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Find the slope of the normal to the curve $x = acos^3\theta, \: y = asin^3\theta$$\; at\;$ $\theta =\Large {\frac{\pi}{4}}$.

$\begin{array}{1 1} 1 \\ -1 \\ \sqrt 3 \\ \frac{1}{\sqrt 3} \end{array}$

Can you answer this question?

Toolbox:
• $\large\frac{1}{\big(\Large\frac{dy}{dx}\big)_P}$=Slope of the normal to $y=f(x)$ at point $P$.
Step 1:
Given : $x=a\cos ^3\theta,y=a\sin ^3\theta$
Consider $x=a\cos^3\theta$
Differentiate w.r.t $\theta$ we get,
$\large\frac{dx}{d\theta}$$=a.3\cos^2\theta(-\sin \theta) \quad\;=-3a\sin \theta\cos^2\theta Step 2: Consider y=a\sin^3 \theta Differentiate w.r.t \theta we get, \large\frac{dy}{d\theta}$$=a.3\sin^2\theta.\cos \theta$
$\quad\;=-3a\sin^2 \theta\cos \theta$
Step 3:
Therefore $\large\frac{dy}{dx}=\frac{dy}{d\theta}$$\times\large\frac{d\theta}{dx}$
$\qquad\qquad\;\;\;=\large\frac{3a\sin^2\theta\cos \theta}{-3a\sin \theta\cos^2\theta}$
$\qquad\qquad\;\;\;=\large\frac{\sin \theta}{\cos \theta}$
$\qquad\qquad\;\;\;=\tan \theta$
Step 4:
The slope of the normal at $\theta=\large\frac{\pi}{4}$ is
$\large\frac{1}{\big(\Large\frac{dy}{dx}\big)_{\Large\frac{\pi}{4}}}=\frac{-1}{\tan\big(\Large\frac{\pi}{4}\big)}$
$\qquad\qquad=-1$
Hence the slope of the normal to the curve at $\theta=\large\frac{\pi}{4}$ is $-1$
answered Jul 10, 2013