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If $x=a \sin \theta$ and $y=b \cos \theta,$ then $\frac{d^{2} y}{d x^{2}}$ is
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The correct answer is:
$\frac{-b}{a} \sec ^{2} \theta$
(b) Given $x=a \sin \theta$ and $y=b \cos \theta$
$\Rightarrow \frac{d x}{d \theta}=a \cos \theta$ and $\frac{d y}{d \theta}=-b \sin \theta$
$\therefore \quad \frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}=-\frac{b}{a} \tan \theta \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-b}{a} \sec ^{2} \theta$
$\Rightarrow \frac{d x}{d \theta}=a \cos \theta$ and $\frac{d y}{d \theta}=-b \sin \theta$
$\therefore \quad \frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}=-\frac{b}{a} \tan \theta \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-b}{a} \sec ^{2} \theta$
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