Question

If x = a (1 + cos θ), y = a(θ + sin θ), prove that $\frac{d^{2}y}{d{x}^2}=\frac{-1}{a}\mathrm{at}\mathrm{\theta }=\frac{\mathrm{\pi }}{2}$.

Answer 1

Here,

$$\begin{gathered} x=a\left(1+\cos \theta \right)\mathrm{and}y=a\left(\theta +\sin \theta \right) \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\theta ,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}x}{\mathit{d}\theta }=-a\sin \theta \mathrm{and}\frac{\mathit{d}y}{\mathit{d}\theta }=a+a\cos \theta \\ \therefore \frac{\mathit{d}y}{\mathit{d}x}=\frac{a+a\cos \theta }{-a\sin \theta }=\frac{1+\cos \theta }{-\sin \theta } \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{d}{d\theta }\left\{\frac{\mathit{d}y}{\mathit{d}x}\right\}\frac{d\theta }{dx} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=-\left\{\frac{-{\sin }^2\theta -\cos \theta -{\cos }^2\theta }{{\sin }^2\theta }\right\}\frac{d\theta }{dx} \\ =\frac{1+\cos \theta }{{\sin }^2\theta }\times \frac{-1}{a\sin \theta } \\ =\frac{-\left(1+\cos \theta \right)}{a{\sin }^3\theta } \\ \mathrm{At}\theta =\frac{\mathrm{\pi }}{2}:\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{-\left(1+\cos {\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{a{\left(\sin {\displaystyle \frac{\mathrm{\pi }}{2}}\right)}^3}=\frac{-1}{a} \end{gathered}$$