Problema Solution
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 33 miles from the dock traveling at a speed of 60 miles per hour and the second boat is 54 miles from the dock traveling at a speed of 26 miles per hour.
Answer provided by our tutors
Let 'z' represent the distance between the boats after t hours.
We need to find the rate the distance is changing that is: dz/dt.
If the boat that heads due North is at (0,y) and the boat that heads due East is at (x,0), then the distance z between the boats is:
z^2 = x^2+y^2 (Using the Pythagorean Theorem).
At the given moment, z=√(54^2+33^2) = √4005
2z dz/dt = 2x dx/dt + 2y dy/dt
Now plug in the numbers (after canceling out all those useless 2's):
√4005 dz/dt = 54*26 + 33*60
dz/dt = 3384 / √4005 = 53.47 mph
The rate at which the distance between the boats is changing is 53.47 mph.