At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 23 knots. How fast (in knots) is the distance between the ships changing at 3 PM?

Let the distance between them be D n miles.

Let the time passes since noon be t hours.

Did you make a diagram?
I see a right-angles triangle with sides
(23t), (17t + 30) and D so that
D^2 = (23t)^2 + (17t + 30)^2

2D(dD/dt) = 2(23t)(23) + 2(17t+30)(17)
dD/dt = (529t + 289t + 510)/D

when t = 3, (3:00 pm)
D^2 = 4761+6561
D = 106.405

dD/dt = (529(3) + 289(3) +510)/106.405
= 27.86 knots

To find the speed at which the distance between the ships is changing at 3 PM, we can use the concept of relative velocity.

First, let's calculate the positions of both ships at 3 PM.

From noon to 3 PM, 3 hours have passed. Ship A has been sailing west at a constant rate of 17 knots for 3 hours, so it has traveled 17 knots/hour × 3 hours = 51 nautical miles due west.

Ship B has been sailing north at a constant rate of 23 knots for 3 hours, so it has traveled 23 knots/hour × 3 hours = 69 nautical miles due north.

Now, we have two right-angled triangle sides: the distance traveled by Ship A to the west (51 nautical miles) and the distance traveled by Ship B to the north (69 nautical miles). The distance between the ships will be the hypotenuse of this right-angled triangle.

Using the Pythagorean theorem, the distance between the ships at 3 PM is √(51^2 + 69^2) ≈ 85.83 nautical miles.

Next, differentiate this distance with respect to time to find how fast the distance between the ships is changing at 3 PM.

Let's call the distance between the ships D(t), where t is the time elapsed since noon. Then we can write D(t) = √(x^2 + y^2), where x is the distance traveled by Ship A and y is the distance traveled by Ship B.

Differentiating D(t) with respect to t:

dD/dt = (1/2) * (x^2 + y^2)^(-1/2) * (2x*dx/dt + 2y*dy/dt)

Plug in the values we have:

dD/dt = (1/2) * (51^2 + 69^2)^(-1/2) * (51 * 17 + 69 * 23)

Simplifying this expression:

dD/dt = (1/2) * (2601 + 4761)^(-1/2) * (867 + 1587)

dD/dt = (1/2) * (7362)^(-1/2) * (2454)

dD/dt ≈ 0.069 knots/hour

Therefore, at 3 PM, the distance between the ships is changing at a rate of approximately 0.069 knots per hour.