Question

The position vector of a particle at time 't' is given by $\overrightarrow{r}=t^2\hat{t}+t^3\hat{j}$ . The velocity makes an angle $\theta$ with positive x-axis. Find the of $\frac{d\theta }{dt}$ at $t=\frac{\sqrt{2}}{3}$

Answer 1

$\overrightarrow{r}=x\hat{i}+y\hat{j}=t^2\hat{i}+t^3\hat{j}$

$\Rightarrow x=t^2$ & $y=t^3\Rightarrow v_x=2t$ & $v_y=3t^2$
But $tan\theta =\frac{v_y}{v_x}=\frac{3t^2}{2t}=\frac{3}{2}t$

$\Rightarrow {\sec }^2\theta \frac{d\theta }{dt}=\frac{3}{2}$
$\Rightarrow \frac{d\theta }{dt}=\frac{3}{2{\sec }^2\theta }=\frac{3}{2(1+{\tan }^2\theta )}$

$=\frac{3}{2[1+\frac{9}{4}{t}^2]}=\frac{3}{2[1+\frac{9}{4}\times \frac{2}{9}]}=\frac{3}{2[1+\frac{1}{2}]}=1$