Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/182

Substituting (D) in (C), we get
(E)	${\frac {d\tau }{d\theta }}$	$={\frac {\rho ^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}+2\left({\frac {d\rho }{d\theta }}\right)^{2}}{\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}}}$. Also
(F)	${\frac {ds}{d\theta }}$	$=\left[\rho ^{2}\left({\frac {d\rho }{d\theta }}\right)^{2}\right]^{frac{1}{2}}$. From (30), §91
Dividing (E) by (F) gives
	${\frac {\frac {d\tau }{d\theta }}{\frac {ds}{d\theta }}}$	$={\frac {\rho ^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}+2\left({\frac {d\rho }{d\theta }}\right)^{2}}{\left[\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}\right]^{\frac {3}{2}}}}$.
But	${\frac {\frac {d\tau }{d\theta }}{\frac {ds}{d\theta }}}$	$={\frac {d\tau }{ds}}=K$. Hence
(41)	$K$	$={\frac {\rho ^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}+2\left({\frac {d\rho }{d\theta }}\right)^{2}}{\left[\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}\right]^{\frac {3}{2}}}}$.

Illustrative Example 1. Find the curvature of the parabola $y^{2}=4px$ at the upper end of the latus rectum.

Solution.	${\frac {dy}{dx}}={\frac {2p}{y}};{\frac {d^{2}y}{dx^{2}}}$	$=-{\frac {2p}{y^{2}}}{\frac {dy}{dx}}=-{\frac {4p^{2}}{y^{3}}}$.
Substituting in (40),	$K$	$=-{\frac {4p^{2}}{(y^{2}+4p^{2})^{\frac {3}{2}}}}$,
giving the curvature at any point. At the upper end of the latus rectum (p, 2p)
	$K=-{\frac {4p^{2}}{(4p^{2}+4p^{2})^{\frac {3}{2}}}}$	$=-{\frac {4p^{2}}{16{\sqrt {2}}p^{3}}}=-{\frac {1}{4{\sqrt {2}}p}}$.[1] Ans.

Illustrative Example 2. Find the curvature of the logarithmic spiral $\rho =e^{a\theta }$ at any point.

Solution. ${\frac {d\rho }{d\theta }}=ae^{a\theta }=a\rho ;{\frac {d^{2}\rho }{d\theta ^{2}}}=a^{2}e^{a\theta }=a^{2}\rho$.

Substituting in (41), $K={\tfrac {1}{\rho {\sqrt {1+a^{2}}}}}$ Ans.

1. While in our work it is generally only the numerical value of K that is of importance, yet we can give a geometric meaning to its sign. Throughout our work we have taken the positive sign of the radical ${\sqrt {1+\left({\tfrac {dy}{dx}}\right)^{2}}}$. Therefore K will be positive or negative at the same time as ${\tfrac {d^{2}y}{dx^{2}}}$ that is (§85), according as the curve is concave upwards or concave dx downwards.