At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 18 knots. How fast (in knots) is the distance between the ships changing at 6 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Question

At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 18 knots. How fast (in knots) is the distance between the ships changing at 6 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Answer 1

To find the rate at which the distance between the ships is changing, we can use the concept of rate of change.

Let's break down the given information and variables:
- Ship A is 50 nautical miles due west of ship B.
- Ship A is sailing west at 15 knots.
- Ship B is sailing north at 18 knots.

We need to find the rate at which the distance between the ships is changing at 6 PM. To do this, we can consider the speeds and directions of both ships as vectors.

Let's assume that the initial position of ship A is represented by the point (0, 0), and the initial position of ship B is (-50, 0) (since ship A is 50 nautical miles due west of ship B).

We can define the position of ship A at time t as A(t) = (15t, 0), and the position of ship B as B(t) = (-50, 18t).

Now, let's find the distance between ship A and ship B at time t. Using the distance formula, we have:

d(t) = √((x2 - x1)^2 + (y2 - y1)^2)

Substituting the positions of ship A and ship B, we get:

d(t) = √((15t + 50)^2 + (18t - 0)^2)

Now, let's find the rate at which the distance is changing at 6 PM, which means t = 6 hours. We can find the derivative of d(t) with respect to t and evaluate it at t = 6.

d'(t) = 1/2 * (2*(15t + 50)*(15) + 2*(18t)^2)^(1/2-1) * (2*(15t + 50)*(0) + 2*(18t)*(18))

Evaluating at t = 6:

d'(6) = 1/2 * (2*(15*6 + 50)*(15) + 2*(18*6)^2)^(1/2-1) * (2*(15*6 + 50)*(0) + 2*(18*6)*(18))

Simplifying the expression, we get:

d'(6) = 1/2 * (2*(135)*(15) + 2*(108)*(18))^(1/2-1) * (2*(135)*(0) + 2*(108)*(18))

d'(6) = 1/2 * (4050 + 3888)^(1/2-1) * (0 + 3888)

d'(6) = 1/2 * 7938^(1/2-1) * 3888

Now, let's calculate d'(6):

d'(6) = 1/2 * 7920^(1/2-1) * 3888

d'(6) = 1/2 * 7920^(-1/2) * 3888

d'(6) = 1/2 * (1/√7920) * 3888

d'(6) ≈ 1/2 * (1/√89) * 3888

d'(6) ≈ 1/2 * (1/9.43) * 3888

d'(6) ≈ 0.0553 * 3888

d'(6) ≈ 214.8104

Therefore, at 6 PM, the rate at which the distance between the two ships is changing is approximately 214.8104 knots.