Question

If $x=a\left[\cos t+\frac{1}{2} \log \left(\tan ^{2} \frac{t}{2}\right)\right]$ and $y=a \sin t$, then find $\frac{d y}{d x}$ at $t=\frac{\pi}{4}$.

Answer 1

$x=a\left[\cos t+\frac{1}{2} \log \left(\tan ^{2} \frac{t}{2}\right)\right]$
$\Rightarrow x=a\left[\cos t+\log \left(\tan \frac{t}{2}\right)\right]$
$\Rightarrow \frac{d x}{d t}=a\left(-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\right)$
$=a\left(-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right)$
$=a\left(-\sin t+\frac{1}{\sin t}\right)$
$=a\left(\frac{1-\sin ^{2} t}{\sin t}\right)$
$=a\left(\frac{\cos ^{2} t}{\sin t}\right)$
$y=a \sin t$
$\Rightarrow \frac{d y}{d t}=a \cos t$
$\Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a \cos t}{\frac{a \cos ^{2} t}{\sin t}}=\tan t$
$\Rightarrow\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{4}}=\tan \frac{\pi}{4}=1$