Find an angle θ
Question:

Find an angle $\theta$

(i) which increases twice as fast as its cosine.

(ii) whose rate of increase twice is twice the rate of decrease of its cosine.

Solution:

(i) Let $x=\cos \theta$

Differentiating both sides with respect to $t$, we get

$\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{\mathrm{d}(\cos \theta)}{\mathrm{d} t}$

$=-\sin \theta \frac{\mathrm{d} \theta}{\mathrm{d} t}$

But it is given that $\frac{\mathrm{d} \theta}{\mathrm{d} t}=2 \frac{\mathrm{d} x}{\mathrm{~d} t}$

$\Rightarrow \frac{\mathrm{d} x}{\mathrm{~d} t}=-\sin \theta\left(2 \frac{\mathrm{d} x}{\mathrm{~d} t}\right)$

$\Rightarrow \sin \theta=-\frac{1}{2}$

$\Rightarrow \theta=\pi+\frac{\pi}{6}=\frac{7 \pi}{6}$

Hence, $\theta=\frac{7 \pi}{6} .$

(ii) Let $x=\cos \theta$

Differentiating both sides with respect to $t$, we get

$\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{\mathrm{d}(\cos \theta)}{\mathrm{d} t}$

$=-\sin \theta \frac{\mathrm{d} \theta}{\mathrm{d} t}$

But it is given that $\frac{\mathrm{d} \theta}{\mathrm{d} t}=-2 \frac{\mathrm{d} x}{\mathrm{~d} t}$

$\Rightarrow \frac{\mathrm{d} x}{\mathrm{~d} t}=-\sin \theta\left(-2 \frac{\mathrm{d} x}{\mathrm{~d} t}\right)$

$\Rightarrow \sin \theta=\frac{1}{2}$

$\Rightarrow \theta=\frac{\pi}{6}$

Hence, $\theta=\frac{\pi}{6}$.