Question

Find an angle θ
(i) which increases twice as fast as its cosine.
(ii) whose rate of increase twice is twice the rate of decrease of its cosine.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}x=\cos \theta \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ \frac{dx}{dt}=\frac{d\left(\cos \theta \right)}{dt} \\ =-\sin \theta \frac{d\theta }{dt} \\ \mathrm{But}\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\frac{d\theta }{dt}=2\frac{dx}{dt} \\ \Rightarrow \frac{dx}{dt}=-\sin \theta \left(2\frac{dx}{dt}\right) \\ \Rightarrow \sin \theta =-\frac{1}{2} \\ \Rightarrow \theta =\mathrm{\pi }+\frac{\mathrm{\pi }}{6}=\frac{7\mathrm{\pi }}{6} \\ \mathrm{Hence},\theta =\frac{7\mathrm{\pi }}{6}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let}x=\cos \theta \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ \frac{dx}{dt}=\frac{d\left(\cos \theta \right)}{dt} \\ =-\sin \theta \frac{d\theta }{dt} \\ \mathrm{But}\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\frac{d\theta }{dt}=-2\frac{dx}{dt} \\ \Rightarrow \frac{dx}{dt}=-\sin \theta \left(-2\frac{dx}{dt}\right) \\ \Rightarrow \sin \theta =\frac{1}{2} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{6} \\ \mathrm{Hence},\theta =\frac{\mathrm{\pi }}{6}. \end{gathered}$$