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Q.
If $x=3 \cos \theta-2 \cos ^3 \theta$ and $y=3 \sin \theta-2 \sin ^3 \theta$. Then $\left(\frac{d^2 x}{d y^2}\right)_{\theta=\pi}$ will be
Continuity and Differentiability
Solution:
$ \frac{ dx }{ d \theta}=3 \sin \theta \cos 2 \theta ; \frac{ dy }{ d \theta}=3 \cos \theta \cdot \cos 2 \theta$
$ \frac{ dx }{ dy }=\tan \theta $
$\Rightarrow \frac{ d }{ dy }\left(\frac{ dx }{ dy }\right)=\frac{ d }{ dy } \cdot(\tan \theta) \Rightarrow \frac{ d ^2 x }{ dy ^2}=\frac{ d }{ d \theta}(\tan \theta) \cdot \frac{ d \theta}{ dy } $
$\Rightarrow \frac{ d ^2 x }{ dy ^2}=\frac{\sec ^2 \theta}{3 \cos \theta \cdot \cos 2 \theta}=\frac{1}{3 \cdot \cos ^3 \theta \cdot \cos 2 \theta} $
$\left(\frac{ d ^2 x }{ dy ^2}\right)_{\theta=\pi}=\frac{1}{3 \cdot(-1) \cdot 1}=-\frac{1}{3} $