Question

If $x$ and $y$ are connected parametrically by the equations $x=a(\cos t+\log \tan \frac{t}{2}),y=a\sin t$ without eliminating the parameter, find $\frac{dy}{dx}$.

Answer 1

Given: $x=a(\cos t+\log \tan \frac{t}{2}),y=a\sin t$
Finding $\frac{dy}{dt}:$
$y=a\sin t$
$\frac{dy}{dt}=\frac{d\left(a\sin t\right)}{dt}$
$\frac{dy}{dt}=a\cos t$

Finding $\frac{dx}{dt}:$
$x=a(\cos t+\log \tan \frac{t}{2})$
$\frac{dx}{dt}=\frac{d\left(a(\cos t+\log \tan \frac{t}{2})\right)}{dt}$
$\frac{dx}{dt}=(\frac{d\left(\cos t\right)}{dt}+\frac{d\left(\log \left(\tan \frac{t}{2}\right)\right)}{dt})$
$\frac{dx}{dt}=a(-\sin t+\frac{1}{\tan \frac{t}{2}}.\frac{d\left(\tan \frac{t}{2}\right)}{dt})$

$\frac{dx}{dt}=a(-\sin t+\frac{1}{\tan \frac{t}{2}}.{\sec }^2\left(\frac{t}{2}\right).\frac{1}{2})$
$\frac{dx}{dt}=a(-\sin t+\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}.\frac{1}{{\cos }^2\frac{t}{2}}.\frac{1}{2})$
$\frac{dx}{dt}=a(-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}})$
$\left|\begin{array}{c}\because \sin t=2\sin \frac{t}{2}\cos \frac{t}{2}\end{array}\right|$
$\frac{dx}{dt}=a(-\sin t+\frac{1}{\sin t})$
$\frac{dx}{dt}=a\left(\frac{-{\sin }^{2}t+1}{\sin t}\right)$
$\frac{dx}{dt}=a\left(\frac{1-{\sin }^{2}t}{\sin t}\right)$
$\frac{dx}{dt}=a\left(\frac{{\cos }^{2}t}{\sin t}\right)$

Finding $\frac{dy}{dx}:$
Now,$\frac{dy}{dt}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\cdots \left(i\right)$
Substituting the value of $\frac{dy}{dt}$ and $\frac{dx}{dt}$ in (i),
we get,
$\frac{dy}{dx}=\frac{a\cos t}{\frac{a{\cos }^{2}t}{\sin t}}$
$\frac{dy}{dx}=a\cos t\times \frac{\sin t}{a{\cos }^{2}t}$
$\frac{dy}{dx}=\frac{\sin t}{\cos t}=\tan t$