Higher Order Derivatives — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{x}=\text{a}(1+\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$ prove that
✓
Answer
Here,
$\text{x}=\text{a}(1+\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=-\text{a}\sin\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=\text{a}+\text{a}\cos\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{a}+\text{a}\cos\theta}{-\text{a}\sin\theta}=\frac{1+\cos\theta}{-\sin\theta}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{d}\theta}\Big\{\frac{\text{dy}}{\text{dx}}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big\{\frac{-\sin\theta\cos\theta-\cos^2\theta}{\sin^2\theta}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$=\frac{1+\cos\theta}{\sin^2\theta}\times\frac{-1}{\sin^2\theta}$
$\frac{-(1+\cos\theta)}{\sin^3\theta}$
At $\theta=\frac{\pi}{2}:\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-(1+\cos\frac{\pi}{2})}{\text{a}(\sin\frac{\pi}{2})^3}=\frac{-1}{\text{a}}$
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