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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta$ Show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
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Answer

Given,
$\text{x}=\text{a}\cos\theta\dots\text{ eq. }1$
$\text{y}=\text{b}=\sin\theta\dots\text{ eq. }2$
To prove: $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, let's first find $\frac{\text{dy}}{\text{dx}}$ using parametric form and defferentiate it again.
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}\cos\theta=-\text{a}\sin\theta\dots\text{ eq. 3}$
Similarly, $\frac{\text{dy}}{\text{d}\theta}=\text{b}\cos\theta\dots\text{ eq. 4}$
$\Big[\because\frac{\text{d}}{\text{dx}}\cos\text{x}=-\sin\text{x}\tan\text{x},\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}\Big]$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=-\frac{\text{b}\cos\theta}{\text{a}\sin\theta}=-\frac{\text{b}}{\text{a}}\cot\theta$
Differentiating again w.r.t. x:
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}\Big(-\frac{\text{b}}{\text{a}}\cot\theta\Big)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{b}}{\text{a}}\text{cosec}^2\theta\frac{\text{d}\theta}{\text{dx}}\dots\text{ eq. 5}$
$\Big[$Using chain rule and $\frac{\text{d}}{\text{dx}}\cot\text{x}=-\text{cosec}^2\text{x}\Big]$
From equation 3:
$\frac{\text{dx}}{\text{d}\theta}=-\text{a}\sin\theta$
$\therefore\frac{\text{d}\theta}{\text{dx}}=\frac{-1}{\text{a}\sin\theta}$
Putting the value in equation 5:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}}{a}\text{cosec}^2\theta\frac{1}{\text{a}\sin\theta}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\text{a}^2\sin^3\theta}$
From equation 1:
$\text{y}=\text{b}\sin\theta$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\frac{\text{a}^2\text{y}^3}{\text{b}^3}}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}\dots\text{proved.}$

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