At noon, ship A is 110 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM?

2d dd/dt = 2(110-25t)(-25) + 2(15t)(15)

That's not right...

Yeah, I originally got 18.9 km/h, but the website is not accepting my answer. Is there something else?

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

Step 1: Determine the positions of the ships at 4:00 PM.
Since ship A is sailing east at a constant speed, it will have traveled for 4 hours (from noon to 4:00 PM) at 25 km/h. Therefore, ship A would be 4 hours * 25 km/h = 100 km east of its noon position.

Ship B, on the other hand, is sailing north at a constant speed and has also been traveling for 4 hours at 15 km/h. Therefore, ship B would be 4 hours * 15 km/h = 60 km north of its noon position.

Step 2: Use Pythagoras' theorem to calculate the distance between the ships at 4:00 PM.
The distance between the ships can be determined using the right triangle formed by the ships' positions. The vertical side represents the northward distance (60 km), and the horizontal side represents the eastward distance (110 km + 100 km = 210 km). Applying Pythagoras' theorem, we have:

Distance^2 = Vertical distance^2 + Horizontal distance^2
Distance^2 = 60 km^2 + 210 km^2

Simplifying, we get:
Distance^2 = 9000 km^2 + 44100 km^2
Distance^2 = 53100 km^2
Distance = sqrt(53100) km ≈ 230.39 km

Step 3: Calculate the rate at which the distance between the ships is changing.
To find the instantaneous rate of change at 4:00 PM, we need to differentiate the distance equation with respect to time (t). Differentiating Distance^2 = 53100 with respect to t, we get:

2 * Distance * (d(Distance)/dt) = 0
(2 * 230.39 km) * (d(Distance)/dt) = 0
460.78 km * (d(Distance)/dt) = 0
d(Distance)/dt = 0 / 460.78 km ≈ 0 km/h

Therefore, the rate at which the distance between the ships is changing at 4:00 PM is approximately 0 km/h.

let the time passed since noon be t hours.

So their paths form a right-angled triangle, with sides
(110 + 25t) and 15t
let d be the distance between them, then
d^2 = (110+25t)^2 + (15t)^2
2d dd/dt = 2(110+25t)(25) + 2(15t)(15)

when t = 4 , (4:00 pm)
d^2 = 210^2 + 60^2 = 47700
d = √47700

dd/dt = ( 50(210) + 30(60) )/(2√47700) = 28.16 mph

You are right, I see my error.

I have ship A going west instead of east.

let the time passed since noon be t hours.
So their paths form a right-angled triangle, with sides
(110 - 25t) and 15t
let d be the distance between them, then
d^2 = (110-25t)^2 + (15t)^2
2d dd/dt = 2(110-25t)(25) + 2(15t)(15)

when t = 4 , (4:00 pm)
d^2 = 10^2 + 60^2 = 3700
d = √3700

dd/dt = ( 50(10) + 30(60) )/(2√3700) = 18.9 mph