If the solve the problem


If $x=a \sec \theta, y=b \tan \theta$, prove that $\frac{d^{2} y}{d x^{2}}=-\frac{b^{4}}{a^{2} y^{3}} .$


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


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

$y=b \tan \theta \ldots \ldots$ 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 $\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}$

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

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

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a} \sec \theta=\mathrm{a} \sec \theta \tan \theta \ldots \ldots$ equation 3

Similarly, $\frac{d y}{d \theta}=b \sec ^{2} \theta$ equation 4

$\left[\because \frac{d}{d x} \sec x=\sec x \tan x, \frac{d}{d x} \tan x=\sec ^{2} x\right]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{\mathrm{b} \sec ^{2} \theta}{\mathrm{asec} \theta \tan \theta}=\frac{\mathrm{b}}{\mathrm{a}} \operatorname{cosec} \theta$

Differentiating again w.r.t $x$ :

$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{b}{a} \operatorname{cosec} \theta\right)$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{b}}{\mathrm{a}} \operatorname{cosec} \theta \cot \theta \frac{\mathrm{d} \theta}{\mathrm{dx}} \ldots . .$ equation 5 [ using chain rule]

From equation 3:

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a} \sec \theta \tan \theta$

$\therefore \frac{\mathrm{d} \theta}{\mathrm{dx}}=\frac{1}{\operatorname{asec} \theta \tan \theta}$

Putting the value in equation 5 :

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{b}}{\mathrm{a}} \operatorname{cosec} \theta \cot \theta \frac{1}{\mathrm{a} \sec \theta \tan \theta}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{-\mathrm{b}}{\mathrm{a}^{2} \tan ^{3} \theta}$

From equation 1:

$y=b \tan \theta$

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

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