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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{x}=\text{a}(\theta+\sin\theta)\ \text{and}\ \text{y}=\text{a}(1+\cos\theta)$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{a}}{\text{y}^2}.$
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Answer

Here
$\text{x}=\text{a}(\theta+\sin\theta)\ \text{and}\ \text{y}=\text{a}(1+\cos\theta)$
Differentiating w.r.t.$\theta$, we get
$\frac{\text{dx}}{\text{d}\theta}=\text{a}+\text{a}\cos\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=-\text{a}\sin\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{-\text{a}\sin\theta}{\text{a}+\text{a}\cos\theta}=\frac{-\sin\theta}{1+\cos\theta}$
Differentiating w.r.t.$\theta$, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big\{\frac{(1+\cos\theta)\cos\theta+\sin^2\theta}{(1+\cos\theta)^2}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$=\frac{-\cos\theta-\cos^2\theta-\sin^2\theta}{(1+\cos\theta)^2}\times\frac{1}{\text{a}+\text{a}\cos\theta}$
$=\frac{-(1+\cos\theta)}{\text{a}(1+\cos\theta)^3}$
$=\frac{-1}{\text{a}(1+\cos\theta)^2}$
$=\frac{-\text{a}}{\text{y}^2}\ [\because\text{y}=\text{a}(1+\cos\theta)]$
Hence proved

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