• # question_answer   Find an angle $\theta ,\,\,0<\theta <\frac{\pi }{2},$ which increases twice as fast as it sine.

Let $y=\sin \theta$ On differentiating w.r.t.$'t'$, we get $\frac{dy}{dt}=\frac{d\,(sin\theta )}{dt}=\cos \theta \frac{d\theta }{dt}$ According to the question, $\frac{d\theta }{dt}=2\frac{dy}{dt}$ $\therefore$      $\frac{dy}{dt}=\cos \theta \times 2\frac{dy}{dt}$ $\Rightarrow$   $\cos \theta =\frac{1}{2}$         $\Rightarrow$$\theta =\frac{\pi }{3}$ Hence, the required angle is $\frac{\pi }{3}$.