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

Question:

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

$\theta=-\pi / 2$

Solution:

Given:

$x=a(\theta-\sin \theta) \& y=a(1+\cos \theta)$ at $\theta=\frac{-\pi}{2}$

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 $\frac{\frac{d y}{d x}}{\frac{d x}{d \theta}}$ and we get our desired $\frac{d y}{d x}$.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \cdot \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow \mathrm{x}=\mathrm{a}(\theta-\sin \theta)$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}\left(\frac{\mathrm{dx}}{\mathrm{d} \theta}(\theta)-\frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}(\sin \theta)\right)$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1-\cos \theta) \ldots(1)$

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})=\cos \mathrm{x}$

$\Rightarrow \mathrm{y}=\mathrm{a}(1+\cos \theta)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}\left(\frac{\mathrm{dx}}{\mathrm{d} \theta}(1)+\frac{\mathrm{dx}}{\mathrm{d} \theta}(\cos \theta)\right)$

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}(\cos \mathrm{x})=-\sin \mathrm{x}$

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}($ Constant $)=0$

$\Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \theta}=\mathrm{a}(0+(-\sin \theta))$

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} \theta}=\mathrm{a}(-\sin \theta)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \theta}=-\mathrm{a} \sin \theta \ldots(2)$

$\Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-a \sin \theta}{a(1-\cos \theta)}$

$\Rightarrow \frac{d y}{d x}=\frac{-\sin \theta}{(1-\cos \theta)}$

The Slope of the tangent is $\frac{-\sin \theta}{(1-\cos \theta)}$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{-\pi}{2}}=\frac{-\sin \frac{-\pi}{2}}{\left(1-\cos \frac{-\pi}{2}\right)}$

Since, $\theta=\frac{-\pi}{2}$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{-\pi}{2}}=\frac{-\sin \frac{-\pi}{2}}{\left(1-\cos \frac{-\pi}{2}\right)}$

$\therefore \sin \left(\frac{\pi}{2}\right)=1$

$\therefore \cos \left(\frac{\pi}{2}\right)=0$

$\Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\frac{-\pi}{2}}=\frac{-(-1)}{(1-(-0))}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\frac{-\pi}{2}}=\frac{1}{(1-0)}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\frac{-\pi}{2}}=1$

$\therefore$ 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 $=\frac{-1}{\left(\frac{d y}{d x}\right)_{\theta=\frac{-\pi}{2}}}$

$\Rightarrow$ The Slope of the normal $=\frac{-1}{1}$

$\Rightarrow$ The Slope of the normal $=-1$