# If x and y are connected parametrically by the equation,

Question:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.

$x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$

Solution:

The given equations are $x=a\left(\cos t+\log \tan \frac{t}{2}\right)$ and $y=a \sin t$

Then, $\frac{d x}{d t}=a \cdot\left[\frac{d}{d t}(\cos t)+\frac{d}{d t}\left(\log \tan \frac{t}{2}\right)\right]$

$=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \frac{d}{d t}\left(\tan \frac{t}{2}\right)\right]$

$=a\left[-\sin t+\cot \frac{t}{2} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{d}{d t}\left(\frac{t}{2}\right)\right]$

$=a\left[-\sin t+\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \times \frac{1}{\cos ^{2} \frac{t}{2}} \times \frac{1}{2}\right]$

$=a\left[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right]$

$=a\left(-\sin t+\frac{1}{\sin t}\right)$

$=a\left(\frac{-\sin ^{2} t+1}{\sin t}\right)$

$=a \frac{\cos ^{2} t}{\sin t}$

$\frac{d y}{d t}=a \frac{d}{d t}(\sin t)=a \cos t$

$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \cos t}{\left(a \frac{\cos ^{2} t}{\sin t}\right)}=\frac{\sin t}{\cos t}=\tan t$