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Question:

If $x=a\left(1-\cos ^{3} \theta\right), y=a \sin ^{3} \theta$, Prove that $\frac{d^{2} y}{d x^{2}}=\frac{32}{27 a} a t \theta=\frac{\pi}{6}$

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

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

Given,

$x=a\left(1-\cos ^{3} \theta\right) \ldots \ldots$ equation 1

$y=a \sin ^{3} \theta, \ldots \ldots .$ equation 2

to prove : $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{32}{27 \mathrm{a}}$ at $\theta=\frac{\pi}{6}$.

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 $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$

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

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

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a}\left(1-\cos ^{3} \theta\right)=3 \operatorname{acos}^{2} \theta \sin \theta \ldots \ldots$ equation 3 [using chain rule]

Similarly,

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a} \sin ^{3} \theta=3 \mathrm{a} \sin ^{2} \theta \cos \theta \ldots \ldots$ equation 4

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{3 \operatorname{asin}^{2} \theta \cos \theta}{3 \operatorname{acos}^{2} \theta \sin \theta}=\tan \theta$

Differentiating again w.r.t $\mathrm{x}$ :

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(\tan \theta)$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\sec ^{2} \theta \frac{\mathrm{d} \theta}{\mathrm{dx}} \ldots . .$ equation 5

[ using chain rule and $\left.\frac{d}{d x} \tan x=\sec ^{2} x\right]$

From equation 3:

$\frac{d x}{d \theta}=3 \operatorname{acos}^{2} \theta \sin \theta$

$\therefore \frac{\mathrm{d} \theta}{\mathrm{dx}}=\frac{1}{3 \operatorname{acos}^{2} \theta \sin \theta}$

Putting the value in equation 5 :

$\frac{d^{2} y}{d x^{2}}=\sec ^{2} \theta \frac{1}{3 \operatorname{acos}^{2} \theta \sin \theta}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1}{3 \mathrm{a} \cos ^{4} \theta \sin \theta}$

Put $\theta=\pi / 6$

$\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)$ at $\left(\mathrm{x}=\frac{\pi}{6}\right)=\frac{1}{3 \mathrm{a} \cos ^{4} \frac{\pi}{6} \sin \frac{\pi}{6}}=\frac{1}{3 \mathrm{a}\left(\frac{\sqrt{3}}{2}\right)^{4} \frac{1}{2}}$

$\therefore\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)$ at $\left(\mathrm{x}=\frac{\pi}{6}\right)=\frac{32}{27 \mathrm{a}} \ldots$ proved