The current velocity of river grows in proportion to the distance from its bank and reaches the maximum value `v_0` in the middle. Near the banks the velocity is zero. A boat is moving along the river in such a manner that the boatman rows his boat always perpendicular to the current. The speed of the boat in still water is u. Find the distance through which the boat crossing the river will be carried away by the current, if the width of the river is c. Also determine the trajectory of the boat.

Explanation

Case I : If `v_R lt v_B`, the boat can cross river along a path perpendicular to flow.
Case II : If `v_B lt v_R`, drift cannot be zero, apply calculus in this case
Case I : If `v_R lt v_B`
Shortest path : `(D)/(v_B sin theta) = 3 T_0` ….(i)
Quickest path : `(d)/(v_B) = T_0` ...(ii)
Also `v_R - v_B cos theta = 0` for shortest path. ....(iii)
Thus, `sin theta = (1)/(3)` from (i) and (ii) or `v_R = v_B cos theta`
=`(d)/(T_0) sqrt(1 - (1)/(9)) =(2 sqrt(2 d))/(3 T_0)`
Case II : If `v_B lt v_R`
`x = ((d)/(v_B sin theta))(v_R - v_B cos theta)`
=`(d)/(v_B)(v_R cosec theta - v_B cot theta)`
For minimum `x, (d x)/(d theta) = -0`
`v_B(- cosec^2 theta) + v_R cosec theta cot theta = 0`
`cos theta = (v_B)/(v_R)`
Time taken in this case is given by `3 T_0 = (v_R d)/(v_B sqrt(v_R^2 - v_B^2))`
Also `v_B = (d)/(T_0)`
On solving, we get `v_R = sqrt((3)/(2)) (d)/(T_0)`.
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