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If $x=k(\theta+\sin \theta)$ and $y=k(1+\cos \theta)$, then what is the derivative of $y$ with respect to $x$ at $\theta=\pi / 2 ?$
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The correct answer is:
$-1$
Let $x=k(\theta+\sin \theta)$ and $y=k(1+\cos \theta)$
Differentiate both the functions w.r.t. ' $\theta$ '
$\Rightarrow \quad \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{k}(1+\cos \theta)$
and $\frac{\mathrm{dy}}{\mathrm{d} \theta}=-\mathrm{k} \sin \theta$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d} \mathrm{y} / \mathrm{d} \theta}{\mathrm{d} \mathrm{x} / \mathrm{d} \theta}$
$=\frac{-\mathrm{k} \sin \theta}{\mathrm{k}(1+\cos \theta)}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}=-\tan \frac{\theta}{2}$
$\left(\because \sin 2 \theta=2 \sin \theta \cos \theta\right.$ and $\left.\cos 2 \theta=2 \cos ^{2} \theta-1\right)$
$\Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}=-\tan \frac{\pi}{4}=-1$
Differentiate both the functions w.r.t. ' $\theta$ '
$\Rightarrow \quad \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{k}(1+\cos \theta)$
and $\frac{\mathrm{dy}}{\mathrm{d} \theta}=-\mathrm{k} \sin \theta$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d} \mathrm{y} / \mathrm{d} \theta}{\mathrm{d} \mathrm{x} / \mathrm{d} \theta}$
$=\frac{-\mathrm{k} \sin \theta}{\mathrm{k}(1+\cos \theta)}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}=-\tan \frac{\theta}{2}$
$\left(\because \sin 2 \theta=2 \sin \theta \cos \theta\right.$ and $\left.\cos 2 \theta=2 \cos ^{2} \theta-1\right)$
$\Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}=-\tan \frac{\pi}{4}=-1$
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