
and fo the Fluent of this will be the ſame Ex-
preſſion as the Fluent of the Arch AM.

If Aa be Sa, and CP x, we ſhall have
the Length of the Arch Me =

2.
z2aa 3a

5 M. 1 |
2 d 1 I 5 55 *

b h 2
for — X 2a —xx==y); and * —xx*. = y.

D bb – xx
Whence = x == r
a Fre aa aa—xs

*
aa — xx
a =j +56″ And /the Fluxich of the
Arch

XxX =. Conſequently I 1 — 25

109

110

F 1. 41.

APPENDIX,
Arch fe vill be 3 T .Wir
aa a4 — AN
thrown into an infinite Series will be = 1 + +
| 1 hs
rien
aa aa xx a aa— 9
—— ; “xa

8
OM: XxX 8 ;

— 0/3 i007, x ,
edc
7 „ =
F 2 gen” IN” [7
n —

And the Fluent of this will be

Al þ* xe 2.x” x?
** * a TS top ce
WP 2 i x*
—— T N * — + =p &c
5 þ$ | x?
— To XC
being = to the Series firſt propoſed.

EXAMPLE X.

T- reftify the Hyperbola or find the Length
4 if any Arch A Mitheraaf, = x
| ct

APPENDIX.

Let C be the Centre, a A(2a) the tranſverſe
Axis, Cc () the Semi-conjugate, AP (x) any
Abſciſs, and PM (y) the correſpondent Ordi-
nate. Now the Nature of this Curve is

aPxAPxCc
a 4
=yy. And throwing both Parts of the E-

quation into Fluxions we have 2 ax + 2 x x =

— . 55
= EM; that is, Zan

and ſquaring both Sides x* =

b Xa +*
ay” 3 ©
Tia Ta And ſince (from the Na
ture of the Curve) bb x 2ax + xx = aayy, and
Ax 2ax + xx =a”byy; if a be put for its
Equal 2 ab +64xx, we ſhall have 3* =
.
a’Vi+ab‘yy y :
each Side, and & + j* will be = LL

Then adding 9* to

FFP TP
8 a* SK 12
5* = —— x z*, Conſequentl
: 1 5 in
i +) —_ FT = Mm, the

Fluxion of the Arch 4M; being the very
ſame Expreſſion as that of the Ellipſis in the
laſt Example, only with this Difference, that
in the Ellipſis the Sign of the Term b*yy is ne-

tive, and here it is affirmative. Conſequent-
y the Fluent here will be the fame as the Flu-
ent for the Ellipſis; only with the Altcration

of the Signs, v:z. here a +1 =c,, and c—1
=},

Note,

111

112

4 + 2ax+xx

F16. 38.

Fic. 26.

APPENDIX.
Note, If the Hyperbola be an Equilateral

one, then 2ax E =yy expreſſes the Nature

of the ſame. This thrown into Fluxions will
be 2ax+2xx=2y5, or ax+xx =, and x =

Which ſquared, and then & =

a-.

2 and ſubſtituting yy (=2ax+xx)

1
e
Conſequently y/x* f = 3 / 457 = Flu-

xion of the Arch AM; and making a==1, the

L229 . The Fluent of
I +

which will be | EE 1D

in the Denominator, & will be =

the ſame will be /

SCHOLIUM:

104
in the Hyperbola, QF = 49+ = —
0

AP, and the Secant FME be drawn; then
the Tangent AE will be nearly equal to the
Arch 4 M of the Ellipſis or Hyperbola, if the
ſame be not very large.

EXAMPLE XVI.

7047 0 refify the Cycloid, or find the Length
of any Arch AM thereof. Let
2

APP E ND IX. 113

Let AQ=x, AB=1; then 9g =MS=
x, and PD =y/x—xx from the Nature of the

Circle: therefore A P = /x=x*. Conſequent-
ly becauſe the Triangles 4 P.9, Mm are ſi-

milar 4 Q (): AP ():: MS (8) : Mm =
„ =Fluxion of the Arch 4M; the Flu-

ent of which will be 2x*=2 /?2—Arch AM.
So the Length AD of ; the Cycloid is = to
twice the Diameter AB of the generating
Circle, which we know to be true trom other
Principles, |

LL X A M$: 2 AL

51.7 find the Length of any Arch AM of
the Cifſoid of Diocles A MI.

Let AB the Diameter of the generating p. s. 42.

Circle be a, and call PP, x, and the Ordinate
PM, y); then APS -x. Now from the

Nature of the Curve PM (3) is = A Px 5

a — A — 3
that is, y =a—x = a —x +

And throwing this Equation into Fluxions, we
3 1

1 T
get ) -x N = —ix KANN

— —— „ k F
= * * Which ſquar’d, and

Pan GIANTESS NNE
4x

3 .
e x*. Whence j T =

a* + 3aax
4.x?

— x

KA and fo FA A N=

8 8 —ä]— > —p—
Dax *xXy/a+3x.

Now

— j——U—U—ñäẽ—öhã ! — Vee FAS — 7 not II > _— — =

114

APP EN DI X.
Now this muſt be compar’d with Mr. Cotes’s
4th Form 42 , ez. And

making z , d a, | =—1, a=1I, e==a,

f=3, we get P WEE = SERV
=/3- W/E )=y —.— nd A

a
R
oy

And ſo the Fluent — 12 7 — : dq R

becomes the Fluent of the given Flu-

a+3x
/ 3 +
—.— – +ay3 2 br =
92
Y 1 ee
But this 8 muſt be alter’d before it can

expreſs the Arch AM, becauſe x begins at B
and not at A.

In order to this, make ==, and the ſaid
a KY.

ian 7
from which if the other Fluent be taken, we
a + 2x

xion — 5

Fluent becomes =— 2 a + 3/324

ſhall get 2 „ met 4 ob

— 24 |
Zy 544 PETTY — — (remembring that
rheSubſtraftionoftheRatio/ = — rom
A
Va + 14

the Ratio 2 is the Diviſion of the latter

by che former) N the Length of the
Arch AM.

Now

APPENDIX.

Now for the Conſtruction of the Fluent.

Draw AC (V cutting the Aſymptote in C
ſo, that CAB be 3 of a right Angle; and let
BD (yax) be a mean Proportional between

AB (a) and PB (x); and draw C Dax).
Alſo draw AE biſecting PM, which will be

2a 5 Z Then the Arch AP of the Ciſ-
ABT AC
BD+DC
a+y3Faa
ara.

ſoid will be =2 AE—2 AB ＋＋ 30

e —24＋3 7144

116

F 16, 7.

APPENDIX,

SECTION V.

Of the U/o of Fluxions in the Cuba.
tion of Solids, and in the Quadra-
ture of their Surfaces.

P R O .

283 78 cube or find the Solidity of any Solid
generated by the Rotation of a plain Figure
AMN about the Axis AQ.

Draw the Ordinate pm infinitely near PM;
then the Parallelogram PMrp may be taken
for the Trapezium Pmp; and conſequently
the Cylinder deſcribed by the ſaid little Paral-
lelogram PMrp, while the Figure AN. re-
volves about the Axis 4, may be taken for
the Increment or Fluxion of the Portion of
the Solid, generated by the Rotation of the
Portion AMP of the plain Figure; and the
Fluent of that Fluxion will be equal to the
Solidity of the ſaid Portion, conceived as made
up of an infinite Number of Cylinders of in-

flinitely ſmall equal Altitudes.

Now let 4 P=s, PM=y, and the Ratio of
the Radius of a Circle to the Circumference,

be, expreſſed by =. Then will the Circumfe-

‘ rence of the Circle deſcribed by the Radius

APPENDIX. 117
PMbe = E, and the Area of the ſaid Circle
will be 2 . Which multiplied by P (q) is

5 ) = Solidity of the Cylinder aforeſaid, or
.

Fluxion of the Portion of the Solid. And if
for y* in this Expreſſion you ſubſtitute its Va-
lue gained from the Equation of the Curve
AMN, you will have a fluxionary Expreſſion
affected with only one unknown Quantity x
and the Fluent thereof will expreſs the Soli-
dity required.

1 Þ Ex AM IL E I.
7 53. 1 cube a right Cone. |

A right Cone may be deſcribed by the Ro- F 1 c. 43.
tation of a right-angled Trang ABC about
the Side or Axis AB. Now let A B—a, BC
=, AP==x, PM==y. Then becaule the Tri-
angles APM, ABC are ſimilar, 4P (x) : PM

07A (a): BC ); therefore = =y; and

1 V DS wo”

yy my FF

ſquaring both Sides = y*. Therefore

(= Fluxion * of the Solid) £& Br 1 = * Art. 52.
27 24 T7

SE by ſubſtituting’— for y) = the Fluxi-
; on of the Solidity of the Part of the Cone
| generated by the Triangle APM. The Flu-
ent of which will be = 25 = Solidity ofthe

Part of the Cone; and if you ſubſtitute à for

— —— —ͤ—e— —

118

Fre. 44.

APPENDIX.
x, the Solidity of the whole Cone will be

Pt laprx tas 2p 34; that is, the Baſe

is to be multiplied into F of the Altitude.

EX AMP L E II.

71 cube a Sphere, or any Segment thereof.

A Sphere is generated by the Rotation of a
Semicircle AB a about the Axis or Diameter
ACa. Let the Radius A Cr, AP=x, PM
=; then from the Nature of the Circle ge-

25
nerating it, 9y=2rx—xx. Whence 2
px — p
1

Sphere generated by the Rotation of the Se-
mi- ſegment of the Circle A M P; and the

Fluent thereof will be 22 — = Solidity
of the Segment 4 Mm. And if the whole

= Fluxion of the Segment of the

Diameter 27 be put for x, the Solidity of the
whole Sphere will be = 2 = apr t—

6r
gr =Fpr*=27px7Fr;, that is, the Rectangle
under the Diameter 27, and Circumference p,
is to be multiplied by © r) a third Parr of the
Radius, or ſixth Part of the Diameter. And

if the Diameter 27 be = x, then the Solidity
of the Sphere will be > p.

5-04 05:1.

H/ NCE a Sphere is equal to a Quadrangu-
lar Pyramid, whoſe Baſe is the Rectangle
under the — of the Sphere 2x, * the

L 5 eri-

ale

APPENDIX. 119
periphery deſcribed by the ſame, and Altitude ;
equal to the Semidiameter of the Sphere.

Cox OL. II.

A? becauſe the Solidity of a Cylinder cir-
cumſcribing the Sphere is pr*; therefore it

is to the Sphere as pr to; p, or as 3pr* to

2pr*, or as 3 to 2.

EXAMP L E III.

EDBF be a right Cylinder, and the Part
DB MAF be cut off from the ſame by a
Plane DF A paſſing thro GC the Centre of the
lower Baſe, and F the Extremity of the Diame-
ter FD of the upper Baſe; it is required to cube
the ſaid Part DBMAPF: which is called the
Ungula or Hoof.

Let J be at right Angles to BE=2r, Fic. 45.
draw CF, any where in 4D take the Point
P, draw PM parallel to CB, and PN paral-
lel to C; join the Points M, N, and call the
Altitude FB, a. Then any Triangle PMN
(formed thus) right-angled at M, will be ſimi-
lar to the right-angled Triangle CBF. This

being allow’d, call CP,x; then will PM* =

rr—xx, and CB (r): B F(a):: PM (yri—xx):
MV = yrr—xx; and the Area of the Tri-
angle PM N will be? PMx MN = rux.

Which multiplied by Pp (x) will be —

= the Fluxion of APN the Part of the
0 3
Ungula; and the Fluent will be _ ——=

120

F 1G. 46.

APPENDIX.
Solidity of the ſaid Part APM N, and when
x becomes r, then will this laſt Expreſſi-
on be ar, = Solidiry of ⁊ the Ungula, and fo
the Solidity of the whole will be Tar.

EX AMP LE IV.

76 of bs cube a Solid generated by the Rotation of
the Part Fm M of the Lunule FAD of
Hypocrates about the Radius E. D as an Axis.

Draw the Tangent Aa. Call the Radius
CE, 1; then the Radius C 4 will be /2. Allo
let EP, the Ordinate PM=y, and Pm==z.

Now CA —TCP —PM* from the Nature of
the Circle; that is, 2—1—2x—xx (=I—2x
—x) =yy. And ED —EP = Pm that
is, I —xx=22. Whence the Area of the Cir-

cle deſcribed by PM will be — ==; and

the Area of the Circle deſcribed by Pm will

be 2 — ned and the Difference of. theſe Areas
will be = = Anulus deſcribed by Mm. And

this drawn into the Fluxion æ will be E & =

1
to the Fluxion of the Part of the Solid deſori-
bed by the Part Vm M of the Lunule; and the

Fluent thereof will be = = Solidity of the
ſaid Part.

E x A M-

N x F +a et 222 , S OEM -7
x pf * 4 9 5

mes R ra

| a, BD=r, 4P=x, PM=y.

APPENDIX, IZI

EXAMPLE V.

„ cube a Parabolical Conoid generated by
the Rotation of a Parabolick Space of

any Kind about its Axis.

Let the Parameter be = r; let AD be =F 1«. 1.

Then the
Nature of all Parabola’s will be expreſſed by

yx. Whence y=x”; and ſo yy = x”;

therefore e = Fluxion of the Solid
generated by the Rotation of the Portion APM
of the Parabola. And the Fluent thereof will
mp m +2 a

x z and ſubſtituting 5* x for

m+2, mpy* x

Equal, there will come out
. – 4 + 2mxr
= Fluent of the ſaid Solid.

And if 4 the Altitude of the whole Conoid
be put for x, and 27 the Diameter of the Baſe
for 2); then the Solidity of the whole Conoid
will be — —— X aprt pr x mock

a+ 2mw 472M 2 Amn

Hence if the generating Parabola be the

=— . Whence the Baſe is to be drawn

into ⁊ the Altitude; and conſequently the ge-
nerated Conoid will be + a Cylinder of the
lame Baſe and Altitude.

| I i Ex A M-

_ – 10 ů —
— 7 .
—

— — — eg ed
2

— — >, .

nn, pay” ue” ES CGR” 41
88 1 —

— „ w . – ;
. — —
— e_—_ aud.

— —

— ä — CECESSEInT
“a a –

_—

—

—ä + —_

—
ES — ————

3 _—
—

—— —
— –

122

APPEND IX.

EXAMPLE VI.

. 78. ſas cube a Spheroid generated by the Ro-

F 1G. 40.

a Ari. 52.

4
will expreſs the Nature of the Ellipſis. There.

tation of a Semi-elliptick Space about its
tranſverſe or conjugate Axes.

If you make 4 P Dx, PM=y, and A a=,

and the Parameter be — 4; then will PM be

2 AP 5 that is, N
A a

27 21 21 2ra
will be the Fluxion of the Solid generated by
the Rotation of the Part of the Ellipſis APN;

fore # BY g N ; _ poxs pots
27

and the Fluent of the ſaid Fluxion will be

or D —the ſaid Solid.

44 Gar

And if the whole Axis à be ſubſtituted for
x, the whole Spheroid will be Laß — 22 —

6r
—
ide

Hence if the conjugate Axis Ce be 2
Then will 47* be Sab; and fo the Solidity of
the whole Spheroid will become 3 par; that
is, the Elliptick Spheroid is equal to a Cone
of the ſame Height with à the tranſverſe Axis
of the Ellipſis, and Diameter of the Baſe equal
to four times the conjugate Axis of the Ellip-

ſis, viz. 4r. And becauſe the Altitude of 2

Cylinder circumſcribing the Spheroid yg 0,

and Diameter Sr, the Solidity of that Cylin-

der is apr; therefore a Spheroid, as well 2
a Sphere, is 7 of the circumſcribing 9
Ot ber-

APPEND IX.

Otherwiſe :

This Example may be effected ſomething
ſhorter thus: Let r7==1, in the general i

ſion * A

— 7 *
Make . AC Da, and PC x, and pt
Nature of the Ellipſis will be 1 — 15 =y,;

then will the ſame become = 2

and ſubſtituting 1 — 2 for yy in the ſaid gene-

br bt

—— — — —

ral Expreſſion 1 and — will be
2 2aa

= Fluxion of the “he of the Spheroid gene-
rated by the Rotation of the Part MPcC of
the Ellipſis, and the f thereof will be

== Soldity of the ſaid
2 6Gaa 2 Ry

Part of the Spheroid. And ſubſtituting à for

„ we ſhall ny 24 Sig is $2

2 3 2 3
36 E but 2 = Circle deſcribed by the Ro-

tation of Cc; therefore the Solidity of the
Spheroidis = + of a Cylinder of the ſame Baſe
and Altitude. |

EXAMPLE VII.

.
by the Rotation of a Semi-Hyperbola a-
bout the tranſverſe Axis.

Let AP=x, PM=y, the Parameter = 5,

and the tranſverſe __ d. Then the Nature
FE

Ari. $3

O cube an Hyperbolical Conoid generated F 1 6. 41.

= „
— ——— — — — 1 _— — —
. — —ꝛ —— — —— ů — 8 2
%
%

OOO Ours en A Co Gee OA ̃¶ Qn… A IP eons .
“
x

|
ot
on
4
| 3
|
|

124

APPENDIX.

of the Curve will be APN 3 that

is, be += Therefore* 222 der +
82

— is = Fluxion of the Conoid generared

by. “the Rotation of the Part AM of the
Hyperbola about the Axis 4a; the Fluent of

which will be 2 += 775 — Solidity of the

4 Gra
aid Conoid. And when the Altitude of the
Conoid is equal to ; that is, when x is = a,

the Solidity of the Conoid will become
wt I . i ba*. And if 27

be = conjugate 52 then will 27 . V,

and 4 Aj; which Value being ſubſtituted
in the laſt Expreſſion, and the Conoid will be

If the Hyperbola be an Equilateral ‘one, then
the Nature of it will be expreſſed thus, y*=ax

Tex; whence 25 E, and the

Fluent thereof will be — ＋ And ſo ſince

here 27 , and h, the Solidity will be 59.

Hence the Solidity of the circumſeribing
Cylinder will be +par;z which therefore vil
be to the Solidity of the Conoid as + par to

3 or as 3 to 103 and in the Conoid gene-
_ rated by the Equilateral Hyperbola, the cir-
Lutter bing Cylinder is 3 pa*. Whence *

ſame will be to the Conoid as tp to; pa- or
as 3 to 10. .
By making P G=x, and Cr, and AC=4,
a5 before in the Ellipfis; then the Nature of che
Hyper-

APPENDIX. 125
t Hy perbol will be Jy =—=—1. Whence — I

e and the Fluent thereof will be

ted by to Part AMP of the Hyperbola.

EX AMS LE VIII.

1 cube a Solid generated by the Rotation F 1 a. 47.
of the interminate -byperbolical Space

CABED about one of the Alymptotes CD.

Let AB=a, AC=b, C P=x, PM; and

let P; chen let the Circumference of the

Circle deſcribed by the Radius AC be =p, and 1

the Circumference deſcribed by the Radius PC |

wiltbeZ Wich drawn into PR) will give us |

£2, theSuperficies of a Cylinder deſeribed by |
the Parallelogram C PMR. This again drawn
into P (i) will give us the Solidity of the
little Cylinder PpqglM, viz. _ = Fluxion

of the Solid; but the Nature of the Hyper-
bola with regard to the Aſymptotes is x Dab.

Whence y , 2, and E lle mph;

the Fluent of * 5 * ald) of the
interminate Space CAB ED; and if à be put
for x, the whole Solid will be pba.

2 E x A M-

126

APPEND IX.

E x AMP LE IX.

of bas cube a Conoid generated by the Rotation
of the by — Space AMB DC about

FI o. 48. CD the half of the conjugate Axis of the Hy-

perbola AM B.

Call C 4,0, CD,b, CP, x, PM, y, BD, r.
Now 7 – will be che Periphery deſcribed by

the Point M, and 222 – 2 2 = = the whole Circle,

having PM for a Radius; ; which multiplied
by x, and — 5 will be the Fluxion of the So-

lid. But yy 2 57 _ from the Nature of

the Curve; and r this Value in

A, we have — wor the Flu-
27 Zh or

xion of the Conoid, the Fluent of aug is

„, boyy
2 + . and ſubſtituting 55 for
8 D I 1 *
e n 6rxx+6bbr + 2rxx + 2bbr ©
Solidity of the Conoid formed by the cs
AMC; and when x becomes b, and y to

7, then the whole Conoid will be == 2.

Conor.

AY C ylinder —— by the Rotation of the
Parallelogram ACSB about the Axis CS

Is :pba; and ſo it is to the ſaid hyperbolick

Solid

w OO

Ts wes

APPENDIX. I27
Solid as Z pa to pba, that is, as r to 1, or 1
to 2. |

ExXAMreLE:&A

70 cube a Solid generated by the, Rotation
of a Parabola about a” Semi-ordinate
CB.

Let AB, B C=, A Px, PM=zy; then FC. 49.
if the Parameter be 1, this Equation will ex-
preſs the Nature of the Parabola y*=x. But
it is manifeſt *, that the Fluxion of the Solid * At. 52.
is the Circle deſcribed with the Radius MD
drawn into Dd=y. Let the Ratio of the
Radius to the Circumference be as r to p; then
MD=BD=AB—AP==r—x. And the
Circumference of the Circle deſcribed by MD

=p—E=, and the Area of the ſaid Circle

vil be 2. — 5 +25. Whence the Flu-

xion of the Solid will be 5 j—pxj +£ 5

Now if in this Expreſſion for x and x* you
ſubſtitute )“ and y their Equals (from the E-

quation of the Curve) we ſhall have 2 DJ

+2 equal to the Fluxion of the indefinite

Part of the Solid generated by the Rotation
of the Portion HCD about the Axis BC;

the Fluent of which will be #py — 357 ——

PI” aid Part of the Solid.
I Or

But if for y* you put x in the general Ex-
preſſion *, then the ſaid Solid will be py— . 4. 52.
pou |

a zor

128

APPENDIX.
CN) mn —2 SET And if for

307 lor 307 lor
you put &, and for x you put 7, the whole

Solid will be px Tr r rebr=3O— o 6

555 =5>pbr =Fprx:i76; that is, the Baſe or

Circle deſcribed by the Radius AB is to be

drawn into i of the Altitude BC.
Becauſe a Cylinder of the ſame Baſe and Al-

titude is 2 pr; therefore it will be to this pa-

F Ic. 28.

FIC. 27.

rabolical Solid as 3 pbr to T NI; that is,
as I to x, or as 15 to 8.

EXAMPLE XI.

T O cube a Solid generated from the Rota-
Lion of the Logarithmetical Curve about

the Aſymptote AH.

Here the Subtangent being always = a, yx

is g)); and ſo i= Whence 1

= Fluxion of Part of the Solid generated by

the Rotation aforeſaid; and the Fluent thereof

will be 2 And putting A for), then

will the whole Solid be por. =74Þr. 1

Hence becauſe a Cylinder, whoſe Altitude
is = 4, and Radius of the Baſe = 7, is T apr:
therefore it is to the Solid as + 4 to 4 a, or as
2 to 1. |

Ex A M P l. E XII.

To cube a Solid generated by the Rotation
of the Ciſſoid about the Line AB as an

Axis.
Let

APPENDIX. 129
Let AB=1, AP=x, PM=y; then the

Nature of the Curve will be) — and

DK x3x : :
oat = = z that is, (making 214

27 ZII —Xx

=1)= = The Fluent of which will be

=7zpxt TTD TTD Y D*, &c. = Por-
tion of the Solid deſcribed by APM. And
putting A B==1, for x, we ſhall get 4p fp
2 % +752, &c. or px3+++#& +73, &c. But
this Series is infinite, as may be eaſily demon-
ſtrated from the Hyperbola; therefore the ſaid
Solid is infinite alſo.

Otherwiſe by the Meaſure of a Ratio.

If the Area 4 PM of the Ciſſoid of Diocles Fo. 50.
revolves avout the Baſe AB as an Axis, it is re-
quir d to cube the Solid generated thereby.

From the Nature of the Curve, calling AP, x,

; ; and PM, y, we have & — =y. Whence
DIX =D; > = Fluxion of the Solid re-
27 27 a—x

quir’d. Which compared with the firſt Form
in Mr. Cotes’s Tables, by making 4 = 28,

| | x? axx
1=1, ea, f= I; and we get == _==

41
—— _ ed” [nat The Fluent of the given ·˙¹

px? pax __paax , fall a

; 6r ar 27 ‘ 21x

& Since the Logarithm of the Ratio @ to a—x

= with an affirmative Sign, is the ſame as the
: K k Loga-

Fluxion; or —

Ts, ER I EEEFPTITY a
— er — — — — – cn _ äñä —— — ne
_ * — — ne —
w = — eb” wecs

*.
OT 1
—— = — = w—_—

—— — > a”
— woe ens

— —
” om — 4
2 [2-0 4 *

PN. uy ws ns —— — —
:
wa; —— — — w
_ + — —— otros 1 Ap mera. ——

—

APPENDIX.

Logarithm of che Ratio ax to @ with a nega
tive Sign.

Now this Fluent may be thus conſtructed :
aa a5
Let 4 PGO, AB(a), AR (=) AS (=

AT ( 2 be continual Proportionals. Then
to the Module 78 (==) aſſume P

equal to the Meaſure of the Ratio between
AB (a) and PB (ax); that is, make PX
Sn = z and from & towards B, lay off
X equal to & R 7 +=8 we. ICs +

add -A , G—=Xx
—+

2 z and the Solid will be equal to
a Cylinder, whoſe Baſe is PM, the Value of

which is = and Altitude PZ, che Va-

Aa—Aax Aa—x

lue of which is —

—
XX ZX 3

„„ 42 — 4 „ | 2 x3 *

x? 2— 27 a2— K&K ᷣ —

al— a da- àa— 4 — N a “IM

3
r , the Flu-

ent to be conſtructed.

SCHOLIUM.

65 Tu Valueof the infinite Solid generated
by the Rotation of the Ciſſoid AMI
about the Aſymptote 5 OH may be had thus:

Let

Fay bas

APPENDIX,

Let AN be the generating Semicircle Circle,
AP, x, AB, a; then all the Ordinates PM do

‘ deſcribe cylindrical Surfaces. Whence 7:p::

PB x PM (xyax—xx): = Vaa-xx = Sur-

; face deſcribed by PM; which multiplied by

x, and 2 x x /ax—xx is the Fluxion of the So-

lid generated as above.

Now to get the Fluent of that Fluxion, let
us conceive the Generation Semicircle to re-
volve about an Axis parallel to the Aſymptote
BO H, and paſſing thro’ the Point A; then all
the Ordinates PN will alſo deſcribe cylindri-

cal Surfaces; and 0 Van xx will be the
Fluxion of the Solid generated by this Mo-

tion, which is equal to the Fluxion of the for-
mer Solid to be cubed. W hence the infinite
Ciſſoidal Solid formed by the Revolution afore-
ſaid, is equal to this latter Solid generated by
the Revolution of the generating Semicircle
about a right Line parallel to the Aſymptote,
and paſling thro’ the Point A.

P R O p. II.

131

TO meaſure the Superficies of a Solid gene- p 1 g, 5.

rated by the Rotation of a Figure
AMNQ about its Axis AQ.

Let the Ratio of the Circumference to the Dia-
meter of any Circle be”, let P, PMA; then

will Pp. MAN x, qm=z ; and ſo Mm=y/x+),
the Circumference deſcribed by the Radius

Kk2 PM

132

APPENDIX.
PA =. Which multiply’d by, or drawn

into Mm, will be = 2/ x+j5=PFluxion of

Part of the Superficics of the Solid generated
by the Rotation of the Part 4M of the
Figure AMV; and finding the Value of x-
from the Equarion of the Curve thrown into
Fluxions, and ſubſtituting the ſame in the ge-
neral Expreſſion; the Fluent thereof being
afterwards found, will be the Superficies
ſought. |

SCHOLIUM.

I F N be any Fluxion of a Superfi-
cies generated by the Rotation of a plain Fi-
gure about the Line * as an Axis; and if – be

the Ratio of the Radius to the Circumference

of a Circle, and y repreſents any perpendicular
Ordinate to the Abſciſs deſcribing a Circle du-
ring the Generation of the Superficies; and if

= 2z be ſuppoſed the Fluent of that Fluxion,

viz equal to a Circle whoſe Radius is z; then
will the Fluent of 2 y4/xx + jj be = zz, the
Square of the Radius of the ſaid Circle. Con-
ſequently if inſtead of finding the Fluent of

the Fluxion – 1/xx +77 of any Superficies ge-

nerated, as above, you find the Fluent of
23y/xx +5yz this Fluent will be equal to the
Square of the Radius of a Circle equal to the

Fluent of the Fluxion – Va + Jy ,or toa Cir-
cle

ApPEND IX.

cle equal to che Superficies, whereof this laſt
Expreſſion is the Fluxion.

KW 5

3 find the Super ficies of a right Cone.

Becauſe a right Cone is generated by the F1e. 43.

Rotation of the right-angled Triangle A5 C
about the Axis AB, the Value of x* muſt firſt
be gotten from the Fquarion of the Triangle
thus: Let A B=a, BC=r, A P=x, PH y.
Now fince AP (x): PM(y)::AB(a): BC(r):
therefore x % Which thrown into Flu-

. ; . . . A
xions will be v / Whence x = 2 , and
rn | “7

REY TT” nk.
6 and ſubſtituting —— for x* in the

general Expreſſion _ Ve) we ſhall get
EFF =EVO FTP = ven

fl
= Fluxion of Part of the Superficies of the
Cone generated by the Triang e APM. The

Fluent of which will be = a aid

Part of the Superficies; and ſubſtituting y for
1, the Superficies of the whole Cone will be
=rpya+r pA; that is, equal to the
Rectangle under + the Circumference of the
Baſe into the Side AC of the Cone.

ExXAMPL E II.

7 find the Super ficies of a Sphere, or of F 1 6 45.

am Segment of it. N

134

APPENDIX,

If the Diameter of the generating Circle be
=1, then the Fluxion of the Arch Mm will

be = – 8 Which drawn into the Cir-
X- Xxx

2 |
cumference deſcribed by the Radius PM, will
give px the Fluxion of Part of the Superficies
of the Sphere generated by the Semi-ſegment
AMP; and the Fluent px of it will be the
Superficies of the Segment of the Sphere, ha-
“_ p for the Periphery of the circular Baſe,
and x for the Height; and if you put the Dia-

meter 1 for x, the Superficies of the whole

Sphere will be equal to 2, or (making 1 a) =
ap.
” Whence any Segment of the Superficies of
a Sphere is to the whole Superficies of the
Sphere as px to p or x to 1; that is, as the Al-
titude of the Segment to the Diameter of the
Sphere.

EX AMG LE III.

T O find the Superficies of a Parabolical
Conotd.

The Equation expreſſing the Nature of the

Parabola is A PXA SPM (9), or ax ).
Which thrown into Fluxions, and a =2yy;

whence & . Conſequently 7 +f

== x/an 7 Fa =2/47 Fo” = Flu

xion of the Superficies of the Part of the Co-
noid generated by the Portion APM of the
Parabola. Which comes under the firſt Calc
of the third Form in the little Table of ſimple

Curves that may be ſquared, Art. 8. or r
the

APPENDIX. 135
the third Form in the Tables of Mr. Cotes. For
if in the little Table you make ==, a,

1=2, aa, f=4, we ſhall have the Fluxion

5 e = ;; and the Flu-

ent 7 of it, by making R (=y/e+fz) =
— R _
a4 will become —— x 45 bas =

Fluent of the Fluxion of the Superficies of the
Conoid. In like manner, in Mr. Cores’s zd Form,

the Fluent anſwerable to f=r „vi. 2 – ap,

making P (Se N) = y4y* +a*; and
ſubſtituting the ſame Values for d. u, e, f, as be-

248 ＋ Y E ITS =
„

E rr. :
L— 4 ＋4 = to that found by the little

Table of Quadratures.

This Fluent may be conſtructed thus: If r
be made = y, and you draw PCD Z), and
make PRS a, and join BC; then will BC be

=y/4yyz+aa. Call this , and the Fluent to be

conſtrued will be 1 n*. Let z be the Dia-

meter of the Circle equal to this Fluent; the

Area of this will be — Therefore 1
87 I 2a)

fore, will become

F1c. 51.

=2″. Whence z = 73 * = * Fluent of Art. 67.

2) xx +3). *
ow

136

Fic. 52,
$3»

ture of the Curve. Whence the Periphery —

Ap PEN DIxX.

Now make PH = BC, and PE BC,
join B, E. From H draw HF parallel to BE. ;
Make GP=PF, biſect CH in D; from which,
as a Centre, deſcribe the Semicircle GK H b *
cutting PM continued out in the Point K; 1
and then the right Line PX will be the Dia- a
meter of a Circle equal to the Superficies of 1 74

the Conoid. 3
4
Ex AMP IL. E IV. I

70 find the Super ficies of a Spheroid gene- ni
rated by the Rotation of any Part AM tl

of an Ellipſis about the Part DC of the Semi- at

axis BC. N

Call the Semi-conjugate Axis AC, a, and
the Semi- conjugate B C, b; the Abſciſs PM Fx

and the correſpondent Semi-ordinate M P, ).

bh
Now — x 2ax—xx=yy, and 24—2 * * =

= XyJ, Or 4a v Whence x =

1
55

„ |
—— and 4. yr Conſequently the

Fluxion of the Arch AM (= will
be = ERS. 2 e
AC a); is greater chan BC ), and you put cc
for a -l) or == ——= – —_ when 4 C (a) is

m—

(when

Again, D Mis =/, from the Na-

APPENDIX.

the Circle deſcribed by DM will be 5 5.

Which drawn into the Fluxion of the
Arch before found, and there * ariſe
ö coy — e

; is greater than BC and == VF=ceon,
when AC is leſs than BC; 9 being the
Fluxion of the Superficies of a Spheroid, ge-
nerated by the Rotation of the Part AM of

N the Ellipſis about the Part CO of the Axis;
and may be referred to the fourth Form in the

Tables of Mr. Cotes.” For making d’= £53
| 2 z, 9 o, eb, Er the Fluxion
N 05 Z ay 9e Y will 1

5 ET). And the Fluent =dP+ < .
Ke: — by making P (=y/ Ez) =

ee (Sf) 6 rut
| 7082 . 28

99 SES)
BY pabt

= 2, W ill — = 4 —
7 1 come – 7 Tec * .
0 cc

3 2 — = It being N 000, and

the Meaſure of a Ratio, when AC (a) is great-

er than BC (5); and ech, and the Mea-

ſure of an — when AC (a0 is leſs than
BC(d).

L. I Now

137

— 2×0″ == 5 * Pt
— _—

138

Fic. 52.

APPENDIX.

Now to conſtruct the Fluent in the firſt
Caſe, we may proceed thus: Let F be one of

the Foci. Make CF (Jaa = cc c,
from the W of the Ellipſis): CB (b):

CB(b):CE= ==, and draw the right Line

DE. Alſo make cx(®):pz(: 7 3+)

:CD(9):KL=L FF. To which

if you add LM, made equal to the Meaſure of

the Ratio between DE (= /J*Froy) +DC

( and CE (=) to the Module C E (>);

that is, if you aſſume LM = 7 2 vio,

I + y/ teen MICE),

c gow”
then will CA (a) be on KM=KL =& L + LM

7 n
(A jg Paco + [= — —

Circle deſcribed with C 4 (a), as a Radius,

(which Circle 15 = = 2 = )
of the Spheroid 1 Ry — Rotation of
the Part IMof the Ellipfis = = 255 7 Fo

is to the Superficics

7 77 z which is the Fluent

to be conſtructed. |
Now let the Radius of a Circle, equal to

the Superficies of a Spheroid, be z. Then
we ſhall have this Proportion, CA (a): K M::

Circle deſcribed by CA, viz. . Superficies
of

AP PEN D IX.
of the Spheroid ſuppoſed equal to a Circle —

| whoſe Radius is 2. And multiplying the Ex-
tremes and Means, there ariſes PEER = K M-.

27
paa Ns 1
2 and ſo za æ Mx a; that is, a mean

Proportional between XK M and AC (a) will be

the Radius of a Circle equal to the Superhcies

of the Spheroid generated by the Rotation of
the Part AM of the Ellipſis.

When DC (y) becomes equal to BC (),
then will DE become BE; and ſo if you

bb hb ——
make CE (ECE) αο
XLS VU c = AC Sa, and LM be =
Meaſure of the Ratio between A4 C (a) + BC
(Y, and CE (= ) to the Module CE =) z

we ſhall have CA to XM, as a Circle having
CA for a Radius is to f the Superficies of the

whole Spheroid.
The &

ter Caſe is thus: In the Ordinate DM aſſume
the Point E in ſuch manner, that the Line CE

being drawn be equal to – or a third Pro-
portional to CF (c) and CB(b); and make CE

bb 12
=) *DE (= CO- |

= Foy. Then if to KL you add LM
equal to the Meaſure of the Angle DCE to
the Module CE (E); CA willbero K,

LI as

139

onſtruction of the Fluent in the lat- F i a, 53}

— — _ = ou — — pay

—

—
— b
— EE eee woes Dag —cqccu——c—_-—-— —
”
— –
—
7 pan
— —

140

APPENDIX.

as a Circle, whoſe Diameter is CA (a), is to
the Superficies of the Spheroid generated by
the Rotation of the Part AM of the Ellipſis
about the Part DC of the Axis; and fo 3

mean Propartional between X M and CA, will

be the Radius of a Circle equal to the faid
Superficies.

When CD (y) becomes equal to CB (b),
draw BG perpendicular to CB. In which

aſſume the Point e ſuch, that Ce being drawn,

be = EC (A =): Then if you make CE or

Ce (2): Be (= Vcc): AC MKL
cc a. And to the ſame you add LM
equal to the Meaſure of the Angle eCB to
to the Module 00 ) we ſhall have CA to

KM as a Circle having CA for Radius is to
the Superficies of the whole Spheroid.

rern .

wy: O find the Superficies of a Hyperboli cal
Conoid generated by the Rotation of any

Part AM of an Hyperbola about the ide 55 ſe
Axis AP. |

1 Let C be the — Cc an Aſymptote, {AC
b one of the Semi- axes, and Ac s the

| _ Let CP be y, and PH x. Now

5 ; = x Y xx, from the Nature of the Curve,

and 2aayy = 2bbxx 5 and fo x = 4 Con-

ſequently * K = _ And the F luxion of
E the

. pa

APPENDIX.

the Arch AM (= jj) +xx) will be
1/2 wry — _9 LE, þ4
Ge 4, yy——bb ?

cc for a +b*, Again, PM is = 7

by putting

from the Nature of the Curve; and therefore
the Periphery deſcribed by PM will be

me = \/ yj—420 aa. This drawn into the Fluxion of

the Arch before found, and there will ariſe
pay -e pay

[- mY m—bb rb Veh, being

the Fluxion of the Superficies of an hyperbo-
lical Conoid generated by the Rotation of the
Part AM of the Hyperbola about the Axis
AP. Which comes under the ſame Form in
the Tables of Mr. Cotes, as the Fluxion of the
Superficies of the Spheroid in the laſt Exam-

ple. And making d—= =, Z=)y, = o, n,

GOES f=cc, P ff R =c,

;
T= oF, 5, we ſhall have the
Flcn of that a thus expreſſed 3

_ y hes at the Centre C, and not at A
the u we muſt make Y =y, Then the

firſt Part Z 7 Ve = of the Fluent will be

=4/ — a; which therefore muſt be ta-
ken from the faid firſt Part. _ Moreover, the

Logarithmical Part — ELSE by ma-

king

AP PEN DI X.

king 5 =, will become — EEE Which

———

EE
” bb

the true Logarithmical Part. Now becauſe
the Logarithm of the Ratio of 4b to ab be
having an affirmative Sign, is the ſame as the
Logarithm of the Ratio of ab-+bc to bb
with a negative Sign (by Def. 1. Se. 2. Schol. 2.)

therefore the Sum of — IIs | and of

= — will be equal to the Difference
fought. But the Ratio of cy+,/ coy —b4 to
bb, and the Ratio of bb to ab + bc, do com-

pound the Ratio of cy4+ 4/ccyp—-44 to ab & be.
Whence the Sum now mention’d will be

bb Ty coy — Therefore the true Fluent
cl ab+bc |

muſt be taken from , to have

to be conſtructed will be = Vc – 75

pab fh Vc y- 3
r And this may be done
thus:

Let F be the Focus of the Hyperbola.
Make CF (aa bb c, (from the Nature of

the Hyperbola): C4 ():: CAM: CE =Z.
C

Draw E & perpendicular to CA meeting the

Aſymptote in &. In the Angle C EG, inſcribe
the right Line CH CP (ö), which conti-
nue out to meet PM alſo continued out in the
Point J. Then aſſume KI. equal to PI Ac

0 * 1 8 5
= v ea, ſince from the ſimilar Tri-

4 angles

APPENDIX, 143
angles CE 17, CPI, we have CE (DD

(Lic). == | |
Again, aſſume LM equal to the Meaſure |
of the Ratio between CH (53) + EH

(ab and CG(B)+ EG (S) to the |
Module C E (= =). Then the Superficies ge- |

nerated by the Rotation of the Arch A M a- | |
bout the Axis AP, will be to a Circle deſcri- |
bed with the Semidiameter 4c (a), viz. — | |

as KM to Ac. 15 a |
Ex 4 MF Ls

pb O find the Superficies of a hyperbolical ll

Conoid generated by the Rotation of any |
Part AM of an Hyperbola about the conjugate _ |
Axis CPB. *

Let C be the Centre, CA Da the Semi-

tranſyerſe Axis, CBS the Semi- conjugate,

F the Focus, PM x, any Ordinate to CA,
and C P/, the correſpondent Abſciſs.

F 1 0. 55

Now 72 * xx — ag =yy, from the Nature | |
| |

of the Curve. Whence ce; that is, |;
bh

__ 4ayy
— * — — –
= x;==yp, and bbxx==aay), and æ = 7 — A, and | i
£5 =, and ſubſtituting < aoyFaab for ar ij

aa aN wy Þ x bb+yy |
| W hence I!

5 Which ay be thus conſtructed.

APPENDIX.

| 7 2 2 2 4 2
Whence Y — ETC” But

bv 7 +39
x is = 3s 73, which multiplied by, and

257 bb + Jy will be the Periphery deſcribed by

the Point M; z and if the Fluxion of the
Arch be drawn into nato this, we. ſhall get

paj 2 ee — from ry?
3 bb + yy N * +

And ſubſtituting cc for 0a? 7 the ſame will

become Va FR. Which is the Fluxion

of the Superficies generated by the Rotation
of the Arch AM about the Semi-conjugate
Diameter CP.

Now this Fluxion being the ſame as that in
the firſt Caſe of Example 4. aforegoing, the
Fluent * muſt be the ſame as the Fluent

of that, VIZ. 1 io — 7 4 Thott

Make CF (aaf): CB(b)::CB (b)

OP, and draw the right Line PE

(> Viey+#.) Then make C E (20 PE

Cc OEL = ef.

And aſſume LM equal to the Meaſure of the

Ratio berwoen PE E (> Vion) +PC(9);

that is, bergen CEPOTE + 2 and CE
C

af 9 A —

APPEND IX.

18 to the Module C E (=) And the

Superficies generated by the Rotation of the
Arch AM of the Hyperbola about the Semi-
conjugate Axis C B, will be to a Circle deſcri-
bed with – Semidiameter C 4 (a), (which

Circle is A =) as the Sum of the Lines K L
and LM 10 = ſaid Semidiameter CA.
Ex AMP LE VII.

T® find the Superficies generated by the
Motion of the Arch CM of an. equila-
teral Hyperbola about the Ahmptote ABP.

145

Let A be the Centre, and let AB be equal FC. 5:

to BC, drawn parallel to the other Aſymptote

aſſume AP, and draw PM parallel to the o-
ther Aſymprote. Now call 4B or BC, a,
JP: x, and PM,y. Then from the Nature

of the “+ aa=xy. Whence x = 75 and

_ Sy. Which thrown into F Cas and

= Conſequently jy== — + Again,
x =£Z thrown into Fluxior _ = =;
I, N ona Aejlfer.

bo. of * Arch C M. Which drawn into
25 and the Product ve + = j | & els
M m will

146

APPENDIX.

will be the Fluxion of the Superficies gere-
rated by the Rotation of the Arch CM as
aforeſaid. This Fluxion can be compared to
that of the third Form of Mr Cotes’s Tables,
and the Fluent will be had by free =

I, os beat, , PV N.
Raa, SE and ren for then
the ſame will be = £ 2 2 — 5 ad
——

. But this Fluent muſt be cor-

| ated) 8 the Abſciſs AP (x) increaſes,

while the correſpondent Ordinate PM ( y) de.
creaſes; therefore the Signs muſt be changed;

4% T —— I
that 15, it will be —2 5 Va ＋ * + 1 7 ad

1 =, Moreover, ſince the ſaid Ab-
ſciſs AP (*) hayins at A, and not at B, the
faid Fluent muſt undergo another Correction,
as to Magnitude. And this is done thus:
Make ya, and then the firſt Part of the

Fluent will become 3 pr aay/z. And this mull
| 15 3
be added to — 5 ya* -+ 5*, and the Sum will

be 35 ½—25 P Vas, the true firſt Part

of the Fluent. Again, the Logarithmick
Part = 24 e —
y

— muſt be alter d;
ch]: may be 2 5 making 2 255 for
. Whichbe

ing

then it will become – – = 2 —

n
4 Fn = * +.
2 £ *

ESE

APPENDIX.
2 EY 5)

aa yat+ 54
and then we ſhall have aa 2 * 77

ing ſubſtracted from – : 6 a a

which is the true Logarithmical Part. And
conſequently the true | Fluent will be

14 yaa +5 Vas +”

4
= Eo ENF + Lee ”

which may be thus conſtructed.

Draw AM and AC, and from the Point C
draw. C & parallel to A M, meeting the Aſym-
ptote 4P continued out in EG. Now becauſe
of the ſimilar Triangles A PM, A BF, AP

E 05 PM (9): l B＋ = 2 ? Whence

– Va 75 Ar. . becauſe G CC is pa-
ll to AM, the T riangles APM, GB C are
ſimilar; therefore PM ():4P 22 N CB (a)

PERS
20 Whence 00 = V e + *.
Then if you make AH=AC (4/2) —AF

(- /a*F3 4?) + the Meaſure of the Ratio be-

tween B& (=) +60(S/7F5), and
AC (av2) 4 5 AB (a) which Ratio is =

7 ) to the Module AB 01 z and

“ny 2 2 1
the 8 generated by the Rota-
Mm 2 tion

1 ff 1 Id : RE
App FE N DI X.

tion of the Arch GM about the Aſymptote
AP, will be to a Circle whoſe Semidiameter

in AB (, viz. == as A H (a Va —

+ ya +3 + *

LD . N J 2 .
| ayz+a – |

and therefore the Value of AH drawn into

2 will be equal to the ſaid Superficies.

E x AMP LE VIII. |

7 O find the Superficies generated by thi
Rotation. of the. infinite Arch PZ of the

Logaritbmical Curve about its Aſymptote AX.

Let AP be an Ordinate at Tight \Aboſes to
the Aſymptote, which call y; let TP touch
the Carve in P, and let the invariable’Subtari-
gent AT be = 4. Draw pm parallel to AT,
and infinitely near the Point P.

Then becauſe the Triangles ATP, mp Parc

dne O. 7 ( +39)::mP(3):pP

== y/aa +7 + »y=F luxion of the Arch PZ.

This ; multiplied by 2 9 and the Product
25 yaa +51 yy will be the Fluxion of a Square,
whoſe Side is equal | to the Diameter of a Cir-

cle equal to the Superficies ſought by Scho-
Tium, Art. Gy. which Square let = 404.

Now this Fluxion may be referr’d to the
fourth Form of Mr: Cotes’s, s, by writing 9, 2,0

25 aa, 1, for z, 10 d, e, f. And ann
for P, R, ＋. 5 1 . 157 . 7p, and

APPENDIS.

E, we ſhall have the Fluent of the Fluxion

J
27 Y equal to v Ju [yes = LA

In order to conſtruct this Fluent, ‘we may

| obferve that the right Line 1 O is a mean Pro-

portional between a and 7 7 aa +35 +

4 —— an + 2 . the Quentics 2 U 424 Y

is equal to the right Line EP at right Angles
to the Curve in P, and bounded by the A-

149

ſymptote in E. For becauſe of the ſimilar

Triangles TP, APE, TA (a): Ti PH 0

P ( 3): PE, = vas as +3) , 5 And the other

Quantity a — — is the Mexfure of the

Ratio — 4 P 4 of of and 4 T to the
Modiie AT. Or (becauſe of the ſimnilar Tri-
angles A PE, APE) the Meaſure of the Ra-
tio between AE EE, and AP to che fame
Module AT. .

Hence the F os Is thus conftrudted, Draw

PE 02 —yaa+ aa +») perpendicular to the Curve
in P terminating at the Aſymptote at E. Con-

Tinue « out the Ordinate A to L, fo that AL

—

be = AE + EP (2 ＋ 2 — aa 31) and draw

LX the fame way 3s the T endency of the
Curve, viz. towards Z; Which make = £ P

GN 503 and let the ſame cut the Curve

in M. Laſtly, between X A (2 Naa

150

Fn 6. 42.

and the Fluent 5% 2 RIES 7
by making P (Ve == — 3% Q

APPENDIX.

„N yaa T7 ) and AT, find a mean Propor.

tional A O, (LM from the Nature of the
Curve being the Logarithmick Part;) which
will be the Semidiameter of a Circle equal to

the Superficies generated by the Rotation of

the infinite Arch P Z.

1 find the Superficies generated by the Ro-

tation of the infinite Arch M L or AT,
of the Cifſoid of Diocles about its Aſymptote
AG.

Draw the Ordinate M at right Angles to
the Aſymptote, and mp infinitely near to it.
Then AB being Sa, and AN=x, MN be-
ing at right Angles to 4B, the Fluxion Mn

of the Arch MZ will be i

2 ee

which multiplied by “x M P= N x4 , and

the Fluxion of the Superßcies e by
the aboveſaid Rotation of the Arch M * will

be had, viz. pou ATA:

217 A—X

Now this may be Faun INDE “he ele-

venth Form of Mr. Cotes’s. For making 2=»,
bf=1r, n=1, 4 , f= 3, Dl, and

5 2 the Fluxion 3 in/ the eleventh Form

2 e will become r, A ,
Lt | ax
RT

APPEN D IX.
2— * ?

A — = /——— or/—

x LI being the Fluent
ya

V3
of the given Fluxion = EL 3 wherein
the negative Signs of the two Parts may be
alter’d.

Now this Fluent may be conſtructed thus:
Biſect M in F, and draw AFE meeting the
Aſymptote in E. Make D a mean Propor-
tional between AB and NB. Alſo let the An-
gle CAB =3 of a right Angle, and continue
out the Aſymptote towards B meeting 4 C in

the Point C. Then BC == bor fr

=2BC. Whence BC 4- 1B =4 BC ,and
ſo 4A — Conſeqtently BC 222.

3 v3
bs 1 1 4— 232 —
D Again, AF “of I” And be

” of the ſimilar Triangles ANF, ABE,
AN(#):4F(Z =) :: NB (a- x):
FE I 4/44—3x X Va- x. And DC =
L. Now the Superficies generated
by the 3 of the infinite Arch MA, will
be to a Circle ( = =) whoſe Semidiameter is

I AB

152 APPENDIX.

AB (a), as 2 E F (,/ 4a—3xx Va- x) + the
Meaſure of the Ratio between BD (,/ aa—ax)

＋ D — and BC (55) to
the Module BC (=):

Now making x o in the aforeſaid Fluent,
x 4 » 4 & , A @ 0-8
and it will become – * 2 4 —-
Ty

_ – w 77 IT ALLY, Which is the Fluent

expreſſing the Quantity of the infinite Arch A.
Being to a Circle whoſe Semidiameter is AB
(a), as 2 AB -+ the Meaſure of the Ratio be-

tween BA (a) + AC (= e and BC 2 )
to the Module BC. And the Difference of theſe
Fluents, omirting = viz. 2 AB (2a) — EF

– z 4/a—x) + Meaſure of the Ratio
BA(a)+40( =) to BD (y/a—as) +

DC ( “4 9 the Module being ÞC

wary, will give us the Quantity of the finite

Arch AM. For it will be to a Circle having
BA tor a Semidiameter, as that Difference is
to the ſame Semidiameter B A. |

When the Curve revolves about the Baſe
AB, Mr. Cotes in Harmonia Menſurarum gives
the Quantity of the Superficies generated.
But the Operation is long and troubleſome, on
account of the Fluxion’s not directly coming
under any of his Forms. —

—

SECT.

ä * 2 WER * NT

44.4 aw

APPENDIX, 153

STEG Tro FL

of the V/e of Fluxions in finding the
. Centres of Gravity of Figures.

PRO B.

2 y 76. TO find the Centre of Gravity of a plain
Figure.

Let AB be the Axis, and MM an Ordinate Fc. 58.
to it; and let n be another infinitely near

MN, and ſuppoſe C to be the Centre of Gra-

vity.

if the infinitely ſmall Parts, as Min nN
of the Figure be conceived as ſo many Weights
hung on the Axis AB, at the ſeveral Points
P, P, P, &c. and the Point of Suſpenſion be
in 4 the Vertex of the Figurez the Diſtance
Sh ** — 9 *
FF

of the Centre of Gravity C from A the Point

of Suſpenſion, will be equal to the Quotient
of the Diviſion of the Sum of the Momentums
„of all the ſaid infinitely ſmall Parts or Weights
Mn N by the Sum of all thoſe little Parts

or Weights; that is, by the Area of the whole
Figure. This is plain from common Books
of Mechanicks. |
Therefore calling 4 P, x, MP, y, Pp, à, the
infinitely ſmall Part or Weight Mm N will
be yy, and the Sum of them Fluent of
2) XK. And the Momentum of one of thoſe in-
N n finitely

APPENDIX,

finitely ſmall Parts or Weights will be 2 yx
multiplied by AP (x) =2y9xx. The Fluent
of which divided 4 the Fluent of 2 will
be = AC, the Diſtance of the Centre of Gra-
vity from 4, the Vertex of the Figure.

EXAMPLE I.

1 find the Centre of Gravity of à Ti.
angle ADE.

Draw the right Line AB biſecting the Baſe
DE in B. Then becauſe the Triangle A BD
ABE, each of them may be reſolved into
an infinite Number of little Parts or Weights
MmypP, PNnp, each equal to one another on
either Side the Line AB taken as an Axis;
and conſequently the Centre of Gravity C mult
be ſomewhere in the ſaid Line AB.

Now call A B, a, DE,b; AP,x, MN, ;
and draw AF perpendicular to D E, which
call c. Then fince the Triangles AN N, ADE
arc ſimilar, A B (a): DE (b):: AP (Y): MN=

. Allo becauſe the Triangles APQ, ABF
are ſimilar; therefore AB (a): AF(c):: AP (

: AD = -. And as AP (x) : 42 2

Pp (x): 2g = => Whence the Momentum

9x x will be — 55 ; the Fluent of which is

=== Which being divided by = _ — the A-

rea of the Triangle ADE, and hi Quotient
will be 3 x == Diſtance of the Centre of Gra-

vity

APP E ND IX.

vity of the Part AMNMN of the Triangle ADE
from the Vertex; and ſubſtituting a for x, the

Diſtance of the Centre of Gravity. of the

whole Triangle ADE from the Vertex 4 will
be = 3 A ( a).

78.7 determine the Centre of Gravity C in
the common Parabola.

155

Calling 4 P, x, MN,), and AB, a, and the Fre. 58.

Parameter p; then will px be = yy; and lo y
= px. Whence yx=x,/px; and ſo the

Momentum y x X =x* / px =xxXpIx%, The
1 4
Fluent of which is f px ʒ but the Fluent of

3
Ji is = = px =Area of the Portion AMN

of the Parabola; therefore dividing : #7 7

by = x”, and the Quotient will be = 2 x =

| 5
AC. And ſubſtituting à for x, the Diſtance

of the Centre of Gravity C trom the Vertex
A will be = AB.

354A 3×28;

18 determine the Centre of Gravity C in
all Parabola’s of any higher Kind.

Here pn = expreſſes the Nature of all

Curves of this Kind. Whence ) S x, and conſequently y x =p” x x, and the Momentum
| | n 2 * 9 *

l :

a
„
! 7
_
|
!
‘
i
= {
1
6
1 *

—— — – =
. ——— — — OOO
4 ha — ——

—
1
— — 4

—

a
_
*
_ —
—ͤ— —— ns

—_ ow

156

F 1 G. 60.

APP E ND IX.

n mr

2 1 . JE |
SEE 9 x Which divided by the Flu-
n mn

* _
ent of & = „ and the Ouoti—
_ 5 the Quoti

m + Yr

ent will be = u theDiſtance of the
m + 2Y

Centre of Gravity of the Portion MAN from
the Vertex A; and ſubſtituting a for x, there

4 AC.

comes out

SS
m 5-24
EXAMPLE IV.

TO find the Centre of Gravity of the Space

ADE contained under two equal Para-
bola’s AD, AE, touching one another in their
Vertex A, “and the ſtraight Line D E parallel to
the common Axis of the Parabola’s.

Make AP =x, PM Jz al and let the Para-

meter be = 1 : then will AP (x-) be Ow
=y; therefore the Momentum xyx = x

The Fluent of which is __ but the Fluent

of is 59 Whence f =2 = AC, the

Diftance of the Centre oft Gravity from A.

If the Parabola’s AME, AN 5 be of any
Kind whatſoever, this Equation will expreſs
the Relation of AP to PM, viz. xm =”.

Therefore y x = x x, and the Momentum xx

m + 71

=x a, The Fluent of which is
1 I m2

but

APPENDIX. 157

but the Fluent of y x 1s

—X * , Whence
m u

the Quotient of the Diviſion of the former of

_ Expreſſions by the latter, vig. E * |

will be = 4C.

e EX AML V.
m | 81. TS determine the Centre of Gravity of any
e Arch of a Circle.

Let BE the Chord of the given Arch BDE Þ 16. 61.
be parallel to the Diameter FG; which being
conſider’d as an Axis, to which the ſmall
| Weights B M are ſuſpended; and fo the Mo-
mentums of them as B Mx PB. And ſince the
Numbers and Momentums of the faid little
| Weights on each Side the Radius A biſect-
% ing the Arch BDE are equal, the Centre of
| Gravity will be in 4D.
| Now let AB a, AP HB x, and

Py B Nd x. Then PB (/aa—xx): A

.: Bu (: BM= nd BH BP

X. ; af 44 == Xx
| ax .

nt = HOW: a — — :
Vaa —XX % — a A BKB M;
is and the Sum of the Momentums, or Fluent of
this Fluxion, is ax ABB H. Which di-
„5 vided by the Arch BD and — . S the
ls Diſtance of the Centre of Gravity C from the
*. Centre A of the Circle. And ſubſtituting the

T: Quadrant FD for BD, and the Radius FH or

4B for BH, then will Ei be the Diſtance |

of |

158

FI 62.

APPENDIX.

of the Centre of Gravity C of the Semicircle
FDG from the Centre A.

Ex AMP L E VI.

2 determine the Centre of Gravity of a
Sector of a Circle ABE.

The Centre of Gravity will be in the Ra-
dius AD, which biſects the Arch BE. De-
ſcribe the Arch M PM with any Diſtance AP,
and another Arch pm infinitely near it; then
the Momentum of the Arch 1 PM drawn in-
to Mm or Pp, will be the Momentum of the
annular Segment mM PM, or the Fluxion

of the Momentum of the Sector.

Now let AB be =a, IF ==, and BD
Sc; then the Momentum of the Arch BDE
(2ab) is to the Momentum of the Arch M PM,
as the Triangle AB E to the Triangle AM,

or as AB to AM ; ſince the Triangles ABE,
AM M are ſimilar. W hence the Momentum of

the Arch M PM= 2008″ = =, and the Mo-

mentum of the annular Segment mM PH =
2bxx RY)
— The Fluent (or Sum) of which *

divided by the Sum of the Weights, or the

3
Area ac of the Sector, and the Quotient . |

is the Diſtance of the Centre of Gravity of
the Sector of the Circle M PM.

ExAMPLE VII.

f £5. find the Centre of Gravity of any Seg-

Making

ment MA m of an Hyperbola.

APPENDIX.

159

Making AC,a, CB, ö, 4 and PM, y, Fic, 63.

we have from the Nature of the Curve

25 x 2ax+xx =5yy. Therefore the Fluxion of

—
the Momentums will be = & /2a—+x, and

the Fluxion of the Weights 7 * * 72 7 + x.

Each of which may be compared with that of
the fourth Form in the Tables of Mr. Cotes ;
and ſo making 9 = 2, „t, Sc. we ſhall
have the Sum of the Momentums =

—z aa TAX + 2×8 2 inn E.
— xy 4 x 2.4

And again, putting 9 = 1, Sc. the Sum of the
Weights will be = a+x x y +246 —

and ſo dividing the former Fluent by the lat-
ter, we have

—3aa & ax + 2xxxy + 24*

7—

f/x+ 4 2a +x
y/ 24

ln A2 + x
Va

= Diſtance of the Centre of Gravity from
the Vertex D.
Now to conſtruct this Expreſſion, make

CB(b):CA(4)::PM (3):CF= =. And a-
gain, PM ()): CB (H):: CA (a): CG =
Then take CH equal to the Meaſure of the

Ratio between C Aa) and FP (a+x — 7)

Which Ratio is = duplicate Ratio of y/24
2 to

160

APPENDIX.
to /yx—y/2a+x. When this is done, if you

, a
make as 3 P H (3a + 38 —— 2
. 249 ® 0 a) o pom
2
44x + 2:xx

This fourth Proportional will be = Diſtance
of the Centre of Gravity z from the Centre
C. For if CA (a) be taken from it, we ſhall

have — Bai _ — 42 2

3-43 —4.—

/xÞ+V2a + x a

—3aabkaxt+2xxx3 + 240

PNY T 245 [ET .
the Expreſſion firſt found.

SCHOLIU x.

＋ HE. Diſtance of the Centre of Gravity

of the Segment of an Ellipſis or Cir-
cle from the Vertex will be expreſſed thus:
vx v24—x

—JAG—AX2XX XY — 24˙⁰—

a2 —

a—xXy — 2ab | —

V2.4

it being the ſame as that for the Centre of
Grayt-

Q * = &

APPENDIX. 161
Gravity of the Hyperbola only with the Alte –
ration of ſome Signs, and the Meaſure of the
Ratio there being here changed into that of
an Angle, becauſe x (=, =y/—1 is ne-
gative.

EXAMPLE VIII.

12 find the Centre of Gravity of the eu- Fre. 64.
ternal hyperbolick Space A MmaA.

Let the Semi-conjugate Diameter B C ,
the Semi-tranſverſe 40 Da, the Abſciſs C P
=, and the Semi- ordinate PM x.

Then from the Nature of the Curve,

; 2. 25; and ſo = .

Whence = 5h is the Fluxion of the

Weights, and = Vb that of the Momen-

tums. And from the Compariſon of this Fluxion
with the 3d Form of Mr. Cotes’s Tables, & being
=1,n=2, c. the following Fluent will be had,

VIZ. rl EINE and making y=0, the ſame
will become zal; which muſt be taken from

. 2.4 1 2
that Fluent, and the Remainder I bb-yyy 2a

—ͤ— **
—
——
— — –

willbethe Fluent of the Fluxion _ % N.

Alſo the Fluent of — JV +33, by compa-

ring it with the fourth Form, 9 being = 0,

| TV.
b

—— ———
OT — =

; —
2 — —

*

, Ec. will be Vi + ab

By which dividing the Fluent juſt now found, , l
O o and ill

— — EE — ————_—
—— — — —

162 Ar PEN D IX.

x Fg 245 g
and the Quotient 7

7 ＋e Tab LL
4 =

2 * 77 —25¹

= — — 72 will be =
39/0b+39 + 306 i DIL a
Diſtance of the Centre of Gravity Z from the |

Centre C.
Now to conſtruct the Expreſſion: Make

AC (a): PA Vs =x)::BC(b):C R=
y/bb + yy. And again, PM J % +39): AC
(a):: B CO: CS

; s. And Wy ta N

PM (% N)! a bs of == ==) F
þ3
; CT = = And let C R, CT, tend both

the fame way, viz. from C towards PM, but
CS the contrary way. Then take CN equal
to the Meaſure of the Ratio between C 0

and BE (MDD) to the Module CS

= _ | Which will be thus expreſſed,
8

N (33+

This 225 done, make

==): A
0 = 77 Y 175

bb 75 7 ) :CR (Y 11
a fourth

18 1

the Vertex A.

APPENDIX,

2 fourth Proportional, which will be =

2X bb—+- — 2 b3

| 21/3306 2 =

ſtance ſought.

Note, The Conſtruction of theſe two Ex-
amples are the ſame as Mr. Cotes has given in
Harmonia Menſurarum, p. 25, 26. Part I.

Ex AML E VIII.

TO find the Centre of Gravity of a right
Cone and Pyramid.

| The Centre of Gravity will be ſomewhere Pe. 43.
in the Axis AB.

Now ſif AP be = x, and B C a, and 4B
=6, we ſhall have PM = J. And the Ra-

tio of the Radius to the Periphery being that

of 7 to p, the Fluxion of the Weight will be

» * 3 *
Has, and the Momentum of it — Con-

2zrbb 27

ſequently the Sum of the Momentums LED .

divided by the Sum of the Weights =

will give us : x = AG, the Diſtance of the
Centre of Gravity of the Part 4M Pm of the

Cone; and ſo 2 4B is the Diſtance of the

4 | |
Centre of Gravity of the whole Cone from

Ooz Much

— ——— —
— z: ĩʒ — —

164

F c. 44.

APPENDIX.

Much after the ſame way you will find the
Diſtance of the Centre of Gravity of a Pyra-
mid from the Vertex tobe z of the Axis from
the Vertex.

Ex AM I. E IX. 5

11 find the Centre of Gravity of any Sep-
ment of a Spbere.
Let AC=r, AP=x; then the Is of

the Weights will be y — £22 — , and that of

the Momentums p * 2 and the Sum of
2 x+ 27

the —_— 5 which being divi-

dea by ES, the Sum of the Weights

and the Quot! lent . hole e villbethe Diftance

22
of the Centre of Gries from A, of the Seg-
ment of a Sphere generated by the Semi-ſeg-
ment {4M P — a Semicircle about 4 P.

Co n 01.

„Hine the Centre of Gravity of 1 the
Sphere will be ß : for — x be-
comes == 7.

E Xx A MSP L E X.

T® find the Centre of Gravity of à para-
bolick Conoid formed by the Revolution of
a Parabola 1 its Axis.

Here 2 = is the Fluxion of the Weights

and 2 — X the Fluxion of the Momentums, r be-

ing

APPENDIX,
ing the pap wp Rectum. The Sum of the

ys will be 2 * Which is the 1
of the Centre of Gravity from the Vertex 4

of the common Parabola.

Ex AMy L E XI.

| 90. T2 find the Centre of Gravity of a Solid

formed by the Revolution of the parabo-
lick Space A 215. B D about the Line BT Parallel

to the Axis.

165

Let DB or AT be Sr, AP =xX, PM 3 F1 0. 6s:

then will Z— py be == Circle deſcri-

bed by M.; and conſequently * & we 22,
will be the Fluxion of the Weights, and

— 2 7 that of the Momentum. The

Fluent of the r Expreſſion, or Sum of
the Weights, is 1 —E , and the Sum of

the Momentums 2 — Ps JT _ Which divided

by the Sum of the Weights, and the Quoti-

oat will be 2X 1 D., be.
40-17 x 401 7

ing the Diſtance of the Centre of Gravity
of the Part of the Solid generated by A *

166

F 16, 40.

APPENDIX.
and when x becomes = a, and fo y r, the
ſaid Diſtance will be = Th

Ex AMP L E XII.

7075 find the Centre of Gravity of an hyper-
bolick Conoid AB G generated by the Ro-

tation of the hyperbolick Space ABG about the
tranſverſe Axis aC AP.

Now calling AC, 2b, A B, a, and the Or-
dinate B C, r, the F luxion of the Weights will
be Tg, and the Fluxion of th
24a Fqab
Momentums. A being the Centre of Motion,

will be = —— — — the Fluent of which
1 prix $4 bprs* _ 3 prot + Shore
Sag JI 6a Jaa PGA 24a

Which being divided by * 2 177 – =

dprx* +1 2bprex
IT” the Sum of the Weights, and

the Quotient -der. will be the Diſtance

4x+126

of the Centre of Gravity from the Point A in

the Axis AP of the Conoid form’d by the Re-
volution of the Part AMP of the Hyperbola;

+126
of the Centre of Gravity of the whole Solid

and when x = =a, 21 25 will be the Diſtance

from the Vertex A. Whence za 86: Ales

34a+ 8ab

25 in

Ex Au-

APPENDIX.

Ex AMP LE XIII.

| 92. 7 O find the Centre of Gravity of an hy-

perbolick Conoid form’d by the Revolution

| of the E perbolick Space AMBDC about CD
‘= WW the half of the conjugate Axis.

e | The ſame being ſuppoſed as in Art. 61. the Fic. 4
Fluxion of the Weights is Cor? 175 aabbpx,

„ nd that of the Momentums with regard to

|
” 463 <= — the Fett of which
1 1 aapx* aabbpxx priyy4+2bbpx*y? b 1
4 Wc wh” i; “abbr Fre TND r
J ting — 75 for a a. Now this divided by the
Sum of the Weights, and the Quotient
Prog
= 2 , will be the Diſtance of the Cen-
| 4Xx–1206
1 [MW tre of Gravity of the Conoid generated by the
| Space AC PM from the Line or Axis _ C3
© aud when becomes = 6b, we ſhall have — 5
n W for the Diſtance of the ſaid Centre.
Therefore the Centre of Gravity of the
b whole Conoid is ſo ſituate in the Axis AC, (|
6 that the Part from the Centre C to the Centre Cl
of Gravity isto AC as 9 to 16, | it
; : |

E x A M*

APPEN DI xX.
E x AMX L E XIV.

7 O find the Centre of Gravity of a Semi.
[pheroid generated by the Rotation of x
the ASE — Space AMeca about the Axis Aa.

Make AC a, Ce, AP=x, PM=y,
Now when PC is x, the Fluxion of the
Weights will be 22 3 but from the Nature
of the Ellipſis, (AP being = «) yy: za =
:irr:aa. Whence yy= — — and ſo

ſubſtituting this Value in =—— – 2 . there ariſes

zaprxx pr *
246

Fluxion of. the Momentums will be

pro? pred __ Sapro? —3 EE,

34 * 244 divided by the

Sum of the Weights — and the
“”

Quotient — — Ps will be the Diſtance of the

z and 3 by ; the

I 24—

Centre of Gravity of the partial Solid gene-
rated by the Space APM trom the Point 4;
and when x becomes = a, the ſaid Diſtance

will be 8 a; that is, the Centre of Gravity of
the Semi-ſpheroid will be diſtant from A by

SECT.

4
.
„
“M
K

F
F N
F
5 o
b
1
£
i
4
A
© :
16
2
. ‘
!
.
3g
4

.

H
4

e
j
f

«
2
:&
K.

— ” N W
T2 ä * TAY 7 __ 2 8

2 W

b SECTION VII.

Of the “Ofe of Fluxions in finding the
Centres of Percuſſion of Figures.

DEFINITION.

HE Centre of Percuſſion, or Oſcilla-

tion of a Figure in Motion, is that
Point in which all the Forces of the ſame are
conſider’d as united together in one; fo that if
the ſaid Figure meets any Obſtacle contrary to
the Motion thereof, it ſtrikes the Obſtacle
with a greater Force than any other Point of
the Figure.

169

In order to this, it is neceſſary that the Parts

of the Figure do conſtantly alter their Diſpo-
ſition to move; that they ſeparate their Quan-
tity of Motion, not as in the Centre of Gra-
vity in the Ratio of the Spaces run thro’; but
in a Ratio compounded of their Velocities,
and the Diſtances of that Centre reciprocally
proportional to the ſaid Velocities; or, which
is the ſame thing, into equal Quantities of
Motion on each Side that Point. Therefore
the Centre of Percuſſion is the ſame with re-
gard to Velocities, as the Centre of Gravity is
with reſpect to Weights: and as in finding the
Centre of Gravity, we divide the Sum of the
Momentums by the Sum of the Weight; fo to
find the Centre of Percuſſion, we mult multi-

P p ply

Fi.s. 59.

_
— — —

—
— _
— .
—_

— . n

APP E N DI X.

ply the Sum of the Momentums by ſtraight
Lines equal or proportional to the Spaces mo-
ved thro’, and divide the Product by the Sum
of the Momentums. Whence the |

General Rule for finding the Centre of Per-
cuſſion of a Figure that revolves about a given
Point or Axis, is to multiply all the ſmall Parts
of which the Figure conſiſts, (that is, the A-
rea or Solidity of it) looked upon as ſo many
Weights, by the Squares of their Diſtances
from the Point of Suſpenſion, and divide the
Product by the Product of the ſame Weights
into the Diſtances from the Axis of Motion,
and the Quotient will be the Diſtance of
the Centre of Percuſſion from the Point or
Axis of Motion. 4

Hence if AP be = x, MN =2 9, Pp=x,
the Momentum of the whole ſmall Weight
M Num will be =2yxx. Conſequently the
Diſtance of the Centre of Percuſſion from the
Point A is = to the Fluent of 2 * x divided
by the Fluent of 25 *. Conſequently, if
from the Equation of the Figure you get the
Value, of y, and pur it in thoſe Fluxions, and
then find the Fluents of them, you will get
the Diſtance of the Centre of Percuſſion from
the Point A. This will be evident from the
following Example. |

ExAMPLE I.

70 find the Centre of Percuſſion of a right
. 1 Line A B, moving about one End A
c

Now if the ſaid Line be conceived to be di-
vided into an infinite Number of equal ſmall Parts
; 47 4 341 r p 4 8,5 ; © Pp

APPENDIX.
ht Pp (x) (AB being a, and A P, x), it is mani-
| feſt that in equal times they will deſcribe equal

A Arches of concentrick Circles, that will be to

| each other as the Diſtances from the Point 4:
_ But the Velocities wherewith the ſaid Arches
a are moved through, are proportional to the

ny _
4 mentums, which multiplied by x repreſenting
he the Velocities, and there will ariſe x? z for the
| 3
its WW Fluxion of Forces; the Fluent of which —
n,

of being divided by the Sum of the Momentums

75 and the Quotient Ex will be the Diſtance

*, of the Centre of Percuſſion of the Part 4 P
ht of the Line from the Point 4; and the Centre

of Percuſſion of the whole Line will be 35 by

making x .

LY ds IT.

| 96. TPO fond the Centre of Perouſſion of a Nett.
angle RISH moving about one of the
Sides RI.

Pp be the Fluxion of the Area, and one
of the ſmall Weights will be Sa, and the
Momentum of it will be axx. Whence the
Fluent of a x* divided by the Fluent of a5

Tax : :
(or py and the Quotient x will be the
T4 Xx

Diſtance of the Centre of Percuſſion of the
| Pp 2 Part

171

If RI=SH be Sa, APD x; then will F 1c. 66,

—— — —— . EG

172 APPENDIX.

Part RC DTof the Parallelogram from the
Side RI; and if for x be put the Altitude R$
==b of the whole Rectangle, the Diſtance of
the Centre of Percuſſion from the Side RI will Þ 5

be 25.

EXT AMP L E IH;

7 find the Centre of Percuſſion of an >
| ſoſceles Triangle 8 AH moving about tht |

Line R paſſing thro’ the Vertex A, and parallel
to the Baſe SH. |

FI c. 66. Let the Altitude AE be = ay AP =;
EH, PL=y. Then AP (): PL ()::
AE (a): EHS. Whence ay = x, and

—

L 121 = Now the Fluent of ** = Fluent
of 2 a, and the Fluent of y x x =PFlu-
24 8a

N 3
ent of — is 2 . Whence the Fluent of
2a 64

FER bat…
yu a, divided by that of yx, or gory =q”

=> x.
4

Now if for x you ſubſtitute the whole Al
titude AE Da, the Diſtance of the Centre of
Percuſſion of the whole Triangle AS H from

the Vertex A will become de 40 E.

E x a

RS
ll

— — . — ln .
= —

ArtPENDI® 173
Ris anions iv. 3

8 find the Centre of Percuſſion of an J.
ſoſceles Triangle S AH moving about the

| Baſe SH.

Let all things be as in the laſt Example, Fi c. 66.
then will PE be Sa- x. Whenee the Flu-

ent of 57 4 b Fluent of a (=Flu-

bx® x

ent of Lab = T is 14
2 24 4

5 * + 2 ; and the Fluent of y#4 =Pluent of

8 N

x = (Fluent of 75x —
* 75 . Whence the Quotient of the
Fluetit of y * 4 divided by that of 52K
Lab Lb a* + 3

_— 3 348335 L YA

241″ by—22abx +12bx*

E .

| 064 .
PINT Gare —abs”

e RE
ah —t 6abx* + 6bx* 64 — 84x – 3x?
92 1 2abx—8abx wt: 6a—4x
= Diſtance of the Centre of Percuſſion of
the Segment SZ VA from the Baſe & H.

Now if for x you ſubſtitute a, we ſhall
have the Diſtance of the Centre of Percuſſion

“2
4
b

d
i 8
+2
%
“1
‘f
1
. 4
in
| i
CY
1
Il
TY –
ö N
N
N

—

174

APPENDIX.
of the whole Triangle 8 4 H =
(<=) ited FI,

64—44
EXAMPLE V.

12 find the Centre of Percuſſion of a para-
| bolick Space, moving about à Line paſ-
fing thro’ the Vertex parallel to the Baſe.

Now calling the Abſciſs x, and the whole
Height a, and the Fluxion of the Momentum

will be 1, and that of the Forces pts 3 the

Fluent of which will be 75 Which divi-
ded by of the Sum of the Momentums ; the

Quotient will be ; x =Diſtance of the Cen-

tre of Percuſſion: of the Part of the Parabola
whole Height is x from the Vertex; and when
& becomes equal to a, the Diſtance of the
Centre of Percuſſion of the whole Parabola

will be Ca.
7 15
SCHOLIUM.

too. 1 the Centre of Percuſſion of a Parabo-

la of any Kind be ſought, you will
= a = Piſtance of the Centre of Per-
cuſſion from the Vertex; where m is the Ex-
ponent of the Power of the Ordinate of the

have

Parabola. So that if be = 2, the Diſtance

Will be Sa, as in the common Parabola. If
n be

ad ty Wim

| by 2 — K, and we get ax K
Fluxion of the Momentums; which being a-
gain multiplied by a—x, and the Fluxion of

| F. Zr.
| the Forces will be aa x x ——— 4d

ö the Fluent of which, viz.= aa *

g 2 2
12

ArrENDIX. 175
„ be =3, as in the cubick Parabola, the Di-
ſtance will be 2 a, XC.

EXAMPLE VI.

| 101. T2 find the Centre of Percuſſion of a pa- p 1 6. 38.

rabolic Space moving about its Baſe DD.
Multiply the Fluxion of the Weights & &

— 0
e

*,
e
5 7

e *. 30 K

—

Wee 197
being divided by the Sum of the Momentum

41 T+2
loax* — 6

| = — | , and the Quotient will

| be ZN, the Diſtance of the

7a—21K

Centre of Percuſſion of the Space DMV

from the Baſe D D; and putting x for a, the
| Diſtance of the Centre of Percuſſion of the

whole Parabola from the Baſe DD will be

42 24 4B,
„ nk

EAM L E VII.

102, T2 find the Centre of Percuſſion of a Cy. Fe. 67:
Ader AB moving about the End A

Now

«A

thereof.

— – — =
n 83 — 4

n
-. x ag

2 WIR —
— Fer —
as. _ _— —
— 3 = *

— —
— 2

796

APPENDIX.

_ Now it is evident that the Velocities of al

the ſmall. equal Parts of this Solid will be to

each other, on account of the al times, in

the ſame Ratio as the Spaces 7 #2 thro” that is

as the Arches deſcribed in their Motion, or 2

the Radii or Diſtances from the Point of Sul.
on. Now let the Axis AB of the Cy
inder be = a, any Part AP Sa, the PG. .
phery of the Baſe = p, and the Radius of it
=; then the Fluxion of the Momentums wil 1

be . and the Fluxion of the Forces * 8 .

the Fluent of which is = Which being
divided by . the Sum of the Momentum —
pro „ and the Quotient < 2 x will be the Di
los of the Centre of 2 of the Patt

of the Cylinder, 1 Altitude is A from

FIC. 67.

But the Velocities of the little equal

thePoint A; and 2 a will be the Diſtance of

the Centre of Percuſſion of the whole Cylin-
4 from the ſaid Point 5

E N VIII.

9.4 find the Centre of Percuſſion of a Cy:
| linder moving, about tC Point R in the
His continued our.

Make RB =a, RA=b, 428 then
RP =b+x, and AB =a—b. W hence

b |
2 — — —1 is the Fluxion of the Momentuns:

Weights of the Solid are to one another a
the Arches deſcribed from the Point & by
3 thoſe

APPENDIX. 177
thoſe Weights; ſo the Forces are to each other
as the right Lines that fill up a Trapezium de-
| ſcribed by the Cylinder. Whence multiply-
ing the F luxion of the Momentums by b + x;
boprx+2bprex4pro*x

of the Forces; the Fluent whereof is =

— ＋ 2 +2. Which divided by the

and we get —= Fluxion

Sum of the Momentums = = 4.4 » and

. 6bb+ 6bx+-2Xx .
the Quotient | will be the Di-

ſtance of the Centre of Percuſſion of the Part
whoſe Altitude is AP from the Point &; and
e will be the Diſtance of the
Centre of Percuſſion of the whole Cylinder
from the Point K, ſince then x becomes =
ab. |

Hence it is eaſy to find on what Part of a
Cylindrical Stick a Man ought to ſtrike in or-
der to gain the greateſt Blow poſhble, tuppo-
ſing AR repreſents the Man’s Arm, and A
the Stick,

ExAMPLE IX.

O find the Centre of Percuſſion of à Cone

moving about its Vertex A.
The Fluxion of the Momentums will be Fic. 433

“RIX »
— and the Fluxion of the Forces 7 5

the Fluent ors of which being divi-

al I0a
1 ded by the Sum of the Momentums =
4 = and the Quotient + , will be the |

Q q c Diſtance |

178 APPENDIX.
| Diſtance of the Centre of Percuffion of
the Part of the Cone, whole Altitude is A P

from the Vertex A, and + 4 will be the Di-

ſtance of the Centre of Percuſſion of the
whole Cone from the Vertex 4.

Note, This Centre of Percuſſion is the ſame as
the Centre of Gravity of the Complement of
a cubick Parabola; becauſe the Forces are as
right Lines which fill up that Complement.

rn.

10 1 70 find the Centre of Percuſſion of a Sphere
moving about a Point A in the End of a
Diameter AD.

Let the Radius be = #, the Periphery =p,
and AP = x; then the Fluxion of the Mo-

mentums will be px*x , and the Fluxion

F16. 44.

this will be N=; which being divi-
4 lor
ded by the Sum of the Momentums = 5 —

e, and the Quotient 2 ＋ vill be the
4 497——I5x

Diſtance of the Centre of Percuſſion of the
Segment of the Sphere, whoſe Height is a,

from the Point A; and * will be the Diſtance

if. | J |
of the hemre of Percuſſion of the whole Sphere
from that Point. |

E x Au-

APPENDIKX.
EXAMPLE XL

06.7 O find the Centre of Percuſſion of a Pa-
rabolict Conoid moving about the Vertex.
The Fluxion of the Momentums wall be

= — „and the Fluxion of the Forces 2 * The
27 2r

Fluent of which is = = 5 75 which being divi-

ded b at and 2 x = be the Diſtance of
T’s 4 |

the Centre of Percuſſion of the Part of the
Conoid, whoſe Altitude is x from the Vertex,

and : @ will be the Diſtance of the Centre of

Percuſſion of the what Conoid from the Ver-
tex. £5

ExAMPLE XU

Fo fn the Centre of Percuſſion of a Sphe-
roid moving about one End of the tranſ=
verſe Axis.

The ſame things being ſuppoſed as in Art.

the Fluxion of the Momentums will be

2
Lap x — ad and the F luxion of the For-

244

ces EZ — 2 the Fluent of which, viz.
4 3
ode; — „ being divided by the Sum of the
44 10
dlm = = 2 pr” 9 ‘and the Quotient

34 844
; Qq z

179

APPENDIX,

0 — . will be the Diſtance of the Centre
40 I
of b of the Part of the Spheroid,

whoſe Altitude is x from the Point of Morion,
and Za 4 will be the Diſtance requir’d,

180

APPENDIX.

SECT. VII.

Of the Reſolution of ſome miſcellaneous
Problems by Fluxions.

Mn.

TO find a Line, wherein the Subtangent

is equal to the Semi-ordinate,
In all Caſes it is plain that 25 is an Expreſ-
ſion of the Subtangent; and ſo from the Con-
dition of the Problem 22 =y, and yx =y y;

that is, x ==, and the Fluent of each Side
will be x =y. Whence the Line ſought is
the Hypothenuſe of a right-angled Equicrural
Triangle; a Line biſecting the right Angle
being looked upon as the Axis. But if x be
the Arch of a Circle, then will the Line ſought
| be a Cycloid.

PRO B. II.

log. 7 O find a Curve, whoſe Subtangent is

equal to twice the Square of the Semi-
ordinate divided by a conſtant Quantity : fup-
Poſe a. |

The

181

182

APPENDIX.
The Subtangent is == Whence IE
muſt = — Conſequent! y ax == 2 555
that IS, 21 and finding the Tuents of

each Side, we have ax =yy; therefore the
Curve fought is the Apollonian Parabola.

. 1 RO B. III.

8 O find a Curve, whoſe Subtangent is 1

third Proportional to ſome ſtanding Ouan-
tity a leſſen d by the Abſciſs, and the Semi- ordi.
nate. | hg

Here 4 wn 29:9 2, and multiplying the

Means and Extremes we get — —
and ayz—yxx=5*7 ;. that is, a&—xx==yj ; and
finding the Fluents of each Side, we have
ax – Xx Ty), or za -M iy; therefore
the Curve ſought is a Circle whoſe Radius ;
— .

PR O B. IV.

7 find a Curve, whoſe Subtangent is au
invariable Line.
Here 2 muſt be a, and ſo 2
ay” Y. Whence the Fluent of x, viz. – muſt

be =Fluent of a3 j; and multiplying ay “7
by a, we ſhall have a%y , which is the Flu-

xion of an Hyperbola between the Aſymptotes.

And ſo if y be taken for an Abſciſs, the corte-
{pondentSemi-ordinate x =Fluent of ay “will
be equal to theAſymptotical Hyperbolical Space

Jivi-

div
the

II.

APPENDIX.
divided by the invariable Quantity a, being

| the Side of the Power of the Hyperbola.

P R OB. V.

A Number being given: to find the Loga>

rithm of it.

Let the Ordinate of the Logarithmical F 16. 68.
Curve A B be =1 = Subtangent; then PM
vill repreſent a Number greater than 1, and
Vs Number. leſs than 13 AI the Loga-

rithm of the Number greater than 1, and

| 49 the Logarithm of the Number lefs than 1.

Now let the Difference between AB and

| PM be = y; then will PM be = 1 +7.
| Whence AP, the Logarithm of a Number

| greater than 1, will be the Fluent of I =

= 0 J *
| 5—1j+P3—133 +34, &c. Which Fluent is

+>} — 73%, &c. = Logarithm of
a Number greater than 1.

Again, it the Difference between A and
9N be y; then will 9N=1—zy. And ſo
A9, or the Logarithm of a Number leſs

| than 1, will be = Fluent of _ 1— 95

—;——5z, &c. the Fluent of which will be =—y—23—4y%—£34, &c. being
the Logarithm of a Number leſs than 1.

S cn O l. 1 0 M.

—

183

T* the Side AB or BC of the Power of F. 69.
an Hyperbola be = 1, and BPA);

then will AP =1 +, and the Aſymptotical

Hyperbolical Space will be Y = ν

— P — 2 Ht oc

APPENDIX.
, &c. And if BY be = y, then wil
AD be =1—, and the Fluxion —- wil

be that of the Hyperbolick Aſymptotici
Space. Whence the Space will be = —
—— 7-7″,

Whence Logarithms may be repreſented al.
fo by the Hyperbola: for if the Side 4B of
the Ae of the Hyperbola be = 1, the Ab.
ſciſs AP is a Number greater than 1, and the
Aſymptotical Space BC MP is the Logarithm
of a Number greater than 1; in like manner
the Abſciſs 4 O is a Number leſs than 1, and
the Aſymptotical Hyperbolical Space NVC
is the Logarithm of a Number leſs than 1,
Again, if y=1, then will 1 +y=2; and ſo
the Hyperbolical Logarithm of 2 will be +:
+xy—7, &c. But theſe Series converge very
ſlowly; which may be remedied by ſubſtitu-

ting — for y.

Note, Theſe Hperbolical Logarithms are
the ſame as Napier’s; and ſo are different from
Briggs’s, which we commonly uſed ; but they
may be reduced to Briggs’s: being to his, a
the Hyperbolical Logarithm of 10, viz.
2.392#8#092994, &c. to Briggs’s Logarithm
of 10, that is, 1.000000000000, ec.

rere.

Ir z be an odd Number, whoſe Logi-
rithm is ſought; the Numbers 2 —1

and 2+ 1 will be even; and fo their Logs
rithms and the Difference of their Logarithms
will be given, which let be y. Likewiſe the
Logarithm of a Number being a *
I cal

APPENDIX. 185 |
wil Mean between the Numbers z—-1 and 2 ＋ 1, |
viz, half the Sum of the en The |

181 |

FR I I 7
2 127 Ren Dn + 3602z* T If1202” *

, &.. will be the Logarithm of the

272002
a. Ratio, which the geometrical Mean between |
the Numbers z—1 and z + 1 has to an arith-
Ab. metical Mean, viz. the Number 2.

hm Pos VL

ind 115. IT a Body freely deſcends by its own Gra- F 16. 70.
1 vity from the Point A along two inclined

. Plains AB, AC, to the Points B and C; it is

‘@ WW 7equired to find the Proportion of the Times of

WW Deſcription.

a, Let ADE be a vertical Line, and B D, CE
horizontal ones. Call AD, a, AE, b, DB, x,
and E C, z. Now it is evident, that the Time
of a Body’s deſcribing an infinitely ſmall Part

of any Line, as 4B or AC may be taken

| premiſed, 4 B=y/aa+xx, and AC=ybb+zz.

m The Fluxion of the former will be ===,

2 2
and that of the latter

Again, the Velocity of the Body deſcribing
| the infinitely ſmall Part of AB expreſſed by

4
4 22 Bb, and that of 4 C expreſſed by
ne ZN

117 Cc, will be equal to the reſpective

R r Velo-

186

Fic. 71.

ſcribing Bb and Cc are as 2 and

APPENDIX,
Velocities that the Body falling perpendicular.
ly from 4 will have in the Points D and E,
the Velocities of the Deſcription of the very
ſmall Parts B, Cc being lo oked upon as =

ble: which are proportional to 4/ AD . (Va)
and /AE (V/. Therefore the Times of de.

a + ax

SY: – 22
FI (the Times of the Deſcriptions of
any Spaces being as the Spaces directly, and
the Velocities inverſly.) Conſequently the
Time of deſcribing AB to the Time of de
ſcribing AC, will be as the Fluent of the for:
mer “Oy to that of the latter. But theſe
Fluents are eaſily had from the fifth Form of
Mr. Cotes’s Tables, by making 6=1, — 9 ]

that of the former Fluxion being /= 4 5
and that of the other 9 So that the
c
Time of the Deſcription of 41 B ku that of th the x
2 of AC will be a5 <=) =
8
|
77 4 er ==) * |
P R O B. VII. a
1 16. 2 find the Nature of the Curve BC
being ſuch,” that a Body freely falling
by its own Gravity perpendicularly from the Pont
A to B, and thence continuing to move on along |

the faid Curve, ſpall deſcend equal Spaces in equal

Times.

Let BD be the Axis; then let AB=4,

the Abſciſs B P=x, and the Ordinate n_ .
OW

APPENDIX.
No the Fluxion of the Ordinate x is o
in the Point B the Beginning of the Curve,
becauſe it is convex next to the Axis, and the
Axis is a Tangent to it in B; therefore Mm
(V/x* +3*) is = x at the Point B; therefore
the Time of the Bodies deſcribing the Arch

Bb is as T3 ſince the Times are as the Spaces

directly, and Velocities inverſly : and the Ve-
locities acquir’d in B and M being thoſe ac-
quir’d by the Fall from A to the Points B and
M, (which Velocities are in the ſubdupli—
cate Ratio of 4B to AP), and the Time of
the Deſcription of the ſmall Arch Mm is as

ax
the Problem, during the Deſcription of the.
Curye, the Body falls equal Spaces in equal

But ſince from the Condition of

| Z Ik
Times; therefore — mult be = _

. 5 | – | ** =? 1 b

ſquaring both Sides, there ariſes =o
7 ‘ a

Whence ax – =ax* ; that is, 45
=xx*; and extracting the Root of both

Sides, we get Ya = . Laſtly, finding.

the Fluents, and a) = 4 *, or 2 a9 = x3;

Which is the Equation of the Curve; and ſo
it is a Semicubical Parabola.

Otherwiſe ;

Let AP x, PM y, the Velocity at the
End of the Fall from A to M , and z.=
Time of the Fall to M. Now from the Prin-

Ciples of Mechanicks — is as T, the Time

Rr 2 “0

188

APPENDIX.

of Deſcription of the ſmall Arch Mm, that is,

bes. Whence if 4 be a proper

ſtanding Quantity, we ſhall have ay/z+;=

Butvisasyx, or as V; and fo we
may take v =y/ax. Alſo ⁊ is as æ -a; ſince from
the Condition of the Problem, the Time is as
the Altitudes from whence the Body falls,
W hence x—4a may be taken for z, and x forz;
ſo that ſubſtituting x for z, and / for v in

the Equation a F v, and there comes

out ay/x-+3=xy/ax. hence a

axx*, and a.) =xy/x—a. Then finding the
Fluent of each Side, ( which may be eaſily
done from the little Table of Curves that may
be ſquared page 34- or from the third Table of
Forms of Mr. Cotes) and a) will be =
Lax—24″ : L 2.X—24
— = ax -a, or a) = Van ad.

. 2

Now making a- a, and we have T yan,

1 3 .
or a „* 2, or 2 ay =: therefore the

FIC. 72.

73 4
Curve ſought is a ſecond cubical Parabola, 4}
being = 4, and B Pn.

PR O B. VIII.

70 find the Law of Refraction, admit.
ting this Principle, viz. that Nature it

all its Operations takes the ſborteſi ways.

Becauſe Light cannot move in different Me-
diums with the ſame Velocity; let the Ratio
of the Velocity of the Light during its Mo-
tion from A to B, where it begins to be rc
10 0 74 fractech

APPENDIX.
fracted, to its Velocity, while it is refracted in
irs Motion from B to C, be expreſſed by 73

| then the Times of the Deſcriptions of 5

Lines AB, BC will be as * AB to 2x BC.
Let fall the Perpendiculars 49, CP, and

make AD =a, CP = b, PQ, PB= =x;
then will BY =c x, and | conſequently B C
= xx, and AB=yaa Tec — 2c xx.
Whence the Time in which AB + BC is mo-

ved thro’, is=my/bb+xx+ny/aa+ic—2icxtxx,

which mull be Eu and ſo the Fluxion
nxx

thereof, viz. — + —=

ybb 5 ** yaa Pe — 2

=o. Whence S
y/ bb == Vaart

PB Au NEO 9 25
that is, FT > 7h Make BC As,

then will mx PND BN, and conſequent-

ly m: u:: BO: PB.

Whence if BA or BC be taken for the Ra-
lus, B ©, will be the Sine of the Angle A,
nd PB the Sine of the Angle C; that is, o ſince

19 and PC are parallel to DE, PB is the
ine of the Angle CBE, and B © the Sine of
ABD, viz. PB is the Sine of the refracted
Angle, B © the Sine of the Angle of Inci-
dence. Whence the Sine of the Angle of
Incidence is to the Sine of the refracted Angle
na conſtant Ratio, viz. that of the Velocity
of Light before Refraction to the Velocity du-
ring its Refraction.

PR OB.

189

190 APP END IX.
„ |

Fic, 73. 118. 1 10 find the Angle BCD, in which a Bo-
dy from A obliquely ſtriking a Plain in
C may be reflected to a given Point B, ſo as to
paſs from the given Point A to the given Point C
the ſhorteſt way.

From A, B let fall the Perpendiculars AE,
BD. Let AE S, BD=6b, ED Sc, and
EC x; then CD c – x, and A C
Vaa T xx; alſo CBS Tec cx T xx;
and fo AC+CB muſt be Minimum, that is,
aa + xx + y/bb+c—2cx+xx. Whence the
Fluxion of it muſt be made equal to o;

XX—CX

therefore — =
Vaa – Vic —2cx+x*

and ſo x/b + —2cxÞx* TX -N + x*
=o. Whence b + – 2c + xx x
x Va A; that is, ECXCB=CDxXAC;
and ſo EC: AC:: CD: CB. Conſequently
(by Prop. 7. lib. 6. Eucl.) the Triangles AE C,

BDC are equiangular: Whence the Angle
ACE muſt be = Angle BCD.

one .

JF a thin Fluid conſiſts of equal Particles

= freely diſpoſed at equal Diſtances from
each other; it is required to find that Fruſtum of
a Cone, which of all others of the ſame Baſe
Aa, and Altitude B C, moving in that Fluid ac-
cording to the Direction of the Axis, with the
er Baſe Dd foremoſt, that ſhall have the leaſt

Reſiſtance.
It

APPENDIX.

It is the ſame thing to conſider the Fruſtum
at reſt, and the Particles to move againſt ir
with the ſame Velocity.

Draw DE parallel to B C. Let the given
Altitude BC= 6b, the Radius of the given
Baſe a, and AED x. Now it is well known
that the Effect of any Particle of the Fluid
ſtriking the Surface 4D of the Fruſtum ob-
liquely in the Direction DE to move it ac-
cording to the ſame Direction, is to the Effect

of the ſame Particle ſtriking directly againſt

the Annulus generated by the Line AE (while
AB revolves about) to move it in the Di-
rection D E, as the Square of the Sine
of the Angle of Incidence FEY or ADE
to the Square of the Radius. And ſince here

the Angle of Incidence is invariable, the Ef-

felt of all the Particles ſtriking the Superficies
of the Fruſtum generated by A4 D, will be to
the Effect of all the Particles that can ftrike
the Annulus aforeſaid is in that Proportion,
that is, the Reſiſtance of the Superficies of

the Fruſtum generated by AD, is to the Re-

ſiſtance of the Annulus generated by AE, as

the Square of the Sine of the Angle of Inci-

dence is to the Square of the Radius.
Whence if BC (5) be made the Radius,

the Sine of the Angle of Incidence will be

h |
=, for 4D (VFF#) : AE (H:: BC(b)

9 Now if the aforeſaid Aunulus be

made the Reſiſtance of itſelf, then will the
Circle deſcribed by B E be the Reſiſtance of
itlelf alſo; but ſince Circles are to cach other
as the Squares of their Radii, therefore the
Reſiſtance of the Annulus is to the 8
: 2 O

192

APPENDIX.
of the Circle, as AB — ZIL (2ax—s®) is to
BE (a—2ax+x). Which Quantities now

let repreſent the reſpective Reſiſtances of the
: Hnnulus and Circle. | p

— 2 22 xa 3 2
Then BC (þ * NN ſquare Sine : : AB

— BE (rax—g): = = Reſiſtance of

the Superficies of the Fuſtum generated by
AD. To which adding theRefiſtance of the

lefſer Baſe Dd (being a—2ax T) and the
Reſiſtance of the whole Fruſtum will be

b* + x*
. The Fluxion of
b* +x
which muſt be a Minimum; therefore
2ab%xux+2ab x x—2ab% | | nr 6s — =o. Whence «„ b + x| 2 8 s 1» 2 =4; and fo x=L „ or xt — —. But
4 24 24

becauſe the Triangles AE D, ABV are ſimi-
lar, therefore AE (x) = – VP + x — BC

: 5 |
b)::AB(a): BY = hich
0 e S 43 * k

is = LVA THC; ſince

V +6—TbXt/ 4 + v +T”9 is , as
appears by bare Inſpection: the Sum of any
two Quantities drawn into their Difference be-
ing equal to the Difference of their Squares.
From hence ariſes the following Conſtructt-
on. Biſe& B C () in G, and draw 4G: in Bc
continu-

APPENDIX,

continued out, make GY = AG (4 y/4a* +9″)
and will be the Vertex of the Cone.

P a On. II.

7 find the Duration of a Pendulum of-
cillating in the Curve of the Cycloid.

Let the Diameter of the generating Circle,
or the Altitude of the whole Cycloid be Sa;
and let HB, the Altitude of the Point Q from
whence the Pendulum begins to fall, and de-
ſcribe the Arch QB, be . Alſo let HP
= 2; and ſo PB=2b—z. Now let the
Time of the Pendulum’s deſcribing ©B be ,
and on HB deſcribe the Semicircle ZN B, and
draw PM, pm infinitely near one another, and
perpendicular to H; then will PN be =
VDZ, PP N= Rm, and the Velo-
city in P, and fo in N and © =yZ.

Conſequently ſince the Particle of the Curve
Mm is deſcribed by an uniform Motion, the
Time of the Deſcription of the fame, viz. x

is =” But from the Nature of the Cy-

2
eloid Mm: m R:: BS: BP, and AB: B S:: BS
:BP, from the Nature of the Circle. Whence
BS: B P:: VB: YB, and ſo Mm: m R: :
— — Mm R XJ AB

vAB: HZ; therefore Mm MET; Tae.
2 _— — pr But
VUE 20 ZE -Z

3 Nn; therefore x = 1

UhY—2Z2Z 2b

Now when the Fluent of &, viz. x does ex-
S1ſ preſs

193

Fic. 75.

194

APPENDIX,

preſs the Time of the Deſcent BD of the
whole Arch of the Cycloid, the Fluent of
Nu will be = the Periphery HN; therte-
fore as 2b the Diameter of the Circle to the
Circumference thereof; ſo is 2 y/a to the
Time of the Pendulums deſcribing the Arch

B. Conſequently becauſe 2 74 = 72 de.

notes the Time of the perpendicular Deſcent
thro’ AB, we have the following Theorem,
ig. The Time of an whole Ofcillation thro
any, Arch of the Cycloid, is to the Time of
the perpendicular Deſcent thro’ the Diameter
AB of the generating Circle, as the Periphe-
Ty of a Circle to the Diameter.

Sos r2 4 WH
Hexer the Times of deſcribing all Arches
of a Cycloid are equal.
e. XII. |

7* E Courſe and Difference of Latitude

| of two Places being given: to find the
Difference of Longitude.

FI c. 76. Let P be the Pole ; the Circle ABF, the

Equator; PCB, PD A Meridians; AC@ the
Rhumb Line paſſing thro’ two given Places 4
and C. Draw Pd cb infinitely near P CB, and
with the Diſtance P C deſcribe the Arch CD.
Now make the Radius PA a; the Difference
of Latitude of the two Places, wiz. AD or
BC=y; the Difference of Longitude ſought
AB=x; the Tangent of the given Courſe (0!
conſtant Angle that the Rhumb Line 11
e wit

4a af , ow

hes

APPENDIX,

with any Meridian) viz. of the Angle Ce m;
alſo make the Sine of the Latitude of the
Place Cr, and the Coſine = z. All theſe
are variable Quantities except 4 P (a), and
the Tangent of the given Courſe = in. 0

Now ſince the Arches Bb, Cd are ſimilar; ;
therefore PB (a): Coſine of Lar. of C £58

BG: CA Again, from the Nature of
n

the Circle, P3 — Square of Sine Lat. of C
viz, 47 is , and do(3)w=y/if +20,
as caſily appears. Whence throwing the E-
quation « a. — 1 2. into y luxions, and we

J. 10 1 1 2 7

get 2 — . And ſo —— *
. Peat. Tn
for £ * in. the Equation 5 2 V. =? „and

there ariſes ‘5 ; 5 If j) be made the Ra-

dius, then will Cd be the Tangent of the’ An-

Ced of the e = . cherefore Ph (a)

n: C JN Cd = ===; and drawing the

Meche and Extremes into each other, there a-

riſes my =2x; and ſo x = = Whence ſub-

mar
Qlituring — E for 3 55 and we have # Yn =

Sar” 5 ; * ber the Plaens there ariſes
m__

“E |
24 n . 5+ 7 &c. —

J

Difference of Longitude 0 EA Places A and C.
“The Fldent of — -7 May be had likewiſe

after Mr. Coter’s way in lte Meaſure of a Ra-
4 9 2 3 tio;

195

|
|

1 ˙¹ü—J—ñ ö̃ũ ũ2T A

APPENDIX.

tio; -for it may be referwd to the ſecond Form
of his Tables; whence making 0 o, „,

Sam, ea „=, R 532 =) = = a,

7 * IP, =r, and & Seek. = ao

| 12 2 £ 1 ＋ 1 11—1
F on of the F orm. — J
the uxi N 2 8

become — ang the F luent 2 75 7 * A 2

a +r 1 +; Wars
. = —— = in the Ex.
preſſion for the Difference of Longitude BA

That is, in Words, the Difference of Lon-
gitude is equal to the Meaſure of the Ratio of
the Nadius added to the Sine of the Latitude
of the Place C, and the Sine Complement of
the ſame, the Tangent of the Courſe being
the Module; and if E be ſome other Place
in the ſame Rhumb Line, the Sine of whoſe
Latitude is given; then by the ſame Rule we
can get the – Ls of Longitude A H, and
ſo the Difference of On BH of the
Places C and E.

will be MM

COR LL.

F the Rhumb Line AC be = #, 1 cd

(5) be made the Radius; then will Cc be
the Secant of the Courſe. Whence PB (a)
Ce: :,): Cefn); and ſo au Cc, and

taking the Fluents a =Ccxy, therefore «=

= z – that is, as the Radius i is to the Secant

of * Comte ſo is the Difference of Latitude
AN to the Length AG Y. 5
RO B.

APPENDIX. 197
k P „ Kik „„

4, 122. 1. cube the Solids generated by the Rota- FI d. 27.
tion of the Conchaidal Spaces CP GB,

and BGQ<c about the Line ABC drawn from
the Pole A at right Angles to the Aſymptote BG.

il Draw Ap infinitely near AP; from A de-
ſcribe the ſmall Arches Os, Gr, Pn, with the
Diſtance B Cor Bc, from A deſcribe the —
EF, and from F, 2, Pdraw FH, 9 K, PI pe
«. pendicular to AB” Call c (= 4E 2
BC=9G=CGB) a, AB, b, E Hy x, AH, z,

and HF, . Now from the Similarity of the
n- Trangles 4B G, AH, we have 4 H(z): AF

{Y 9:43); . Whence 42 =
—4, and for the ſame Reaſon 4b ＋ 6.
ginn, ene, ins

1d =D So likewiſe PI T. Now

e the Fhixion of the Arch EF, viz. Ff will be
M and fince the Sectors 4 Ef, A Os are fi-

milar, therefore AFla): Ff (=): 4 2.

JC 25. . 4

F (< 2): D5 =— —az. In like manner
1d = kat —

” NED ada 3

nt Again; the Fluxion of the Solid generated

by the Space 15 the Motion above-
e nentiom d, will be = 1 42x Qs drawn 0
| t

[
[
:
L
f :

; 4 «
7
g 11
k a il
11
7
11’Y
G
N

—

— — — 22 _

*
K ͤ —˙1 . wy
— — — —

——

— —
—

— ——— —
— —
— — — —

— —
— — “IEEE.
— — 22 —ůů —

198

APPENDIX,
the Periphery deſcribed by the Point O, 80

alſo will the Fluxion of the Solid generated
by the Space ACP be = APR Px into the

Periphery deſcribed by the Point P, and the

Phixion of the Cone generated by the right-
angled Triangle 4 U during the aforeſaid
Motion will be; AGxC7 into the Periphery
deſcribed by the Point E; therefore the Flu-
xion cion of the Solid generated | from Ac will be
ab_a a „2 — —yx 225 — 128 (2 beit
3 3,7 . 9

the Ratio of ihe Radius to the Peri iphery of a

Circle) =Z x <= — a 4 . 1 nd

211 7 23 w 2 * 2

the [oy — of the Solid * from Act

1 ** =
hs ban 073 by Js { G. Be
1228142 3 als FEE WY of

theſe two laſt mentioned Fluxions will be =
2 * 5 5 & &=;Fluxion of the Sum of

OA
the two Solids, generated by, the Spaces OS,

and ſo £ SE Ins
BCPc:, an ſo, EX 2 2473/5 is the Flu-
xion of; 2 . + 80s ei Flucnt of which will

——

be l 2 ö * 25 eee for 2)

n 2a 4 ar is
2 . 2 1 2 | . ns — —
2 IE: 1 3 43—=a ** * “|

39045 n bilo3 dn tr 75 *

oy making” S o, — Fluent: become:

2 EP to be ſubſtracted: fo that the true
Fluent

| * Much after the ſame way as the Fluxion of

ATDENDES.

rec: 4 33> 24*Xx 42 **

Fluent i 2 » DET I

the Sum of the Solids was found, we may

get the F luxion of – the Difference : for from
the ſimilar Triangles AHF, AB G, _ ari-
ſes AH (z): AF (a):: AB 00) AG= == and

AH): HF()y): 400. 36 2 Hand fince

che Sectors AFf, A Pr are alſo rw there-

fore IAF (a): Ff (SE): AG (=): Gr==;

and conſequently 8 Fluxion of the Cone de-
ſcribed by the right-angled Triangle 4 BG
73%
will be = = R
„ /\As #1 588)

which en the Fluxion (before found) of the

, from

| Solid deſcribed by the Space 4.5 and the

Fluxion of the Solid kene by the —
ACP,. the Remainder x PE.

is the Fluxion of the . Solid BG PC.

from

— —— — *

200

APPENDIX.
From whick ſubſtracting the Fluxion of the
Solid c Q B, and the Remainder?x L i
the * of the ſaid Solids; fo that
2 „Dis the Fluxion of = the Difference of

the Solids. Which may 7 — compar’d with
that of the firſt Form in the Tables of Mr.
Cotes, (having firſt ſubſtituted a—x for * and

& for 2, the Fluxion becoming 2x: * =)

1 4 —*
For making 2 = x, GI, n=1, d=a’b, =

2
e+ *
2 * LALLY and the Fluent a F becomes

T7 A—X

f=—1, the Fluxion will be =

=—P aab — I = = aab >>, ſince the
r 5 a r ma

Logarithm of the Ratio of a to a— with an
affirmative Sign, is equal to the Logarithm of
the Ratio of a—x to a with a negative Sign.
The Fluents or Quantities of 4 the Sum and

the Difference of the Solids being thus found
we proceed next to their Conſtruction, begin-
ning with that of the half Sum. From what
td dawn already ſaid, it is caſy to find the Flu-
— of the Sector of the Sphere deſcribed by
the circular Sector AE F, and fo the Fluent
or Quantity of that Sector z the Fluxion be-

ing = x £ Le — and the Flu-

4 y
2 22 az _þ aj x
= *-X Ke” * „ by ſubſtituting

a—x for z; and making x=0, the ſaid Flu-
ent will become # ** to be ſubſtrated from

the

APPENDIX. 201

3
of ou. wr deſcribed as aforeſaid.
hat This being granted, make AZ (aa) : 3AB

Ae ZE G = +a0):: Sector Sphe
ir 21 45 : Solidity of + the Sum of the Solids

| t l * * oo! DL — — > being the ſame Ex-
niſin as that before found.

Lafhy To conſtruct the Expreſſion
| Liab , we know that E aa is the Area

7 – Jang, 1 27
| of aCircle, uns B C (a) for a Radius; which
| drawn into þ – will be the Solidity of = the

Difference. “thy becauſe of the ſiinilar Tri-
| angles AHF, AB E, the Ratio of a AF to
a- AA, is equal to the Ratio of 4 to
| AB; therefore the Value of + the Difference
of the Solids to be cubed is equal to a Cylin-
der, the Diameter of whoſe Baſe is 4 C (2 a);
and Altitude the Meaſure of the duplicate Ra-
tio of 4G to AB, the Module being AB (Y.

for La a5 a. — 1s =D. 3ab |£-; becauſe the
27 — r

Logarithm of any Ratio Joubled 1s that of

the duplicate Ratio.

Tt PRO B.

202

MPCRYOLK
Pros. XIV.

115158 ů —

Fic. 78. 123. To, find the Natare of a Cure AMC

being ſuch,” that if a Veſſel be deſcribe
by the W of the. ſame. ahout’ the Axis
AB perpendicular ta the Horizon; and then filled
with Water, which . afterway ds runs out of a
{mall round Hole in the Bottom A, the Sur face

of the Water ſhall deſcend equal Spaces in equal

Times; admitting the Velocity of the Water run-
ming out, to be as the ſquare Root. of the Al-
tude of its Surface above the Hole. |

Let 4 B=a, AP x be any, ai
Altitude of the Water, and let PM=y. Al-
ſo let the Surface of the Hole be = 5, the
Velocity v, and the Time of Deſcent of
the Surface = 7. Which from the Condition

of the Problem is as a—x, but . is as v; that

is, as V or as yan, from the Condition of the
Pr oblem; 3 therefore 9 : Surf. Hole 5:

Surf, Water)“. But {ince ? may be taken =
a—x, therefore: 11 and ſo1:yax::6: T’s
W hence. b – ine =, and ab). Conſe-

quently the urve A s M is a We Pa-
rabola.

Pro B. XV.

F1c. 79. 124% F AC be a horizontal 3 the

Point C of which ſtands an upright
Parallelepipedon CD; one of the plain Sur-
faces of wich is perpendicular to AC; it is
= to find the Augle CAB, in the

= aint A of which ane End A of a long
Solid

Joo. . oP

APP EN PD IX.
Solid AB being ſet, ſo as with its other End B
it may bear againſt the Parallelepipedon CD; the

ame ſpall preſs perpendicularly againſt CD with
a greater Force than if it was ſet at any other

Point befides A.

Let Y be the Centre of Gravity of A B.
From F draw F perpendicular to AC, and
from E, EI perpendicular to AB. Alſo from
B,BH perpendicular to AB, and = A; and
from 1, 11G perpendicular to BD. Make
AB=3, AF=b, and CB x.

Now the Preſſure of the Solid AB in the
Point B, and Direction IIR againſt CD will
be as AE, and ſo may be repreſented by it;
therefore. ince BH = E, the direct Preſſure
againſt CD in the Point ‘B will be = HC.
This is ſhewn in the Principles of Mechanicks.
Whence C Bx7@ is the Effect of the per-
pendicular Preſſure of the Solid AB againſt
CD in the Point B, which muſt be a Maxi-

Mun. Again, AB (a): AC (ya’—x*):: AF
Gb): AE SD Fo =; And ſince the Tri-
angles AE 1, AC are ſimilar, therefore AB
(a): BC(ﬆ):: (5 ya: c:) EI H=

| VC —x, W hence CBxHG = war

74 —.— G

TheFluxion bo which muſt be
24 fa
So; ; therefore 24. , and fo x 2 —

Whence as AB to 75 7B, fois ; the Radius
to the Sine of the Angle C AB $45. 44.

Et 4 PR OR.

203

—— — ͤ ——ä—ĩ —-—ͤtę—U¼ .

—— —

—
— —

l — 2
— . *
— ee — 0 * > —

— — — of
— —

2 — — — — — _— — > 49 FA .
EDS – . 1

ut – — — — – –
a – – — — III ” m — N 2 * — 8
0
<

F 10. 80.

APPENDIX,

XVI.

O find the Nature of a Curve ACEDB,

along which if a Body freely falls by its
own Weight from the given Point A to the given
Point B; the Time of the Deſcent ſhall be le;
than the Time of the Deſcent of a Body from A
to B along any other Curve paſſing thro the
given Points A and B.

PRO B.

Let C and D be two given Points in the
Curve infinitely near each other. Alſo let the
intermediate Point E of the Curve be taken
ſuch, that drawing D and E A perpendicu-
lar, and AK, CL, EM parallel to the Hori
zon, the ſmall Line EL may be = D M.

Now ſince ACEDB is ſuppoſed to be the
Curve along which the Body falls from A to
B in the leaſt Time poſſible; it will fall from

C to D, any Part of the ſaid Curve, in the

leaſt Time poſſible. For it it does not, ſuppoſe
it to fall along CCD in a leſs Time than along
CE; then the Curve 40 D will be de-
{cribed in a leſs Time than the Curve AC EDB.
Which is abſurd.

Becauſe the Points C and D are given in
Poſition, therefore the Lines HE, 1D, EL
=D M, are all given or invariable; and CL,
CE; EM, ED are variable. Make EI. =
DM=m, HE =b, ID y; allo CL =I
EM=2; then the Time of the Deſcription
of the ſmall Line CE (being as CE —_

and the Velocity inverſly) will be as THE

(= ===), the Velocity during the Deſcri-
ption

APPENDIX.

tion of infinitely ſmall right Lines being

b upon as equable; ſo likewiſe the Time

of the Body’s deſcribing the ſhort Line E D is
ED

| 87 (V __ ). Conſequently the Sum

| of theſe Times muſt be a Minimum, viz.

—ů ͤ —— —

a 7 4 + — 2 ; and ſo the Fluxion of

un MEFS T
% bu pi I
muſt be = o, but C Mu +2 is invariable;
whence i. ＋2 — Oz and fo z = —2) therefore

15 2
biin +u piym TA

Now if Abe = x, and HE =y; then
will EM , and MD ); and ſo ED=
/x+3. Conſequently the Fluxion of the

Curve is always as =; that is, in the direct
1

Ratio of the Fluxion of the Abſciſs 477 (x),
and the reciprocal ſubduplicate Ratio of the
correſpondent Ordinate HE (7).

Now a Curve that has this Property
will be found to be a Cycloid, paſſing
through the given Points 4 and B, with the
Vertex downwards, as may be eaſily ſhewn
thus: Suppoſe A and B to be fo poſited, that
ACD be a Semi- cycloid, &c.

Deſcribe the generating Semicircle X PB.
Continue out EM to P and 2; Draw the
Tangent CET to the Curve in E, and from B
draw the Chord BP. Let BK=a; then

the ſame, which is

205

is a noted Property of the Cycloid for the
Tangent E to be parallel to the Chord B H
0

206

FI c. 82.

“pz .
APP⁰PEHND IX.
of tlic correſpondent Arch of the generatin
Circle: W hence the Triangles BO, and EC

are ſimilar; therefore P, (VB

Gu = Y CL GY): bg = SYED =
vVay—y

Which Expreſſion

—

* JAN Ja—y NV
DN wW-

is as A, becauſe V is given or invariable;

8
therefore the Cycloid ACE D B is the Curve
of the ſwifteſt Deſcent.

PR OB. XVII. —

1 O find the . of a Curve D M be-

ing ſuch, that the Solid deſcribed by the
Revolution of it about the Axis AP, moving in a
Fluid (ſuch as that in Art. 119.) in the Diredti-
on of the Avis, ball be leſs reſiſted than any other

| Superficies deſcribed by what Curve ſoever termi-

„„

nating in the given Points D and M about the

ſame Aris AP, and moving in the ſame manner.

Let the right Lines MN, NO be ſuppoſed
to be two aun irdy fall Parts of the
Curve ſought. Now the Surfaces deſcribed
by theſe will meet with a leſs Reſiſtance than
the two Superficies deſcribed by any two
Parts drawn from the Points O and M to any
Point beſides V. This is very plain: for if
the ſame be denied, the Conſequence will be
that Nis not in the Curve fought.
This, being granted; from the Points M,
N, O, draw MP, N 5 O perpendicular to
the Axis; and from M draw M patallel to
the fire; alfo thrs’ draw another *
Paralle

APPENDIX.
Parallel GMI cutting OZ in G. Now the
Reſiſtance of the little Superficies deſcribed by
MN, is to the Reſiſtance of the little Aunu-
lus deſcribed by FN, as the Square of the Sine
of the Angle of Incidence to the Square of the
Radius; that is, (becauſe the Angle TMN is
Angle of Incidence TNM) making MN

the Radius, as FN to TN And if the Re-
ſiſtance of the Aunulus, deſcribed by E M be

repreſented by itſelf; or rather by S +
2 N Fx FN FN mii, DF being the Dit.
ference of ON and 9F ) which Remainder

s29 FxXFN +EN 3 this Difference, being
always as that Annulus. But becauſe FN. 15

infinitely leſs than Q, therefore Nis inft-
nitely leſs than 2 2 FHN; and ſo it may be
cried: And the Reſiſtance of the Annulns
aforeſaid” will be 2 & Fx FN; or 9 Fx FN;
or PMx FN. Whence at length making

AV Fr PEN. NEXPM ang

——

{if MN
this will expreſs. the Reſiſtance of the Super-
ficies deſeribed by MN; fo will E222

be that deſcribed by ON.
Again, let the Points O, M, and the right
Line G be given in Poſition; then we are to
enquire into the Situation of MN, NO, being
ſuch that the Reſiſtance of the Superficics de-
{cribed by them, be leſs than the Superficies
deſcribed by any other Lines O u, 2. M. In or-
der to this, let PM Da, FV =, GO c,
N. Se. Theſe are all given, and. invariable;

and

207

208

APPENDIX,

and let the variable Quantities MN be S,

and NO Dez.
Then will the Reſiſtance of the Superficies

deſcribed by MN be = , as has been ſhewn

above. In like manner 2 will be the Reſiſtance

2
of the Superficies deſcribed by ON. Whence

a TY . ahixs

— muſt be a Minimum; that is, .
x 2 **

ects al*x ec ab;

2 x 23 x3

ect
“bb.

Now in order to get an affirmative Value of
— x, affected with 2; aſſume the Point » in-
finitely near N, and draw the right Lines Ox,
Mn, to which draw the Perpendiculars NS,
NR. Then ſince Nu, and conſequently NS
is infinitely leſs than NM, therefore M N=
MR, and fo the Angle MVR = Angle MRN
== right Angle. Whence the Angle R Ns is
= Angle FN; for either of them added to
the Angle MN # makes a right Angle. Like-
wiſe ſince the Triangles N $’n, »GO are equi-
angular, they having each a right Angle OG,
Nn, and the Angle O NG common; there-

fore the Angle S Ns is = Angle SOG = An-

leNOG; ſince the Angle NOS is infinitely

ſmall. Therefore making Nu the Radius, we

have Rn (— ): $3 (S) : : Sine Ang. F VMM:

Sine Ang. GON. Suppoſing MN the Radi-

us, (that is, making MF = m, and NG =.

and aſſuming NI. MN, and drawing LK

parallel to OG) as MF(m):NC(7); therefore
u

= ==; which being put in the Equati-

0x –
_

99 << 1 Go

» Oh <<

SA. > i.

APPENDIX.

ab m __ ecu
44 24 ;

Now draw AB (a) perpendicular to the
Axis AP, and the right Lines BC, BE pa-
rallel to the two 2 ſmall 8 MN,

NO; then will 44B „AC (4a x = Ic
0 =): :BC (E)- MP (a); and multiply-

ing the Means and Extremes, we have 2
=+, or — 4 In like manner, 4 AB

AP: TA. BE. VN =.

2
Therefore the Nature of the Curve DM is
ſuch, that if 4B be taken in the Line 4X
perpendicular to the Axis a; and drawing
BC parallel to a Tangent to the Curve in wy

Point M, we have always 4 75 B x AC: FT
BC: MP the Ordinate.

Now chis Curve may be deſcribed by the
Logarithmick Curve thus: Aſſume AB 4

in AK; and 1 in A continued out towards A,

take AE SVH ; and thro’ the Point Z de-
ſcribe the Logarithmick Curve FEMN to the
Aſymptote ARA, whole Subtangent is S 72.
Then aſſuming AC (ſuppoſe = 5) at pleaſure,
and drawing – N parallel to A R, — AR =

an
<+i5 + Land AP = — 5

i CN, w — when 48 is greater 8
4 E, and A+ when it is leſs; and draw the
V u rigat

209

210

APPENDIX.

right Lines XM, PM parallel to 4 P, AX;
then will their Point of Interſection M be in
the Curve D M ſought.

For making AP x, PM =), and 4C
=z5; the Property the Curve mult have, gives

AK or PM (y) = bn (hom OS. con-

—

4 24 4a

40 = 3, minus the Fluent of 8

us or minus ſome invariable Quantity. This

Quantity we will take to be Za, which we
48

ſubſtracted, that ſo CN, which from the Na-

ture of the Logarithmick Curve FE N is the

=o allo. Whence, &9:.

When AC AE, the Ordinate PM, which
is then a Minimum, becomes 4D —=4 AE, and
the Tangent in D will be parallel to BE. But
if AC be taken leſs than AE, the Part DO
of the Curve will be deſcrib’d conyex towards
DM, diverging more and more from AP,
A K. Therefore the Solid of the leaſt Reſiſtance
may be convex or concave, or partly convex
and partly concave, and the Point D 1s a Point
of Retrogreſſion.

Note, The Logarithmick Curve may be
deſcrib’d eaſily after this manner. For you
need only take C M to the Meaſure of the
Ratio between ZE and 4 C, the Module be-
ing 4 A3 (a).
| | PRO].

e _ oo: ww 7+

APPENDIX.

Polt.

211

f he find the Angle ABC, which the

Plane of the Sail of a Windmill in Fi-

gure of a right-angled Parallelogram, whoſe given

Breadth is CB, makes with the Axis AB being

ſuch, that a given Wind blowing in the Directi-

on of the Axis, ſhall drive it round with a greater
Force than if it had any other Iuclination to the
Axis.

Draw CD perpendicular to the Axis. Let Fe. 84.

CB=a, and D B=x.

If BC repreſents the Number of Particles
of the Air ſtriking the Sail when it is perpen-
dicular to the Axis A; then will. D be the
Number of Particles ſtriking, when the Sail
is inclined to the Axis in the Angle DB C.

Now it eaſily follows from the Principles
of Mechanicks, that the Force of the Sail in

the Direction DC will be as BD KDC DC

=BDx DC; chat is, the Sail will be
drove round by the given Wind with

a Force that is always as BD x DC »
which conſequently muſt be a Maximum. But

BDx DC is =x x aa—xx=4x—x.Whence
** zx =, and a =3x; therefore 3 a
=x*’, and ſo Va = M. Whence as CB to

+a, ſo is the Radius to the Sine of the An-
gle C; the Complement of which is the An-
gle ABC ſought. Conſequently the Angle

C15 35. 16, and the Angle ABC, J. 44″

Uu2 Seno-

APPENDIX,

SCHOLIU N.

T His Problem, on account of its eaſy So-
lution, ſhould have been antecedent to
ſome of the others, eſpecially the two laſt
Problems; but before I thought of it, all the
others were printed off: yer rather than leave
it out, I have added the fame here. Monſieur
Mariot, in his Hydroſtaticks, has endeavour’d
at a Solution of it; but his Concluſion is
wrong. Monſieur Parent has done it, as we
find in the Hiſtory of the Royal Academy of Sci-
ences at Paris; but we have not his Proceſs,
neither in the Hiſtory or Memoirs, as I can

find.

I
10 A 65

” _ * – 4

K — en SR — * 5 — – – – 2 ik 2 ER \

ERRATA in the APPENDIX.

AGE 3. line 14. for denominates r. Deno-
minators. P.y.1.1.forbr. x. P.7.1.11
and 14. for Surd r. Series. P. 10. I. 3. for PS

r. N. Ibid. 1.7. for 4 (=P=)ay. r. 4 (=

F) . P.23.1.21. for will be x5, r.willbe
xy. P.24.1.27.forar. an. P. 48.1. 11. for
A — 2ax + ub

a a

P. yy. 1.11,12.for

b — —

5 P —— h — Ln
=/aa—aa, and -*y/aa—aa,r. V Naa, and

= F. 78. I. 1. for Art. 1 1. r. Art.

Ibid. 1. 2. del. =. Ibid. 1. 10. for PHr.
CH. Þ. $0. 1. 35. 8

P. 60. 1. 18. for /x—xx r. x V. P. 84.
line 7. for & read z. Page 93. line 17. for
Quadrature read Reftification. P. 104.
|. 21. for Vu X ＋ r. Vr .. P.108.1.7.
for V Ty r. Y t. P. 114. I. 11. dele
becauſe x begins at B, and not at A. P. 115.
. 7. for AP r. AM. P. 118. J. 11. for
. thidem, 1 14. 7:
27 £7” +
. Ibid.L 17: Gr SEES x.

6r
= . P. 119. I. 2. for Semidiameter r.

f of the Semidiameter. P. 126. I. 7. for 5 r.

Ibid. I. 13. for 22% r. . Ibid. I. 15. for
F 8

ATA.

2 r. . Pp. 126. 1. 19. r. . P. 1 27. l. 14.
17 3

for B Dr. PD. P.135.1. 22. for and ſo

I24
r. and ſo = Ibid. 1. 23. for „n r. 1.

P. 136. 1.8. add, when Igfen’d by 4 of a Circle,
whoſe Radius is A. Ibid. I. 14. r. tranſverſe,
Tbid. I. 15. for PM r. PA. P. 139. l. 27 for
DC Er. DEC. P. 140. Art. 71. Fig. 6 want-
ing in the Margin. P. 141. J. 20. ae, but
becauſe y begins at the Centre C. ot at A
the Vertex. P. 145. Art. 73. tor Hg. 7. in the
Margin, r. Fig. 76. P. 146. | 17. 12. dele,
ſince the Abſciſs AP (x) begins at A aaa at
re. ferCr-B. P. 101. 1. 2
dele is.

—

pee N
F
b

Appenitar Plated |

|
E

iſe, entituled, Harmonia Menſurarum, ) from

: Which, tho” it be true and exatt, it is next to

2 –

be Relation of the Spaces EG QFss; ANB, t;

7

* + j* — muſt change from poſitive to ne- 53:

4

311 1 R911 1 * 7711 : 1

” : + 5 *

Q wk bw 4 bd $+4% £# Þ\ .

= . a —

4 wt wb | * 1 541 # +»

* 15 v4 4

will be in the Cauftick A FX.

3 % — one” air 13

\::a16 N. Ae e

[|

C

, becauſe while x and — 004 tf

; we ſhall get an Equation having two Roots

+

T

”

T

*

*

–

of

if

; 2

4

–

*

[ * — 22 9 . — — 7 4 7 *

* p * : ü 82 * 1

7

— — Pans —

:

bove, and we ſhall have —Xax” x Mar

and C are right ones, and the Angle P is

y/ps +px=1x*Xpa+px’x is the Fluxion

3 © ax* axs av a n

4b—tab+%ab—746, 9c. and when «x

8 bbl a+x

l :

.

J

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