Question

If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find $\frac{d^{2}y}{d{x}^2}$.

Answer 1

Here,

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