A person, intending to cross a river by the shortest path, starts at an angle `prop` with the downstream. If the speed of the person be less than that of water current, show that `prop` must be obtuse.

Explanation

Let `v and u` be the velocities of the person and the water current, and `d` the width of the river. If `t` be the time taken by the person to cross the river, then `t = (d)/(v sin prop)` and distance carried downstream in this time
=`(d)/(v sin prop) [u + v cos theta] = x (say)`
Now, eliminating `prop`, we get
`t^2 (u^2 - v^2) - 2 xut + (x^2 + d^2) = 0`
Clearly `t` will have real solutions, if `"discriminant ge 0"`
`rArr 4 x^2 u^2 ge 4 (u^2 - v^2)(x^2 + d^2)`
`rArr x ge (d)/(v) (sqrt(u^2 - v^2))`
Distance swayed away by the water current (x) has to be minimum, for the shortest path.
Again `t = (d)/(v sin prop) and x = (d)/(v) sqrt(u^2 - v^2)=(d)/(v sin prop) (u + v cos prop)`
`rArr (u^2 - v^2) sin^2 prop = (u + v cos prop)^2`
`rArr u^2 (1 - sin^2 prop) + 2 u v cos prop + v^2 = 0`
`rArr u^2 cos^2 prop + 2 uv cos prop + v^2 = 0`
`rArr (u cos prop + v)^2 = 0 rArr cos prop = -(v)/(u)`.