Question
If $\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
✓
Answer
Here,
$\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sec\theta\tan\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=b\sec^2\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{d}\theta}\times\frac{\text{d}\theta}{\text{dx}}=\frac{\text{b}\sec^2\theta}{\text{a}\sec\theta\tan\theta}=\frac{\text{b}\ \text{cosec}\theta}{\text{a}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{b}}{\text{a}}\times-\text{cosec}\theta\cot\theta\times\frac{\text{d}\theta}{\text{dx}} $
$=-\frac{\text{b}}{\text{a}}\times\text{cosec}\theta\cot\theta\times\frac{1}{\text{a}\sec\theta\tan\theta}$
$=\frac{-\text{b}}{\text{a}^2}\times\frac{1}{\tan^3\theta}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\frac{\text{a}^2\text{y}^3}{\text{b}^3}}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
Hence proved
$\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sec\theta\tan\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=b\sec^2\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{d}\theta}\times\frac{\text{d}\theta}{\text{dx}}=\frac{\text{b}\sec^2\theta}{\text{a}\sec\theta\tan\theta}=\frac{\text{b}\ \text{cosec}\theta}{\text{a}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{b}}{\text{a}}\times-\text{cosec}\theta\cot\theta\times\frac{\text{d}\theta}{\text{dx}} $
$=-\frac{\text{b}}{\text{a}}\times\text{cosec}\theta\cot\theta\times\frac{1}{\text{a}\sec\theta\tan\theta}$
$=\frac{-\text{b}}{\text{a}^2}\times\frac{1}{\tan^3\theta}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\frac{\text{a}^2\text{y}^3}{\text{b}^3}}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
Hence proved
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