44 DIFFERENTIAL AND INTEGRAL CALCULUS. 



of the string will be a maximum or minimum when the 

 tangents at P, P' are equally inclined to that at 0; and 

 that it will be a maximum or minimum according as the 

 curvature at is greater or less than the arithmetic mean 

 between the curvatures at P, P'. 



4. A family of curves r = cf(0) is described, the pole 

 being a point 0, and any one of these curves meets the 

 prime radius in A^ and a fixed circle with its centre at and 

 radius a in P, prove that when the area A OP is a maxi- 

 mum or minimum, it is equal to half the triangle POT. 



I give solutions of (2), (3), (4), which I think may be 

 instructive to a student. 



(2) any fixed point on the curve, 6>P=c, 

 6 the angles which the tangents at P, Q make with that 

 at (fig. 8), then if S be the whole arc of the involute 



-r 



J a 



therefore 

 ^ = - ^ (27r) + (FQ)* = perimeter of oval - 27r ~ , 



6tCC CZ5t UOL 



dc 

 or $ is a maximum or minimum if 2?r x y- = perimeter of 



oval (and -=- radius of curvature at P j , a maximum if -^ 



increases from P towards Q, i.e. if the curvature decreases. 



If unwound in the opposite direction from P, the arc of the 



-m+a 

 involute =1 (Ps + c}d6, where Pis the perimeter, 



J a 



and the sum of the two = 2?r P, or is the same for every 

 point, so that if one be a maximum the other must be a 

 minimum. 



(3) The same notation, arc QF= arcP#' = fl, /3, /3' the 

 angles between the tangents at Q, Q' (fig. 9), and that at 

 P, then the whole arc generated 



r x 



J a- ft 



