Question

A particle moves in $x-y$ plane. The position vector of the particle at any time $t$ is $\overrightarrow{r}=\left\{\left(2t\right)\hat{i}+\left(2t^2\right)\hat{j}\right\}m$ . The rate of change of $\theta$ at time $t=2s$ (where $\theta$ is the angle which its velocity vector makes with positive $x$ -axis ) is

• A. $\frac{2}{17}rad{s}^{-1}$
• B. $\frac{1}{14}rad{s}^{-1}$
• C. $\frac{4}{7}rad{s}^{-1}$
• D. $\frac{6}{5}rad{s}^{-1}$

Answer 1

Answer: (A)

$x=2t\Rightarrow v_x=\frac{dx}{dt}=2$
$y=2t^2\Rightarrow v_y=\frac{dy}{dt}=4t$
$\therefore tan\theta =\frac{v_y}{v_x}=\frac{4t}{2}=2t$
Differentiating with respect to time we get,
$\left({\sec }^2\theta \right)\frac{\mathrm{d}\theta }{\mathrm{d}t}=2$
or $\left(1+{\tan }^2\theta \right)\frac{\mathrm{d}\theta }{\mathrm{d}t}=2$
or $\left(1+4t^2\right)\frac{d\theta }{dt}=2$
or $\frac{d\theta }{dt}=\frac{2}{1+4t^2}$
$\frac{d\theta }{dt}att=2sis\frac{d\theta }{dt}=\frac{2}{1+4{\left(2\right)}^2}=\frac{2}{17}rad{s}^{-1}$