Question
If $\text{a}(1-\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$ prove that, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{1}{\text{a}}$ at $\theta=\frac{\pi}{2}.$
✓
Answer
$\text{a}(1-\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$
Differentiating w.r.t.$\theta$,
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}(0+\sin\theta);\ ...\text{Eq}\ 1$
$\frac{\text{dy}}{\text{d}\theta}\ \text{a}(1+\cos\theta)\ ...\text{Eq}\ 2$
Dividing (2) by (1)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{d}\theta}\times\frac{\text{d}\theta}{\text{dx}}=\frac{\text{a}(1+\cos\theta)}{\text{a}\sin\theta}$
Differentiating w.r.t.$\theta$,
$\Rightarrow\frac{\text{d}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\text{d}\theta}=\frac{\sin\theta(0-\sin\theta)-(1+\cos\theta)\cos\theta}{\sin^2\theta}...(3)$
$=-\frac{\sin^2\theta-\cos\theta-\cos^2\theta}{\sin^2\theta}$
$=-\frac{(1+\cos\theta)}{\sin^2\theta}...(4 )$
dividing (4) by (3)
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{(1+\cos\theta)}{\sin^2\theta\times\text{a}\sin\theta}$
putting $\theta=\frac{\pi}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{1}{\text{a}}$
Hence proved
Differentiating w.r.t.$\theta$,
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}(0+\sin\theta);\ ...\text{Eq}\ 1$
$\frac{\text{dy}}{\text{d}\theta}\ \text{a}(1+\cos\theta)\ ...\text{Eq}\ 2$
Dividing (2) by (1)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{d}\theta}\times\frac{\text{d}\theta}{\text{dx}}=\frac{\text{a}(1+\cos\theta)}{\text{a}\sin\theta}$
Differentiating w.r.t.$\theta$,
$\Rightarrow\frac{\text{d}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\text{d}\theta}=\frac{\sin\theta(0-\sin\theta)-(1+\cos\theta)\cos\theta}{\sin^2\theta}...(3)$
$=-\frac{\sin^2\theta-\cos\theta-\cos^2\theta}{\sin^2\theta}$
$=-\frac{(1+\cos\theta)}{\sin^2\theta}...(4 )$
dividing (4) by (3)
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{(1+\cos\theta)}{\sin^2\theta\times\text{a}\sin\theta}$
putting $\theta=\frac{\pi}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{1}{\text{a}}$
Hence proved
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