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:

$x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ at $\theta=\pi / 4$

Solution:

Given:

$x=\operatorname{acos}^{3} \theta \& y=a \sin ^{3} \theta$ at $\theta=\frac{\pi}{4}$

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

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

$\Rightarrow \mathrm{x}=\mathrm{a} \cos ^{3} \theta$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}\left(\frac{\mathrm{dx}}{\mathrm{d} \theta}\left(\cos ^{3} \theta\right)\right)$

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

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}\left(3 \cos ^{3}-1 \theta \times-\sin \theta\right)$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}\left(3 \cos ^{2} \theta \times-\sin \theta\right)$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{d} \theta}=-3 \mathrm{a} \cos ^{2} \theta \sin \theta \ldots(1)$

$\Rightarrow y=a \sin ^{3} \theta$

$\Rightarrow \frac{d y}{d \theta}=a\left(\frac{d y}{d \theta}\left(\sin ^{3} \theta\right)\right)$

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

$\Rightarrow \frac{d y}{d \theta}=a\left(3 \sin ^{3}-1 \theta \times \cos \theta\right)$

$\Rightarrow \frac{d y}{d \theta}=a\left(3 \sin ^{2} \theta \times \cos \theta\right)$

$\Rightarrow \frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta \ldots(2)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{-3 \operatorname{acos}^{2} \theta \sin \theta}{3 \operatorname{asin}^{2} \theta \cos \theta}$

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

$\Rightarrow \frac{d y}{d x}=-\tan \theta$

The Slope of the tangent is $-\tan \theta$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{4}}=-\tan \left(\frac{\pi}{4}\right)$

$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{4}}=-1$

$\therefore \tan \left(\frac{\pi}{4}\right)=1$

$\therefore$ The Slope of the tangent at $x=\frac{\pi}{4}$ is $-1$

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

$\Rightarrow$ The Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{4}}}$

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

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

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