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If $x=\sin t-t \cos t$ and $y=t \sin t+\cos t$, then what is $\frac{d y}{d x}$ at point $\mathrm{t}=\frac{\pi}{2} ?$
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As given :
$\mathrm{x}=\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t}$ and $\mathrm{y}=\mathrm{t} \sin \mathrm{t}+\cos \mathrm{t}$
On differentiating w.r.t. t, we get
$\frac{\mathrm{dx}}{\mathrm{dt}}=\cos \mathrm{t}-\{\cos \mathrm{t}+\mathrm{t}(-\sin \mathrm{t})\}$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\cos \mathrm{t}-\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}=\mathrm{t} \sin \mathrm{t}$
and, $\frac{d y}{d t}=t \cos t+\sin t-\sin t=t \cos t$
Hence, $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{t \cos t}{t \sin t}=\cot t$
$\Rightarrow\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{2}}=\cot \frac{\pi}{2}=0$
$\mathrm{x}=\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t}$ and $\mathrm{y}=\mathrm{t} \sin \mathrm{t}+\cos \mathrm{t}$
On differentiating w.r.t. t, we get
$\frac{\mathrm{dx}}{\mathrm{dt}}=\cos \mathrm{t}-\{\cos \mathrm{t}+\mathrm{t}(-\sin \mathrm{t})\}$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\cos \mathrm{t}-\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}=\mathrm{t} \sin \mathrm{t}$
and, $\frac{d y}{d t}=t \cos t+\sin t-\sin t=t \cos t$
Hence, $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{t \cos t}{t \sin t}=\cot t$
$\Rightarrow\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{2}}=\cot \frac{\pi}{2}=0$
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