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,
$$\begin{gathered} x=a\left(1-\cos \theta \right)\mathrm{and}y=a\left(\theta +\sin \theta \right) \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}x}{\mathit{d}\theta }=a\sin \theta \mathrm{and}\frac{\mathit{d}y}{\mathit{d}\theta }=a+a\cos \theta \\ \therefore \frac{\mathit{d}y}{\mathit{d}x}=\frac{a+a\cos \theta }{a\sin \theta }=\frac{1+\cos \theta }{\sin \theta } \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\left\{\frac{-{\sin }^2\theta -\left(1+\cos \theta \right)\cos \theta }{{\sin }^2\mathrm{\theta }}\right\}\frac{\mathit{d}\theta }{\mathit{d}x} \\ =\left(-1-\frac{\left(1+\mathrm{cos\theta }\right)\mathrm{cos\theta }}{{\sin }^2\theta }\right)\frac{1}{a\sin \theta } \\ \therefore \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}\mathrm{at}\theta =\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\left[\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}\right]}_{\theta =\frac{\mathrm{\pi }}{2}}=\frac{1}{a}\left(-1-\frac{0}{1}\right)=\frac{-1}{a} \end{gathered}$$

Hence proved.