Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/181

Differentiating with respect to x, using XX
(A)	${\frac {d\tau }{dx}}$	$={\frac {\frac {d^{2}y}{dx^{2}}}{1+\left({\frac {dy}{dx}}\right)^{2}}}$. Also
(B)	${\frac {ds}{dx}}$	$=\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {1}{2}}$ From (24), p. 143 [§90]
Dividing (A) by (B) gives
	${\frac {\frac {d\tau }{dx}}{\frac {ds}{dx}}}$	$={\frac {\frac {d^{2}y}{dx^{2}}}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}}$
But	${\frac {\frac {d\tau }{dx}}{\frac {ds}{dx}}}$	$={\frac {d\tau }{ds}}=K$. Hence
(40)	K	$={\frac {\frac {d^{2}y}{dx^{2}}}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}}$.
If the equation of the curve be given in polar coördinates, K may be found as follows:
From (B), §67,
	$\tau$	$=\theta +\psi$. Differentiating,
(C)	${\frac {d\tau }{d\theta }}$	$=1+{\frac {d\psi }{d\theta }}$.
But	$\ \tan \psi$	$={\frac {\psi }{\frac {d\rho }{d\theta }}}$. From (A), §67
	∴ $\ \psi$	$=\arctan {\frac {\rho }{\frac {d\rho }{d\theta }}}$.
Differentiating by XX with respect to $\theta$ and reducing,
(D)	${\frac {d\psi }{d\theta }}$	$={\frac {\left({\frac {d\rho }{d\theta }}\right)^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}}{\rho ^{2}\left({\frac {d\rho }{d\theta }}\right)^{2}}}$