Consider the situation of the previous problem. The man has to reach the other shore at the point, directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.

Velocity of man, $V_m$ = 3 km/hr

BD = horizontal distance for resultant velocity 'R'

X-component of resaultant,

R = 5 + 3~cos~$\theta$

t =$\frac{0.5}{3sin\theta }$

Which is sa me for horizontal compinent of velocity.

H = BD

= $5+3cos\theta \left(\frac{0.5}{3}sin\theta \right)$

= $\frac{5+3cos\theta }{6sin\theta }$

For H to be minimum $\left(\frac{dH}{d\theta }\right)=0$

$\Rightarrow \frac{d}{d\theta }\left(\frac{5+3cos\theta }{6sin\theta }\right)=0$

$\Rightarrow 18(si{n}^2\theta +co{s}^2\theta )-30cos\theta =0$

$\Rightarrow 30cos\theta -18$

$\Rightarrow cos\theta =-\frac{18}{30}=-\frac{3}{5}$

$Sin\theta =\sqrt{1-co{s}^2\theta }=\frac{4}{5}$

$\therefore H=\frac{5+3cos\theta }{6sin\theta }$
=$\frac{5+3(-\frac{3}{5})}{6\times \frac{4}{5}}=\frac{25-9}{24}$

=$\frac{16}{24}=\frac{2}{3}$~km

H = BD

= (5 + 3~cos~$\theta )\left(\frac{0.5}{3}sin\theta \right)$

= $\frac{5+3cos\theta }{6sin\theta }$

For H to be minimum $\frac{dH}{d\theta }$ = 0

$\Rightarrow \frac{d}{d\theta }\left(\frac{5+3cos\theta }{6sin\theta }\right)=0$

\Rightarrow 18 (sin^2\theta + cos^2\theta) - 30~cos~\theta = 0\)

$\Rightarrow$ 30~cos~\theta = 18

$\Rightarrow cos\theta =-\frac{18}{30}=-\frac{3}{5}$

$Sin\theta =\sqrt{1-co{s}^2\theta }=\frac{4}{5}$