I have a related rates problem!!!?

Question

Cars A and B leave a town at the same time. Car A heads due south at a rate of 8km/hr and car B heads due west at a rate of 60 km/hr. how fast is the distance between the cars increasing after 3 hours

Answer 1

revicamc's solution is correct, except for typos in the final two equations. They should read as follows:

SQR(3600*h*h + 64*h*h) = SQR( 3664*h*h)
= 60.531*h

d' = 60.531

Since the velocities are constant (in magnitude and direction), you don't really need calculus here.
At any time t, the two cars and their starting point form the 3 vertices of a right triangle. The lengths of the two legs of the triangle are 60t and 8t (unless you mistyped the question and it should be 80t). The third side (the distance between the cars is the third side, and it can be found by the Pythagorean theorem. If the rate for A should be 80km/hr, then this distance is equal to 100t.
So if a particular distance is 100t km (with t measured in hours), at what rate is that distance increasing? Obviously, it increases by 100 km each time t increases by 1 hour, so it is increasing at a rate of 100 km/hr.
No calculus required.

Answer 2

Distance between points (x,y) and (0,0) are:
SQR((x*x) + (y*y))

(whoops! this is a calculus problem!)

after h hours: A is at (0, 8h )
after h hours: B is at (60h, 0)

distance between A & B after h hours :

SQR(3600*h*h + 64*h*h) = SQR( 3664*h*h)
= 60.531*h^2

For the rate of increase we take the first derivative:

d' = 121.062*h

The rest is left for the student.