Question:
Find $\frac{d y}{d x}$, if $y=12(1-\cos t), x=10(t-\sin t),-\frac{\pi}{2}
Solution:
It is given that, $y=12(1-\cos t), x=10(t-\sin t)$
$\therefore \frac{d x}{d t}=\frac{d}{d t}[10(t-\sin t)]=10 \cdot \frac{d}{d t}(t-\sin t)=10(1-\cos t)$
$\frac{d y}{d t}=\frac{d}{d t}[12(1-\cos t)]=12 \cdot \frac{d}{d t}(1-\cos t)=12 \cdot[0-(-\sin t)]=12 \sin t$
$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{12 \sin t}{10(1-\cos t)}=\frac{12 \cdot 2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{10 \cdot 2 \sin ^{2} \frac{t}{2}}=\frac{6}{5} \cot \frac{t}{2}$