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Find an angle $\theta$,$0$ <$\;\theta<\Large \frac{1}{2}$,which increases twice as fast as its sine.

$\begin{array}{1 1} \frac{\pi}{6} \\ \frac{\pi}{3} \\ \frac{\pi}{4} \\ \frac{\pi}{2} \end{array} $

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  • If $ y = f(x)$, then $ \large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $ x $.
  • $ \bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
Let $ x = \sin\: \theta$
On differentiating w.r.t $t$ we get,
$ \large\frac{dx}{dt}$$=\cos\theta\large\frac{d\theta}{dt}$
But it is given that $ \large\frac{d\theta}{dt}$$=2.\large\frac{dx}{dt}$
$ \Rightarrow \large\frac{dx}{dt}$$=\cos\theta.2\large\frac{dx}{dt}$
$\Rightarrow \cos\theta=\large\frac{1}{2}$
$ \Rightarrow \theta = \large\frac{\pi}{3}$
Hence the angle is $ \large\frac{\pi}{3}$
answered Jul 31, 2013 by thanvigandhi_1
edited Feb 5, 2014 by rvidyagovindarajan_1

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