Question

If \[ x = a\left(\cos t + \log \tan \frac{t}{2}\right), y = a \sin t, \] find the value of $\dfrac{dy}{dx}$.

Hint:

For parametric differentiation, always compute $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$. Simplify step by step to avoid mistakes.

Answer 1

Step 1: Differentiate $x$ w.r.t. $t$. \[ x = a\left(\cos t + \log \tan \frac{t}{2}\right) \] \[ \frac{dx}{dt} = a\left(-\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \frac{1}{2}\sec^2\frac{t}{2}\right) \] \[ = a\left(-\sin t + \frac{\sec^2\frac{t}{2}}{2\tan\frac{t}{2}}\right) \] Now, \[ \frac{\sec^2\frac{t}{2}}{2\tan\frac{t}{2}} = \frac{1}{\sin t}. \] So, \[ \frac{dx}{dt} = a\left(-\sin t + \frac{1}{\sin t}\right) = a\left(\frac{1-\sin^2 t}{\sin t}\right) = a\left(\frac{\cos^2 t}{\sin t}\right). \]

Step 2: Differentiate $y$ w.r.t. $t$. \[ y = a\sin t $\Rightarrow$ \frac{dy}{dt} = a\cos t. \]

Step 3: Find $\dfrac{dy{dx}$.} \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a\cos t}{a\cdot \frac{\cos^2 t}{\sin t}} \] \[ = \frac{\cos t \cdot \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} = \tan t. \]

Final Answer: \[ \boxed{\tan t} \]