Find the The Slopes of the tangent and the normal to the following curves at the indicated points :
$x=a(\theta-\sin \theta), y=a(1-\cos \theta)$ at $\theta=\pi / 2$
Here, To find $\frac{d y}{d x}$, we have to find $\frac{d y}{d \theta} \& \frac{d x}{d \theta}$ and and divide $\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$ and we get our desired $\frac{d y}{d x}$.
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}$
$\Rightarrow \mathrm{x}=\mathrm{a}(\theta-\sin \theta)$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}\left(\frac{\mathrm{dx}}{\mathrm{d} \theta}(\theta)-\frac{\mathrm{dx}}{\mathrm{d} \theta}(\sin \theta)\right)$
$\Rightarrow \frac{d x}{d \theta}=a(1-\cos \theta) \ldots(1)$
$\therefore \frac{d}{d x}(\sin x)=\cos x$
$\Rightarrow y=a(1-\cos \theta)$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}\left(\frac{\mathrm{dx}}{\mathrm{d} \theta}(1)-\frac{\mathrm{dx}}{\mathrm{d} \theta}(\cos \theta)\right)$
$\therefore \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})=-\sin \mathrm{x}$
$\therefore \frac{\mathrm{d}}{\mathrm{dx}}($ Constant $)=0$
$\Rightarrow \frac{d y}{d \theta}=a \sin \theta \ldots(2)$
$\Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a \sin \theta}{a(1-\cos \theta)}$
$\Rightarrow \frac{d y}{d x}=\frac{-\sin \theta}{(1-\cos \theta)}$
The Slope of the tangent is $\frac{-\sin \theta}{(1-\cos \theta)}$
Since, $\theta=\frac{\pi}{2}$
$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=\frac{\sin \frac{\pi}{2}}{\left(1-\cos \frac{\pi}{2}\right)}$
$\therefore \sin \left(\frac{\pi}{2}\right)=1$
$\therefore \cos \left(\frac{\pi}{2}\right)=0$
$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=\frac{(1)}{(1-(-0))}$
$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=\frac{1}{(1-0)}$
$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=1$
$\therefore$ The Slope of the tangent at $x=\frac{\pi}{2}$ is 1
$\Rightarrow$ The Slope of the normal $=\frac{-1}{\text { The Slope of the tangent }}$
$\Rightarrow$ The Slope of the normal $=\frac{-1}{\left(\frac{\mathrm{y} y}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}}$
$\Rightarrow$ The Slope of the normal $=\frac{-1}{1}$
$\Rightarrow$ The Slope of the normal $=-1$