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If $\mathrm{x}=\cos \mathrm{t}, \mathrm{y}=\sin \mathrm{t}$, then what is $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ equal to?
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The correct answer is:
$-\mathrm{y}^{-3}$
Given that $x=\cos t, y=\sin t$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\sin \mathrm{t}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}=\cot \mathrm{t}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=-\frac{\cos \mathrm{t}}{\sin \mathrm{t}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\cot \mathrm{t}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=\operatorname{cosec}^{2} \mathrm{t} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\operatorname{cosec}^{2} \mathrm{t} \frac{1}{-\sin \mathrm{t}}=-\frac{1}{\sin ^{3} \mathrm{t}}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=-\mathrm{y}^{-3}$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\sin \mathrm{t}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}=\cot \mathrm{t}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=-\frac{\cos \mathrm{t}}{\sin \mathrm{t}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\cot \mathrm{t}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=\operatorname{cosec}^{2} \mathrm{t} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\operatorname{cosec}^{2} \mathrm{t} \frac{1}{-\sin \mathrm{t}}=-\frac{1}{\sin ^{3} \mathrm{t}}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=-\mathrm{y}^{-3}$
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