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33. a**

n. log. a 34. i y. log. y +ry ON THE CORRECTION OF FLUENTS only part given by the preceding rules) differs

from the truth in some circumstance of the 40. Though, by the rules which have been problem in which its valne is known; and this given for finding fluxions, the fluxion of any tlu- difference added to, or subtracted from, the variaent may be found, and by a reverse operation the able part, as may be required, will give the fluent fluent may in most cases be found from the truly corrected. fluxion; yet the fluent so found may often require to be increased or diminished by some constant Example 1.-Let j = ax é, then y="; quantity depending on the nature of the problem

where if y = 0, then r = 0; if therefore r and y under consideration. For example, the fluxion

are by the conditions of the problem simulof x”is nx" , and the fluxion of r“ + a is the same quantity; we cannot therefore affirm

taneously = 0, the quantity is the true without reference to the nature of the problem in fluent. which the fluxion n x *--* i arises, whether Example 2., and y commencing together, its fluent is x” or x" + a.

let the true fluent of j = a FI:.i be required. The most direct and simple method of finding By the common rules y = ** , where whether a fluent does or does not require correction, and the amount of that correction, if any, when

w
any; when y = 0,

a t

which should also is to see what the variable part of the fluent (the

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Put AC = 4, CD= b, AP= P, PR = 9, have been o, hence y = -

and PQ = 1, then by conics Q B’= :19+ rected fluent.

The true Auent of this quantity, however, and 93-pr - x) and P B = (P9+91many others may be found without correction; in the present case (j=a+) if atq's pr — 2) + xo. Hence, by putting the fluxion be expanded, we have j = a1 + 3 a ri + = 0, and reducing, we get r=9 . 3 a rac + xyi, whose fluent is yra3X + 3 a 22 er at man

Example 4.—To determine the dimensions of -+ ax +

as before. a cylindrical vessel, open at the top, that shali 2 4

contain a given quantity under the least internal In the preceding examples x and y are sup

are sup- superficies. posed to be equal to nothing at the same time; Let the diameter of its base be r. and its altibut in the solution of problems this will often

en tude y, and put p for the circumference of a not be the case. Thus, though the sine and the

circle, whose diameter is 1, then pr = the cirtangent of an arc are nothing, when the arc itself is nothing, yet the secant and the cosine are then cumference of the base, its area, and psy equal to the radius. We shall therefore add an example or two, in which when y=0, x is equal

the area of the concave superficies of the cylinto a given quantity a.

der. Hence P*= c, the given quantity that Let j = 2* i be the proposed Auxion, then its Auent is y = . Here when y = 0, y = the vessel will hold, or pry= ; therefore on; hence the corrected fluent is y=-=a. **+ is a maximum and consequently –

"+' 4¢i+Pri = 0, 0 = = 2:7 Again, let j = - r* i, then y=-*

; and as "Ti pa = 8 c, and p ro y = 4c, r = 2y; whence nti nt

y is also known, and it appears hence too that which, corrected, becomes y ="_

the diameter of the base must be just double the n + 1.

altitude. AppliCATION OF FLUXIONS TO THE DETERMI- Example 5.—If two bodies move at the same

NATION OF THE MAXIMA AND MINIMA OP time from two given points A and B, and proVARIABLE QUANTITIES.

ceed uniformly with given velocities in given 41. When a quantity is in its maximum or directions AP and BQ; required their positions, minimum state, it neither increases or dimin- when they are nearest to each other. ishes; therefore if the quantity be represented

Let M and N be any two contemporary posialgebraically, and its fluxion put= 0, the result- tions of the bodies, and upon AP let fall the ing equation will give the maximum or minimum perpendiculars NE and BD: produce QB to value of the quantity.

meet AP in C, and draw MN. Let the velocity Erample 1:-Divide a right line a into two in BQ be to that in AP, as n to m, and let AC, such parts, that their rectangle may be a mini- BC, and CD (which are also given) be denoted mum.

by a, b, and c, respectively, and put the variable * Tor- one of the segments, then outis cistance CN=m. Then we have b: 1:16: the other, whence a x - x? is by the hypothesis to be a maximum, and consequently its fluxion, or a i — 2 x 3= 0, when I = r, or the given term must be bisected.

Example 2.-What fraction is that whose neb power exceeds its mi power by the greatest possible quantity ?

CE= ; and n:m:: BN (3-6):AM= Let r = the required fraction, the x*— is to be a maximum, ornz"'-mom-'iso, ma = mb; and hence CM = a +

_m

nn whence <="_?

Erample 3.-From a given point P, in the Hence M N = CM + CN – 2 CM.CE= transverse axis of an ellipse, to draw PB, the shortest line to the curve :

T na
T
b

b. the fluxion of which put= 0, gives —

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and from this all the other quantities may be Erample 1.—It is required to find the length determined.

of the arc of a circle, in terms of its sine, versed

sine, tangent, and secant. ON THE METHOD OF DRAWING TANGENTS

Let C in the preceding figure represent the To CURVES.

centre of the circle; call the radius AC, r; the By this method the tangent and subtangent to versed sine AD, t; the sine DE, y; the tana curve are determined when its equation are gent TE, t; and the secant TC, s. Then by the given, and vice versa.

property of the circle
y = 251 – = 1
and by taking the fuxions of these equations,
and making proper substitutions in the general
fluxional equation, i = 1 + j?, we obtain

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Any of these quantities may be expanded in a series, and the fluent of each term being taken, a

general value of 2 will be obtained. We shall If A E be a curve, let it be required to draw a tangent TE at any point E. Draw the ordinate take as an example i = DÉ. and another dae indefinitely near to it, This form of the series, however, is one of meeting the curve, or the tangent produced in e, and draw E a parallel to the axis À D. Then the very slow. convergency, so that a great many triangles Ea e, and TED are similar, and

terms of it must be collected before a result of therefore e a: a E::ED:DT. Or y: :::y:

sufficient practical accuracy can be obtained.

But it may easily be transformed into series of =DT, the subtangent; x being the absciss almost any required degree of convergency. The

following are amongst the most useful forms that ÁD, and y the ordinate D E.

have yet been discovered ; A representing the Erample 1.—To draw a tangent to a parabole, circumference to radius unity, and a, b, y, &c., whose equation is a r =y.

the preceding terms in each series :Here a i = 2 yj; whence = 2= = 2.r; or the subtangent is double the corres

10 (1 +3:10 + 5:10+ 7.10 donding absciss.

+16 7 Example 2.-Draw a tangent to the cissoid of 1st A-3 Diocles, whose equation is yo = -

3 a x? ; – 2 r i Here 2 y j =yć 242. a 2x3 a 23

12B y =3ări – 2

g r .3arl B L 5 "*3.10 +5.10 +7.10 _ 2.ro - 3

32 – 2 x.
To DETERMINE THE LENGTHS OF CURVES
WHOSE EQJATIONS ARE GIVEN.

I. 16 y
43. In the annexed figure E a, e a, and E e, are
simultaneous in-

By collecting a sufficient number of terms of crements of x, y,

any of the three preceding series, we find the cirand 2, or of the

cumference of a circle, whose diameter is unity, absciss AD, the

to be 3:1415926, &c. ordinate D E, and the curve A E;

Erample 2.-Let it be required to determine and the triangle

the length of any parabola, the general equation Eac is (see arti

for curves of that kind being a" Mtà cle 42) similar to

r=y". TED; it may

Here i =ny" ,i=Vi+j", is theretherefore be considered as a right-angled triangle. Hence, ** ;? + j?, or i = 1 + j?. Therefore substituting for j its value in terms of fore=V ja + x, and taking the Auent, the value of x is obtained.

and the fluent of this in a series gives 2 =y+

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4n - 3

V j3 + j? = i=9 and 3 cy i = ori, - 3.80

whose fluent is a c r, the value of the spherical 67 - 5

surface. But a c is the circumference of the

generating circle; hence the surface of any ses. on &c.

ment is equal to the circumference of a great 6n— 5 · 164°"

circle, multiplied by the versed sine; or height But when 2n - 2 is either unity, or an aliquot the segment; and, when this versed sine is the part of it, this series will always terminate, and whole diameter. the extression is cal or for consequently the length of the arc will be accu- times the area of a great circle of the sphere. rately obtained from it.

Erample 2.-Let the proposed curve surface TO FIND THE AREAS or CURVES, WHOSE Equa- be that of a parabolic conoid. TIONS ARE GIVEN.

Here a r = y’, whence i = 2yy, and there 44. Adopting the previous notation, it is obvious that y ; is the fluxion of the area; and y being from the equation known in terms of r, fore i = V1 + y = 4+ 4 y, and 2 cyi the fluent of this expression is the area of the curve. Example 1.-Required the area of a parabola,

m +n whose equation is a *z* = y*

rected is
rected is c (* + 4 y 1 – 4”).

6

TO FIND THE SOLID CONTENTS OF BODIES. Here y=a n r 1, whence y i can

46. As a curve surface may be conceived to be munn-m generated by the expanding circumference of a de ", whose fluent is na n.rn. plane moving forward, as the solid itself which

n - m

the surface bounds may be conceived to be gene Erumple 2.-Let it be required to find the rated by the plane itself. Hence, if .r, y, and c

represent the same things as they did in art. 44, area of a circle from the equation y=v a rx?.

we have c ya i for the fusion of the solid, and the fluent of this quantity will be the required solid.

Erample 1.-Required the solid content of a

cone, whose altitude is a, and base b. =a.i{1-,.-*- 15 qu – &c. } Here a : 6::::y=, whence cy'i =

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TO FIND THE

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= vāki whose fluent gives r Var

2

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c ba thindar

02 r = , becomes , or one-third of a cylic- 72 21 704** -,&c. for the area of the der, having the same base and altitude,

Erample 2.- Let the proposed body be a semicircle.

spheroid, the tranverse and conjugate of whose

generating ellipse are a and b. TO FIND THE SURFACES OF Sorids.

By the nature of the curve (see Conics) y = 45. A surface may be conceived to be gene- 62

.ax — , whence e yo i= rated by the circumference of a planę moving a forward, and expanding at the same instant; therefore the fluxion of the surface is equal to the whose fluent is (4*-:- ), which when fluxion of the curve, in which the expanding circumference moves forward at any instant, multi- r = a is , the content of the whole spheplied by the periphery of the variable circumference at the same instant; and the fluent of this

roid. And if a = b, the spheroid becomes a fluxion is the value of the generated surface. sphere, whose solidity is

If c = the circumference of a circle, whose diameter is 1, r the abscissor, ythe ordinate, and

Hence a sphere, or a spheroid, is two-thirds of : the curve in which the expanding circumfer- its circumscribing cylinder, for the solidity of the ence moves forward ; then 2 cv = the circum- cylinder whose base diameter is 6, and altitude &, ference, and 2 cyi = 2cy s j? + js = the isab, of which is evidently two-thirds. fluxion of the surface S, and consequently by taking the fluent, S is obtained.

TO FIND THE Points of CONTRARY FLEXUBE OF Example 1.-Let the proposed curve surface

CURVES FROM THEIR EQUATIONS. be a sphere. In this case y = va x -- it'°, 47. It is evident when a curve is concave a - 2

a - 2 r whence j =

2 whence y = ? Va t - 22 * =

i =" 54 i, and towards its area, that the fluxion of the ordinate

2 y + decreases with respect to the fluxion of the ab2 – a? — 4 ya

sciss; and the contrary when the curve is cooconsequently j? = 9 que ;?; therefore 4 ya

vex towards its axis; hence, at the point of

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ted with the former value of f, gives on

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as of D E; hence the equation may be put in contrary fluxion or is a constant quantity, and

this form GCæ-ji GEY-ýj; or GCconsequently its' fluxion is = 0. Therefore, if :–GE 2 + ja ;; and if each of from the given equation the value of oor be the terms of this equation be respectively multifound, the fluxion of that value will give an plied by the equivalent expressions equation, from which the relation of r and y at the point of contrary flexure may be found. Ce it becomes j j - i

; Erumple 1.-Required the point of inflexion in a curve, whose equation is a r2 = a' y + xo y. whencer This equation in fluxions is 2 a x i = a* j +.

de

..; a general expression

jä- * ad + a2

for r in any curve. But as neither x or y may be 2 x y i + x? j, whence = zar-2xw; the considered as varying uniformly, or either 'r ory fluxion of which made =0, gives 2 x i (a r-ry) may be considered as 0, we may have r. = (a + xo): (a 1 - rj - i y), and this against

23

or=_ according as j or i is con* ; which being equa- - j

r

r y ja? — 33 a-y'

sidered constant. as + ira. Example.-Required the radius of curvature

? - or at the joints of an ellipse, at the point correof a tea 1 m i sponding to the absciss and ordinate x and y, the a - y - 23 a-y!

equation of the curve being a’ ya = ca ar — xud? and consequently y = ently a s1 = a.

By taking the first and second fluxions of the

given equation we have 2 a’ y j=co.3. – 28, Example 2.—Required the point of contrary

and 2 a' ja + 2 a' yy=-2 co io, considering flexure in a curve whose equation is a y = a? r + x ?

cia -- 2 r, and

i as constant; whence j =Here y = armd when : _ }a's + fx i

2 a' y
-; which, by substituting

a'y
the values of y and y, become y=

_cia - 2

20V asam 2 vanast 23

- aciz. T 2270 fluction of this put= 0, and reduced, gives r= and y= : a ( 3-1)

4 ao. a r — 23. a cinar 4 ✓ 12 TO FIND THE Radius OF CURVATURE OF

Zaradi hence 2 CURVES.

( 2 + y 2) = conta — 272 48. The radius of curvature is that of a circle having the same curvature as that of the curve at any proposed point; the general method of

a’ + a? — c. 4 ar – 422.

, and, finding the radius of this equicurve circle may

ar 72 be thus explained.

Let AD and De be the absciss and ordinate ( )=(a' c* + a' - c*• 4 a r - 4 x°) to the curve A E, EC

2 a*c the radius of the equi

which when a and c are equal becomes as it curve circle at 'E,

ought simply , the ellipse in that case degeneconsequently perpendicular to the

rating into a circle. curve at E. From

TO FIND THE INVOLUTES AND EVOLUTES OF C as a centre, with

CURVES. radius CE, describe the circular arc B Exe;

49. If a thread wrapped close round a curve draw C B parallel to

were fastened at one end, and unwound from the AD, and let E D pro

, other in the plane of the curve, the thread being duced, meet B C in

always kept stretched, the end of the thread in G; draw Ed, and ed parallel to ED and AD, to winding off will describe a curve which is called represent the fluxions of AD and DE. Put the involute, that from which the thread is unAD=r, ED = v. and AE = 2. Then by wound being the evolute. similar triangles GC : GE :: j : i, or GC.

Now it is obvious that the length of the thread i = GE: j. Whence by fluxing GC: x +

wound off will be the radius of curvature of the

involute at the instant, and also that it will at G:C: GEW +GE.j. But GC - that point be perpendicular to the involute; and BG, therefore, GC - BG · <=GE.y that the evolute will be the locus of the centres +GE. j. Now i is the fluxion of BG, as of the radii of curvature at every point of the well as of AD, and y is the fluxion G E as well curve.

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