Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when
$x=a(1-\cos \theta)$ and $y=a(\theta+\sin \theta)$ at $\theta=\frac{\pi}{2}$
as $x=a(1-\cos \theta)$
$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}[\mathrm{a}(1-\cos \theta)]}{\mathrm{d} \theta}=\mathrm{a}(\sin \theta)$
And $y=a(\theta+\sin \theta)$
$\frac{d y}{d \theta}=\frac{d(\theta+\sin \theta)}{d \theta}=a(1+\cos \theta)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{\mathrm{a}(1+\cos \theta)}{\mathrm{a}(\sin \theta)} \mid\left(\theta=\frac{\pi}{2}\right)$
$=\frac{a(1+0)}{a}=1$
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