Question

If x = a (cos θ + θ sin θ), y = a(sin θ − θ cos θ), prove that $\frac{d^{2}y}{d{x}^2}=\frac{{\sec }^3\mathrm{\theta }}{a\mathrm{\theta }}$.

Answer 1

Here,
$$\begin{gathered} x=a\left(\cos \theta +\theta \sin \theta \right)\mathrm{and}y=a\left(\sin \theta -\theta \cos \theta \right) \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\theta ,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}x}{\mathit{d}\theta }=a\left(-\sin \theta +\theta \cos \theta +\sin \theta \right)=a\theta \cos \theta \mathrm{and}\frac{\mathit{d}y}{\mathit{d}\theta }=a\left(\cos \theta +\theta \sin \theta -\cos \theta \right)=a\theta \sin \theta \\ \therefore \frac{\mathit{d}y}{\mathit{d}x}=\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{\theta }}\times \frac{\mathit{d}\mathit{\theta }}{\mathit{d}\mathit{x}} \\ =\frac{a\theta \sin \theta }{a\theta \cos \theta }=\tan \theta \\ \therefore \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}={\sec }^2\theta \times \frac{\mathit{d}\theta }{\mathit{d}x} \\ ={\sec }^2\theta \times \frac{1}{a\theta \cos \theta }\left[\because \frac{\mathit{d}x}{\mathit{d}\theta }=a\theta \cos \theta \right] \\ =\frac{{\sec }^3\theta }{a\theta } \end{gathered}$$

Hence proved.