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n. log. a 4. iy" log. y++y 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 Aluxions, the fluxion of any flu- difference added to, or subtracted from, the variaeat may be found, and by a reverse operation the able part, as may be required, will give the fluent trent may in most cases be found from the truly corrected. fusion; yet the fluent so found may

often require to be increased or diminished by some constant Erample 1.-Let j=a? x i, then y =

:

2 quantity depending on the nature of the problem

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

are by the conditions of the problem simul"na

a2 * i, and the fluxion of r" + a is the saine quantity; we cannot therefore affirm taneously = 0, the quantity

2 without reference io the nature of the problem in fluent. which the fluxion nr

Example 2.-5 and y commencing together, i arises, whether let the true fluent of j = 2 FIO; be required'

. its fluent is Ła.

atr 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 y = 0,

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

n

of : is

is the true

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Put AC = a, CD=b, AP = P, PR = 9, have been o, hence y= -- the cor- and PQ=1, then by conics QB = (p9+ rected fluent. The true fluent of this quantity, however, and

93-pr - 24) and P B = 109 +91many others may be found without correction; in the present case (j=a + i) if at 's Pr – x) + x?. Hence, by putting the fluxion be expanded, we have j = ai + 3 a' x = + = 0, and reducing, we get x= 3 a rċ + ryć, whose Auent is y=a3x +.

m 2 7?_'. 3 co ro .

Erample 4.-To determine the dimensions of 2T" T4- 4

as before. a cylindrical vessel, open at the top, that shah In the preceding examples x and y are sup- Superficies

contain a given quantity under the least internal posed to be equal to nothing at the same time;

Let the diameter of its base be t, and its altibut in the solution of problems this will often

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 cir

cire tangent 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 pry equal to the radius. We shall therefore add an example or two, in which when y=0, x is equal to a given quantity a.

der. Hence P4, 9=c, the given quantity that Let y = xi be the proposed Auxion, then its Auent is y = . Here when y = 0, = the vessel will hold, or pry=* ; therefore 2; hence the corrected Auent is y=

++ 4

is a maximum and consequently –

T

ti 4ci pri Again, let j = - * , then y=-*. ; 22 +2

* + P* * = 0, 0r == 2: V
=

; and as
"Tip X = 8 c, and p xe y = 4c, 1 = 2y; whence
enter"+'
I

is also known, and it appears hence too that which, corrected, becomes v ='___

- the diameter of the base must be just double the

altitude. APPLICATION OF FLUXIONS TO THE DETERMI- Erample 5.-If two bodies move at the same

NATION OF THE MAXIMA AND MINIMA of 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 A P 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 Example 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 Let r = one of the segments, then a-ris cistance CN=r. Then we have b: .::c: the other, whence a x - x' is by the hypothesis to be a maximum, and consequently its Auxion, or a : - 2 x <= 0, when r=;, or the given term must be bisected. Example 2.-What fraction is that whose med

- C_ P

IMÈD power exceeds its mth power by the greatest possible quantity ?

CE=*; and n:m:: BN (5—b):AM= Let x = the required fraction, the r" - *** is to be a maximum, or nx"--mI" = 0,

_mann whence r = Erample 3.-From a given point P, in the

Hence MN= CM2 + CN – 2 CM.CE= transverse axis of an ellipse, to draw PB, the

2cdi 2cmr shortest line to the curve :

Imotz

b PB the fluxion of which put= 0, gives – 20 m: +

n +

1.

m

=0, ****0.; and hence CM = a + ",0

mo=d_m1 (by substituting a fora + " ")

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and from this all the other quantities may be Example 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 TIE 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, X; 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

ga t g? poza y = 2 r X — = I thg and by taking the Auxions of these equations, and making proper substitutions in the general fuxional equation, i = + j?, we obtain

i=1273 = n ya = no + p = Í A béc

2 s
sv se A.

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

general value of % will be obtained. We shall If AE be a curve, let it be required to draw a tangent TE at any point E. Draw the ordinate take as an example ; =DE, 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,

very slow convergency, so that a great many and draw E a parallel to the axis Å D. Then the very

terms of it must be collected before a result of triangles Eaė, and TED are similar, and te

sufficient practical accuracy can be obtained. therefore e a: a E:: ED: DT. Or j::::y:

But it may easily be transformed into series of V =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 DE.

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

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

| 10 (1 +3:10 + 5.107 7:10 donding absciss.

Example 2.-Draw a tangent to the cissoid of 1st A -Diocles, whose equation is y* = *

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Here 2 y j = _B e , and consequently y = 2you—2_ 270 m

og j 3 2 x - 30x2–2 ju - 21:-3 -34 -2 r. To DETERMINE THE LENGTHS OF CURVES

1 50 1+3•100+ 5.100 +7.100 WHOSE EQ:JATIONS ARE GIVEN.

| 16 y 43. In the annexed figure E a, e a, and Ee, are

(+ g. 100. simultaneous in

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

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

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

...

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

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

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

for curves of that kind being a " s=y". cle 42) similar to 1 TED; it may therefore be considered as a right-angled triangle. Hence, j' = 3 + jo, or = V i + j?.

, na yen–21. Therefore substituting for j its value in terms of fore-V jo +

tas; 3, and taking the fuent, the value of . is obtained.

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

Here i =ny_3,i=vi+j”, is there

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X

2n - 2

a, and there

– 2 ¢ ¥3, c4 + 4 ya $, the fluent of which cor

ren

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4n-3

VIP + jö = i=

4n-4 2n-1 2a"" 4 n - 3:8 a

whose Auent is a cr, the value of the spherical 6n - 5

surface. But a c is the circumference of the

generating circle; hence the surface of any segon 6. &c.

ment is equal to the circumference of a great 6n— 5 · 16a

circle, multiplied by the versed sine; or height of 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 expression is c a', or four 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 OF CURVES, whose EQUA- be that of a parabolic conoid. TIONS ARE GIVEN.

2 v v

Here a x = y, whence i =>
44. Adopting the previous notation, it is
obvious that y; is the Auxion of the area; and

fore
fore i = V i? + y =

- 72 y being from the equation known in terms of x,

_j va? + 4 va

, and 2 cyi the fluent of this expression is the area of the curve. Erample 1.–Required the area of a parabola, m +n

icle+ 4 7j – ). whose equation is a ""=y" "" mtn m

mtn

To find The Solid Contents of BODIES. Here y=a n r , whence yö sa n.

46. As a curve surface may be conceived to be mt n n - m

generated by the expanding circumference of a z ", whose fluent is no n o n . plane moving forward, as the solid itself which

1 - m

the surface bounds may be conceived to be geneE.rumple 2.-Let it be required to find the rated by the plane itself. Hence, if I, y, and c. area of a circle from the equation y = urx.

represent the same things as they did in art. 44,

we have c y? i for the fluxion of the solid, and the

. ,:. rls fuent of this quantity will be the required solid. Here y ö =é Nar - 2 = aox?i:1

a

Example 1.-Required the solid content of a cone, whose altitude is a, and base b.

Here a : 6::x:y= -, whence cy'i= va)

c l?2? į 2 a 8

30 ; which when whose fluent gives r ~

az r = e, becomes , or one-third of a cylin

or the area of the der, having the same base and altitude. - 72 av — 704 am

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

spheroid, the tranverse and conjugate of whose TO FIND THE SURFACES OF Sorids.

generating ellipse are a and b.

By the nature of the curve (see Conics) y: = 45. A surface may be conceived to be gene- b? rated by the circumference of a plane moving a ? au

a 6 - 42, whence c yü i= ari-di, forward, and expanding at the same instant; therefore the fuxion of the surface is equal to the whose fluen

-- ); which when fluxion of the curve, in which the expanding circumference moves forward at any instant, multi- r = aisa, the content of the whole spheplied by the periphery of the variable circum- wnia ference 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 2cv = the circum- cylinder whose base diameter is b, and altitude a, ference, and 2 cyi= 2cy j? + js = the is ab, of which ob is evidently two-thirds. fluxion of the surface S, and consequeutly by taking the fluent, S is obtained.

TO FIND THE Points OF CONTRARY FLEXURE OF Example 1.–Let the proposed curve surface CurveS FROM THEIR EQUATIONS. be a sphere. In this case y = n a r — x?

47. It is evident when a curve is concave whence j = 2 Tahta

= -4-2 . . and towards its area, that the fluxion of the ordinate • 2 y ? decreases with respect to the fluxion of the ab

sciss; and the contrary when the curve is conconsequently j? =

vex towards its axis; hence, at the point of

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;-aj? + c;

contrary Auxion; or is a constant quantity, and this form GC

as of D E; hence the equation may be put in

-i G Ev_jj, or GC. consequently its Auxion is = 0. Therefore, if ;-GE çiz + ja ;; and if each of from the given equation the value of or 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 Y

Erumple 1.-Required the point of inflexion in a curve, whose equation is a r = ay + zoo y. whencer

; a general expression This equation in Auxions is 2 a r i = a* j +

jr-cy

? + ger? for r in any curve. But as neither r or y may be 2 ryi + xj, whence j = 2a x 2 r y

-; the

considered as varying uniformly, or either x or y

consi fluxion of which made = 0, gives 2 x i (a x _ry) may be considered as 0, we may have a =(a? + r) · (i - rj - iy), and this again 22 - 23

... or Dát

according as j or é is congives = a*y; which being equa- sidered

sidered constant.

a? + ted with the former value of

Erample.—Required the radius of curvature

22 - at the joints of an ellipse, at the point correir a? + ca 1 - ; whence r=an

sponding to the absciss and ordinater and y, the a- y 2 x - y

equation of the curve being a? y? = cm a r — x??

By taking the first and second Auxions of the

given equation we have 2 ao yý=ci•a— 27, Erample 2.-Required the point of contrary and 2 a? j? + 2 ay -2 ci?, considering flexure in a curve whose equation is a y' =

cia - 2 r, and a’ r + x)?

i as constant; whence j =

2 a' y Here y = u* r rus whence i – } a' i + fx i

Ja.

2, ; which, by substituting the values of y and y, become j=

Lcia-2.

ya 2a Vara do 2 va az +

Lazci?. a - 2272 fluction of this put = 0, and reduced, gives r= and —y= a( 3-1)

e c.ar-me.ac.vara 4 v 12

caz? TO FIND THE RADIUS OF CURVATURE OF

avar- da CURVES.

1 + 2 + j 2) = * p. - 2x 48. The radius of curvature is that of a circle

4 a? . a r - grad

—tit = having the same curvature as that of the curve at any proposed point; the general method of

az + ai – cz • 4 ar — 4 x3 and , finding the radius of this equicurve circle may

a X be thus explained.

(a? c+ a’ – ca. 40 x – 4 xo); Let A D and D E be the absciss and ordinate to the curve A E, EC

2a* 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 Eve;

49. If a thread wrapped close round a curve draw C B parallel to 1

were fastened at one end, and unwound from the AD, and let E D pro 3

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 Auxions 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 : : ; : i, or GC: Now it is obvious that the length of the thread i = GE j. Whence by Auxing GC*+.

c. wound off will be the radius of curvature of the

involute at the instant, and also that it will at G:C: GE:v+GE · ý. But GC - that point be perpendicular to the involute; and BG, therefore, GC · Ö- BG ċ=GE.y that the evolute will be the locus of the centres

+GE · j. Now į is the fluxion of BG, as of the radii of curvatur: at every point of the well as of AD, and y is the fluxion G E as well curve.

V a a*x + x) – }, and _ 4 + 3.x3.

C 22

i hence 2

4ar

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