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When the axis of the curve is taken as the axis of X, and therefore of revolution, the equation to the curve is

2 a x) dx

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S = 47Y (2ax)} – 47 (2a) sdx (2a x)).

Integrating from X = 0 to x = 2a, we have for the whole surface

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(10) The surface of the solid generated by the revolution of the tractory

dy y

0,

dx (ay') round the axis of x, and taken from x = 0 to x = co is equal to 27 a?.

+

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QUESTIONS of this kind were by the early writers on the Differential Calculus called Problems in the Inverse Method of Tangents, because, as the direct processes of the Differential Calculus were originally invented for the purpose of drawing tangents to curves, so the inverse Calculus had for its object the investigation of the equations of curves from the properties of their tangents and lines connected with them.

(1) Let it be required to find the curve in which the subtangent is a multiple of the abscissa. If

y = f(x)

da be the equation to the curve, y is the subtangent. Therefore we have the condition

dy

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whence log x = m log y + C = log Cy",

and Cy". When m is positive this gives a parabola of the mth order; when m is negative it gives a hyperbola of the same order.

(2) Find the curve in which the area contained between the axis of x, the ordinate and the curve, is a multiple of the rectangle contained by the ordinate and the abscissa. This stated analytically gives the equation

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The integral of this is

2n-m = Cy". When n = 3, m = 2, this gives the common parabola, as is otherwise obvious.

(3) Find the curve in which the perpendicular from the origin on the tangent is equal to the abscissa. The differential equation is

y

1 + dir

dy

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This is a homogeneous equation, and on being integrated it gives

y2 + x2

CX, the equation to a circle, the origin being in the circumference and the axis of x being a diameter.

(4) Find the curve in which the distance from the origin is equal to the part of the tangent intercepted between the point of contact and the perpendicular from the origin. The differential equation is

ydx wdy = ydy + xdx ; and the integral is

(r? + y) log

= tan

C which is the equation to a logarithmic spiral, the constant angle of which is equal to

()

T

4

(5) Find the nature of the curve BP (fig. 58) such that, if from the origin A a line AQ be drawn making an angle of 45° with the axis of X, and meeting the ordinate at any point P in Q, the ordinate PM shall bear to the sub-tangent MT the same ratio which the difference between PM and MQ bears to a constant line (a).

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or

(y - x) dx = ady, is the equation. This, when put under the form

ady - ydx + xdx = 0, is a linear equation of the first order, and its integral is

y = x + a + Co.

Since when y = 0, the curve must pass through the origin, we have C = -a, and therefore

y = c + a - a. This curve at one time attracted much attention, and it appears to have been the first problem involving a differential equation which was solved. It was proposed to Descartes * by De Beaune, after whom the curve is usually called “Curva Beauniana." The solution will be found in the works of John Bernoulli, Vol. 1. p. 63, and p. 65.

(6) Find the curve in which the product of perpendiculars from two fixed points on the tangent is constant.

Let A, B (fig. 59) be the fixed points; take the middle point between them as origin, and the line joining them as the axis of x. Then x and y being the co-ordinates of the point of contact P, the perpendiculars AY and BZ are, if CA = CB = C,

dy

dy y - (x – c)

y-(x + c) dx

and dy

dy 1 +

1+

dx Hence making their product constant and equal to b?, we have

dy (du Y

+

d x

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See his Letters, Tom. ui, No. 79.

dạy

= xy.

Differentiating this we get

dy
0, and {x? (b + c)
dx

dx
The first solution gives the general integral

y = ax + a'.

dy Substituting the value of derived from this in the

dc given equation, we have

y ax = + {b? + a2 (62 + c')}, the equation to two straight lines.

dy The second solution by the elimination of

gives the

d x singular solution, which is

y?
62

62 + c the equation to an ellipse, the minor axis of which is equal to b.

Euler, Mémoires de Berlin, 1756.

+

= 1

(7) Find the curve in which the normal bears a constant ratio to its intercept on the axis of x.

idyl +

dx)

The length of the normal is y {

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do'

and if n be the constant ratio, we have

dy

dy y {1+

x + y lx,

dy Squaring, and solving with respect to y

dx

{

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dx

we have

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