When the axis of the curve is taken as the axis of ar, and therefore of revolution, the equation to the curve is Integrating from a = 0 to w = 2a, we have for the whole surface o (10) The surface of the solid generated by the revolution of the tractory round the axis of a, and taken from a = 0 to a = co is equal to 27ta”. CHAPTER X. GEOMETRICAL PROBLEMS INVOLVING THE SOLUTION OF 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(a) da, be the equation to the curve, y #. is the subtangent. Therey When m is positive this gives a parabola of the m” 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 a, the ordinate and the curve, is a multiple of the rectangle contained by the ordinate and the abscissa. This stated analytically gives the equation The integral of this is a"-" = 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 This is a homogeneous equation, and on being integrated it gives y” + a” = ca, the equation to a circle, the origin being in the circumference and the axis of a 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 vda – a dy = yd y + wala: ; and the integral is which is the equation to a logarithmic spiral, the constant angle of which is equal to 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 w, 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). ot (y – a da = a dy, is the equation. This, when put under the form a dy – y da' + a da = 0, is a linear equation of the first order, and its integral is Since when y = 0, the curve must pass through the origin, we have C = — a, and therefore r y = a + a - a e". 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 C the middle point between them as origin, and the line joining them as the axis of w. Then a and y being the co-ordinates of the point of contact P, the perpendiculars AY and BZ are, if CA = CB = c, Hence making their product constant and equal to bo, we have * See his Letters, Tom. 111. No. 79. y = a a + a'. - - - - dy - • Substituting the value of do derived from this in the o d The second solution by the elimination of # gives the o the equation to an ellipse, the minor axis of which is equal to b. Euler, Mémoires de Berlin, 1756. (7) Find the curve in which the normal bears a constant ratio to its intercept on the axis of a. d Squaring, and solving with respect to yo, we have |