# If the solve the problem

Question:

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

Solution:

Idea of parametric form of differentiation:

If $y=f(\theta)$ and $x=g(\theta)$ i.e. $y$ is a function of $\theta$ and $x$ is also some other function of $\theta$.

Then $\mathrm{dy} / \mathrm{d} \theta=\mathrm{f}^{\prime}(\theta)$ and $\mathrm{d} \mathrm{x} / \mathrm{d} \theta=\mathrm{g}^{\prime}(\theta)$

We can Write : $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}$

Given,

$y=a(\theta+\sin \theta) \ldots \ldots$ equation 1

$x=a(1+\cos \theta) \ldots \ldots$ equation 2

to prove : $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{1}{\mathrm{a}}$ at $\theta=\frac{\pi}{2}$.

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{d^{2} y}{d x^{2}}$

As, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

So, lets first find $d y / d x$ using parametric form and differentiate it again.

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a}(\theta+\sin \theta)=\mathrm{a}(1+\cos \theta) \ldots \ldots$ equation 3

Similarly,

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a}(1+\cos \theta)=-\operatorname{asin} \theta \ldots \ldots$ equation 4

$\left[\because \frac{d}{d x} \cos x=-\sin x, \frac{d}{d x} \sin x=\cos x\right]$

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a(1+\cos \theta)}{-\sin \theta}=-\frac{(1+\cos \theta)}{\sin \theta} \ldots . .$ equation 5

Differentiating again w.r.t $x$ :

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(-\frac{(1+\cos \theta)}{\sin \theta}\right)=-\frac{\mathrm{d}}{\mathrm{dx}}(1+\cos \theta) \operatorname{cosec} \theta$

Using product rule and chain rule of differentiation together:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\left\{\operatorname{cosec} \theta \frac{\mathrm{d}}{\mathrm{d} \theta}(1+\cos \theta)+(1+\cos \theta) \frac{\mathrm{d}}{\mathrm{d} \theta} \operatorname{cosec} \theta\right\} \frac{\mathrm{d} \theta}{\mathrm{dx}}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\{\operatorname{cosec} \theta(-\sin \theta)+(1+\cos \theta)(-\operatorname{cosec} \theta \cot \theta)\} \frac{1}{(-\mathrm{asin} \theta)}$

[using equation 4]

$\frac{d^{2} y}{d x^{2}}=\left\{-1-\operatorname{cosec} \theta \cot \theta-\cot ^{2} \theta\right\} \frac{1}{\operatorname{asin} \theta}$

As we have to find $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{1}{\mathrm{a}}$ at $\theta=\frac{\pi}{2}$

$\therefore$ put $\theta=\pi / 2$ in above equation:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left\{-1-\operatorname{cosec} \frac{\pi}{2} \cot \frac{\pi}{2}-\cot ^{2} \frac{\pi}{2}\right\} \frac{1}{\operatorname{asin} \frac{\pi}{2}}=\frac{\{-1-0-0\} 1}{a}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{1}{\mathrm{a}}$