At noon, Ship B is 80 miles due S of ship A. B is moving S at 17 mph, and A is moving W at 14 mph.?

Question

what is the rate of change of the distance between the ships at 1 pm?

Answer 1

To first order, this can be solved with the information given, bt if the ships are NOT near the equator, then the answer is not exactly correct.
To figure this out on a "flat earth", write the equation for the distance between the ships as a function of time:

d(t) = sqrt(x^2(t) + y^2(t))

where x(t) = the east-west distance between the ships and
y(t) is the north-south distance

x(t) = 14t
y(t) = 80+17t

So d(t) = (14*14*t^2 + (80+17t)^2)^.5

Using the chain rule, setting r = 14*14*t^2 + (80+17t)^2
d(t) = r^.5
d'(t) = .5r^(-.5)*r'(t)
r'(t) = 2*14*14*t + 2*(80+17t)*17 (used the chain rule inside here as well)

entering t = 1, we get

d'(1) = .5/sqrt(14*14 + (80+17)^2) * (2*14*14+2*97*17)
Which is about 19.
This makes sense because ship A is moving away from the n-s line along which ship B is traveling, and ship B is moving away from the east-west line along which A is traveling. The combined speed should be greater than the maximum of the two speeds which is 17. However, not all that much greater because they started out 80 miles apart.