# If the solve the problem

Question:

If $x=a \cos \theta, y=b \sin \theta$, show that $\frac{d^{2} y}{d x^{2}}=-\frac{b^{4}}{a^{2} y^{3}}$

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 $\mathrm{dx} / \mathrm{d} \theta=\mathrm{g}^{\prime}(\theta)$

We can write : $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$

Given,

$x=a \cos \theta \ldots \ldots$ equation 1

$y=b \sin \theta \ldots \ldots$ equation 2

to prove : $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{b}^{4}}{\mathrm{a}^{2} \mathrm{y}^{3}}$

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.

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a} \cos \theta=-\mathrm{a} \sin \theta \ldots . . \mathrm{equation} 3$

Similarly, $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{b} \cos \theta \ldots \ldots$ equation 4

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

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=-\frac{b \cos \theta}{a \sin \theta}=-\frac{b}{a} \cot \theta$

Differentiating again w.r.t $x$ :

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(-\frac{\mathrm{b}}{\mathrm{a}} \cot \theta\right)$

$\frac{d^{2} y}{d x^{2}}=\frac{b}{a} \operatorname{cosec}^{2} \theta \frac{d \theta}{d x} \ldots . .$ equation 5

[ using chain rule and $\frac{d}{d x} \cot x=-\operatorname{cosec}^{2} x$ ]

From equation 3:

$\frac{d x}{d \theta}=-a \sin \theta$

$\therefore \frac{d \theta}{d x}=\frac{-1}{\operatorname{asin} \theta}$

Putting the value in equation 5 :

$\frac{d^{2} y}{d x^{2}}=-\frac{b}{a} \operatorname{cosec}^{2} \theta \frac{1}{a \sin \theta}$

$\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2} \sin ^{3} \theta}$

From equation 1:

$y=b \sin \theta$

$\therefore \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{-\mathrm{b}}{\frac{\mathrm{a}^{2} \mathrm{y}^{2}}{\mathrm{~b}^{3}}}=-\frac{\mathrm{b}^{4}}{\mathrm{a}^{2} \mathrm{y}^{3}} \ldots \ldots$ proved.