Question

Suppose that two boats leave a dock at different times. One heads due north, the other due east. Find the rate at which the distance between the boats is changing when the first boat is 56 miles from the dock traveling at a speed of 35 miles per hour and the second boat is 20 miles from the dock traveling at a speed of 32 miles per hour.

Answer 1

The distance traveled after t hours is:

If boat N is at (0,y)
boat E is at (x,0),
then the distance z from N to E is

z^2 = x^2+y^2
At the given moment, z=√(56^2+20^2) = √3536 = 4√221

2z dz/dt = 2x dx/dt + 2y dy/dt
Now plug in the numbers (after canceling out all those useless 2's):

4√221 dz/dt = 20*32 + 56*35
dz/dt = 2600 / 4√221 = 43.72 mi/hr