Question

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

Answer 1

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

Hence proved.

(ii) Here,

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