Question

If $u=2(t-\sin t)$ and $v=2(1-\cos t),$ then $\frac{dv}{du}$ at $t=\frac{2\pi }{3}$ is equal to :

• A. $\sqrt{3}$
• B. $-\sqrt{3}$
• C. $2\sqrt{3}$
• D. $\frac{2}{\sqrt{3}}$
• E. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is E $\frac{1}{\sqrt{3}}$

Given, $u=2(t-\sin t)$ and $v=2(1-\cos t),$
On differentiating w.r.t. t, we get
$\frac{du}{dt}=2(1-\cos t)$ and $\frac{dv}{dt}=2\left(\sin t\right)$
Therefore, $\frac{dv}{du}=\frac{dv/dt}{du/dt}$
$=\frac{2\left(\sin t\right)}{2(1-\cos t)}=\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2{\sin }^2\frac{t}{2}}$
$\Rightarrow \frac{dv}{du}=\cot \frac{t}{2}$
At $t=\frac{2\pi }{3}$, we have
$\frac{dv}{du}=\cot \left(\frac{2\pi }{3\times 2}\right)=\cot \frac{\pi }{3}=\frac{1}{\sqrt{3}}$