At noon, ship A is 70 km west of ship B. Ship A is sailing south at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 PM

57.6 km/h

Step-by-step explanation:

We are told that At noon, ship A is 70 km west of ship B.

Thus, coordinates of initial position of A and B is;

A(0,0) and B(70,0)

Now, we are told that Ship A is sailing south at 35 km/h and ship B is sailing north at 25 km/h. Thus, the final position of ship A and B after t hours are;

A(0,-35t) and B(70,25t)

Thus, distance between the two ships at time t hours is;

d(t) = √[(70 - 0)² + (25t - (-35t))²]

d(t) = √(4900 + 3600t²)

Now, to find how fast the distance between the ships changing, let's differentiate using the chain rule;

d'(t) = [1/(2√(4900 + 3600t²))] × 7200t

d'(t) = 3600t/√(4900 + 3600t²)

d'(t) = 3600t/10√(49 + 36t²)

d'(t) = 360t/√(49 + 36t²)

Now, at 4pm,it would have been 4 hours from noon. Thus, t = 4.

So;

d'(t) = (360×4)/√(49 + 36(4²))

d'(t) = 1440/√(49 + 576)

d'(t) = 1440/√625

d'(t) = 1440/25

d'(t) = 57.6 km/h