Find the The Slopes of the tangent and the normal to the following curves at the indicated points :

Solution:

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 $=-1$