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If $x=10(t-\sin t), y=12(1-\cos t)$, find $\frac{d y}{d x}$


We have, $x=10(t-\sin t)$ and $y=12(1-\cos t)$

$\Rightarrow \frac{d x}{d t}=\frac{d}{d t}[10(t-\sin t)]$ and $\frac{d y}{d t}=\frac{d}{d t}[12(1-\cos t)]$

$\Rightarrow \frac{d x}{d t}=10 \frac{d}{d t}(t-\sin t)$ and $\frac{d y}{d t}=12 \frac{d}{d t}(1-\cos t)$

$\Rightarrow \frac{d x}{d t}=10(1-\cos t)$ and $\frac{d y}{d t}=12[0-(-\sin t)]=12 \sin t$

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{12 \sin t}{10(1-\cos t)}$

$\Rightarrow \frac{d y}{d x}=\frac{12 \times 2 \sin \frac{t}{2} \cos \frac{t}{2}}{10 \times 2 \sin ^{2} \frac{t}{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{6}{5} \cot \frac{t}{2}$

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