A swimmer `S` is in the sea at a distance `d` `km` from the closest point `A` on a straight shore The house of the swimmer is on the shore at a distance `Lkm` from A He can swim at a speed of `ukm`/`hr` and walk at a speed of `vkm`/`hr(v>u)` At what point on the shore should he land so that he reaches his house in the shortest possible time?

Answer

`L-(ud)/(sqrt(v^(2)-u^(2))` km from his house

Explanation

Let the swimmer lands at the point P,x km from A and when walks from P to the point B to be reached

Given that AB=L km.Thenn PB =(L-x)km
T=Total time from S to B
=(Time taken from S to P) +(Time taken from P to B) `=SP/u+PB/v`
`therefore (dt)/(dx)=(x)/(u sqrt(d^(2)+x^(2))-(1)/(v)`
for maximum or minimum of `t,d//dx=0`
or `x=(ud)/(sqrt(v^(2)-u^(2))`
Therefore t is minimum for this for the value of x `(since (d^(2)t)/(dx^(2))is +ve)`
time if he lands at distane `L-x=L-(ud)/(v^(2)-u^(2))` form his house to be reached