question_answer

  Find an angle \[\theta ,\,\,0<\theta <\frac{\pi }{2},\] which increases twice as fast as it sine.

Answer:

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 \[\frac{\pi }{3}\].