A river flows $3\text{km}{\text{h}}^{-1}$  and a man is capable of swimming$2\text{km}{\text{h}}^{-1}$ . He wishes to cross it such that the displacement parallel to river is minimum. In which direction will he swim?

Let us assumes he swims at an angle $\theta$  with the perpendicular as shown. If river is $\text{'}d\text{'}m$  wide time taken to cross it

$t=\frac{d}{2\cos \theta },v_x=3-2\sin \theta$  horizontal distance

Covered along x direction during this period $x=v_x$

               $t=(3-2\sin \theta )\frac{d}{2\cos \theta }$

For $t$ to be $\min \frac{dx}{d\theta }=0$

$d[\frac{3}{2}\sec \theta \tan \theta -{\sec }^2\theta ]$  $=0$

$\sin \theta =\frac{2}{3}$