At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 19 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour

draw the figure, at noon, then at 4PM.

At 4 pm, I have a right triangle of A,B of sides (base 10+19*4; altitude 15*4)

d^2=(10+va*t)^2 + (vb*t)^2

2d dd/dt=2(10*va*t)va+2(vb*t)vb

so when t=4, solve for dd/dt

9898

khj

To find the speed at which the distance between the ships is changing, we need to use the concept of rates of change and relate the velocities of each ship to the distance between them. Here's how we can solve this problem step by step:

1. Define variables: Let's denote the distance between the ships as "d(t)" where t represents time in hours.

2. Determine the relationships between the variables:
- Ship A is sailing west at a constant speed of 19 knots. Therefore, the rate of change of its position with respect to time will be -19 (negative since it's moving west).
- Ship B is sailing north at a constant speed of 15 knots. Therefore, the rate of change of its position with respect to time will be +15 (positive since it's moving north).
- The distance between the ships, "d(t)," can be calculated using the Pythagorean theorem:
d(t) = √[ (x(t) - 0)^2 + (y(t) - 10)^2 ], where x(t) is the position of Ship A and y(t) is the position of Ship B.

3. Express the rates of change in terms of d(t):
- The rate of change of d(t) (the distance between the ships) with respect to time can be found by differentiating the above distance equation with respect to time.
- Let's call this rate of change d'(t).

4. Calculate the values at 4 PM:
- We need to find d'(t) at 4 PM, with t = 4 hours.

5. Plug in values and solve:
- Substitute the values of t into the equation for d'(t) and calculate the result.

Let's perform these calculations:

The position of Ship A at 4 PM (with t = 4 hours) will be x(t) = 10 - (19 * 4) = -66 nautical miles.
The position of Ship B at 4 PM (with t = 4 hours) will be y(t) = 15 * 4 = 60 nautical miles.

Now, substitute these values into the equation for d(t) to find the distance between the ships at 4 PM:
d(t) = √[ (-66 - 0)^2 + (60 - 10)^2 ] = √[ 4356 + 2500 ] = √6856 = 82.77 nautical miles.

Next, differentiate the distance equation d(t) with respect to time to find the rate of change of d(t) (d'(t)):
d'(t) = [ (-66 - 0) * (-19) + (60 - 10) * 15 ] / d(t) = (-1254 + 750) / 82.77 = -504 / 82.77 ≈ -6.08 knots.

Therefore, the distance between the two ships is changing at a rate of approximately 6.08 knots at 4 PM.