Expert: Alon Mandes - 11/13/2008

Question
Suppose that you go to a lake on a nice summer day. Someone throws a ball into the water and it lands diagonal from you. (The picture given is a right triangle with you at the bottom left, the ball at the top right, and a 90 degree angle at the bottom right). The vertical distance from the ball to land is called d, the horizontal distance from you to the point on land closest to the ball is L.

Suppose that you can run on the beach at a speed of 10 feet per second. The speed at which you can swim is a constant s.

Suppose that you can run along the beach for a while before jumping in and swimming tom the ball. Find a function T (θ) that measures the amount of time it takes you to get to the ball in terms of the angle θ. (this is the angle where the ball is at.

Find any critical points for T (θ) and determine whether they are local maxima or minima.

What route would you take to arrive at the ball in the shortest possible amount of time?


Answer
Let's find the optimal rout (shortest possible amount of time) :
If you observe the drawing attached you can see that :
x is the amount of distance on land.
y is the amount of distance in water.
Tx=x/10.
Ty=y/s.
So, Time=Tx+Ty=x/10+y/10. Let's find a relation between x & y :
According to Pythagoras y=√[d^2-(L-x)^2]. Thus ,
Time=x/10+(1/s)√[d^2-(L-x)^2]. Let's derive :
Time'=1/10+(1/s)(1/2√[d^2-(L-x)^2])*2(L-x)
=1/10+(2/s)(L-x)(1/2√[d^2-(L-x)^2]), Now we set Time'=0 :
=(-1/20)√[d^2-(L-x)^2]=(4/s)(L-x). We raise both sides to power 2 :
(1/400)[d^2-(L-x)^2]=(16/s^2)(L-x)^2
(1/400)d^2=(L-x)^2*[(1/400)+(16/s^2)]
L - x = sqrt { (1/400)d^2 /[(1/400)+(16/s^2)] }
x=L-sqrt{ (1/400)d^2 /[(1/400)+(16/s^2)] }
& that means this is the route.

As for the angle θ , previous section :
Time=Tx+Ty=x/10+y/s & we know that tang(θ)=d/(L-x) , so
x=L-d/tang(θ) , thus :
T(θ) = L/10 - d/10tang(θ) + (1/s)√[d/tang(θ)]
T(θ) = L/10 - d/10tang(θ) + √(d/s^2)*[1/√tang(θ)]
Now derive & set the derivative equal zero.

Alon.