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



Find the curvature of the parabola $$y^2 = 4 px$$ at the upper end of the latus rectum.




 * Solution.
 * style="text-align: right;"|$$\frac{dy}{dx} = \frac{2p}{y}; \frac{d^2 y}{dx^2}$$
 * $$= - \frac{2p}{y^2} \frac{dy}{dx} = - \frac{4p^2}{y^3}$$.
 * Substituting in (40),
 * style="text-align: right;"|$$K$$
 * $$= -\frac{4p^2}{( y^2 + 4p^2 )^{\frac{3}{2}}}$$,
 * colspan="3"|giving the curvature at any point. At the upper end of the latus rectum (p, 2p)
 * style="text-align: right;"|$$K = -\frac{4p^2}{ (4p^2 + 4p^2)^{\frac{3}{2}} }$$
 * $$= - \frac{4p^2}{16 \sqrt{2} p^3} = - \frac{1}{4 \sqrt{2} p}$$. Ans.
 * }
 * style="text-align: right;"|$$K = -\frac{4p^2}{ (4p^2 + 4p^2)^{\frac{3}{2}} }$$
 * $$= - \frac{4p^2}{16 \sqrt{2} p^3} = - \frac{1}{4 \sqrt{2} p}$$. Ans.
 * }
 * }

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


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


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