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inclination to the į axis a; then « = p cos.0, y=p sin. O, and by

substitution in equation (a) we have

psin. 8 = (a pocos.0)

Hence p =

aola

b) cos.'0 Now it is well known that a – b = (dist. between focus and centrele = (eccentricity) = (ae), by supposition. a’ba

62

the equation required. a - ae* cos. O

e cos.28

51. To trace the curve whose equation is

ys = ax + x. Let I = 0, and - a. Then the corresponding values of y are each = 0. Hence, putting AB (Fig. 37), = - a, the curve will pass through A and B.

Again, let I= . Then y = $oo, or we have two infitite branches AQ, Bq lying on different sides of the axis Cc. To find the greatest ordinate of the branch AMB, we have

dy 2a + 3.2
di

3.13(a + x)
3.1

2a
and ::y= (ax8+2037

Hence,

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which gives I =

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3

3

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angles to Cc, we obtain the position and magnitude of the greatest

ordinate required.

To find the asymptotes, we have y=r.(1 + ax=1} = x(1+

a

x" 3

a' x + &c.)

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Hence, the equation to the rectilinear-asymptote, (see Francoeur, Mathemat. Pures, p. 323, or Stirling, Lineæ Tertii Ordinis Newtonianæ, p. 48,) is

y' = x' + , being measured from A.

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Therefore DR being joined and produced indefinitely will give the asymptotes DE, De to the respective branches AQ, Bq.

For the common method of finding the asymptotes to this curve, see Vince's Fluxions, p. 52.

2a+3.7 Since tan. 0 = dy.

O being the inclination of the di 3.x}(a + x) tangent to the axis. By substituting the singular values of x,viz.,0,– a,

2a and 24, the corresponding values of tan. 6, are oo, 00 and 0.

3

Consequently the tangents at A and B are I to the axis, and that at M is parallel to it.

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52. To trace the curve whose equation is

ao za y=+

a+ x? Let y = 0; then x = 0 or = fa. Take therefore AB = + a, Ab =

- a(Fig. 1), in the same straight line, and the curve passes through the points A, B, 6.

Again, B and b are the limits of x, since y is imaginary for every value of x > a. Again, putting x = 0 and + a, in

2a?r?
tan. 0 =

we get
dx
(a? + x2)! V a?

x2
tan. 0 = £ 1 and

tan. 45°

(and tan. 90°. htn, 135°

dy

fand

:: The point A is double, its two tangents being inclined to the axis at the angles 45° and 135° respectively, and the tangents at Bare each I axis.

The greatest ordinates PM, pm are given by putting A-2a*r-**=0, and thence obtaining x=+a 1 2 - 1 = AP or Ap; which being substituted in the given equation afford y=fa N 2 1 for the pairs of maximum ordinates at

Pand p, viz , PM, PM' and pm, pm' respectively.
Again, to find the area, we have

a u’,
a’ + x
substituting, we get

Sydx = sdu / 2a– . Again let
P= u. 2a u, and we finally get
P
2a”

du
Sydı =

+
2a + 1 2a’ + 1 2a

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a
+

x (area of o, rad. = a.)
2a + 1 2(2a + 1)
Hence the whole area AMBM'Am'bm
2a

4a

2 x(Orad.a)

x (a. O – ap). 2a + 1

2a + 1 2a + 1

53. To find the equation to the curve of pursuit. Let T, moving uniformly along the straight line TM with a velocity = v, be pursued by P moving also uniformly with a velocity -; and let PT, be that path of P which is I TM.

ao

n

Let also PT be any other contemporaneous positions of P,T.

Put PT = a, TN = x, PN = y, and PP' = s.

Then, since P'T is evidently a tangent at P and PN is decreasing, we have TT' = 2 + NT = x

ydr

dy
and by the question

TT
s= PP = since they move uniformly.

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2 a'p = - y*;

ya 1. 2a"dx = ay "dy - y*dy, and integrating, on the supposition that dy is negative, we obtain 2a"x =

1-n

nti aan

2artin yl-" + +

the general

1+n 1-n equation of the Curve of Pursuit.

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+ C.

O

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1-1

Let u =

y = 0.

За

and

3 Then TM =

a.

the distance described by T before

1-5 Povertakes it.

The problem admits of being further generalized. T might be made to move in a curve, and with a velocity varying according to a given law, P following it likewise with a variable velocity. The determination of the problem under these circumstances we recommend, as an useful exercise.

54. To transform the equation to the Lemniscata,

(x + y) = a'. (x - y). from rectangular to polar co-ordinates, Make A, (Fig. 1) the origin of the rectangular, the pole of the polar co-ordinates, and put AM =p, and Z MAB = 0. Then, since y = AM. sin. 0 = p. sin. O, and x = AP = pcos. O, by substituting in the given equation, we get

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