# Homework Help: Ship traveling cal word problem i keep getting the wrong answer

1. Mar 12, 2009

### Okie

At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 19 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.)

Note: Draw yourself a diagram which shows where the ships are at noon and where they are "some time" later on. You will need to use geometry to work out a formula which tells you how far apart the ships are at time t, and you will need to use "distance = velocity * time" to work out how far the ships have traveled after time t.

So the way that I did (but it is wrong)
at 4pm 4 hours have past so ship A has travelled 16*4 = 64 knots and ship B has travelled 19*4 = 76 knots.

by pythagoras the distance between them at this time is "r"
r² = 64² + 76²
r² = 9872
r = 4√617

Now:
at any time if A is how far ship A has travelled and B is how far ship B has travelled then
r² = A² + B²
differentiating with respect to time
2r dr/dt = 2A dA/dt + 2B dB/dt
r dr/dt = A dA/dt + B dB/dt
we know dA/dt = 16 knots, dB/dt = 19 knots
at 4 pm
4√617 dr/dt = 64*16 + 76*19
4√617 dr/dt = 2468
dr/dt = 2468/4√617
dr/dt = 617/√617
dr/dt = √617 knots
so at 4PM the distance between them is increasing by 24.84 knots (2dp)

2. Mar 12, 2009

### Staff: Mentor

It looks like you're on the right track, but I don't think you have taken into account that the ships were not at the same point at noon.

Draw two pictures: one showing the ships at noon, and another showing the ships at 4pm.

Also, the distance between them is in nautical miles, not knots, which are units of speed.

3. Mar 12, 2009

### Okie

I am not sure what im suppose to do next? can u guide me please or show me. I would really apreciate it

4. Mar 12, 2009

### MeTh0Dz

This is what I did, I could be wrong though.

First I set up a triangle that described where each ship was at any time.
t = time
Ship A = (10 + 16t)
Ship B = (19t)
Distance from A to B = h

We need to find $$\frac{dh}{dt}$$ when t = 4.

h = $$\sqrt{(19t)^2 + (10 + 16t)^2}$$
h = $$\sqrt{617t^2+ 320t + 100}$$

$$\frac{dh}{dt}$$ = $$\frac{1234t + 320}{2\sqrt{(617t^2 + 320t)}}$$

Now set t as 4.

$$\frac{dh}{dt}$$ @ t = 4, is 24.9 knots.

5. Mar 13, 2009

### Okie

The answer above is wrong. Can anyone else help me out with this

6. Mar 13, 2009

### Okie

7. Mar 13, 2009

### Staff: Mentor

At noon, the ships are 10 n.m. apart, with ship A to the west of ship B.
At any time t >= 0, ship B will be 19t n.m. north of where it was at noon. Ship A will be 16t + 10 n.m. west of where ship B was at noon.

You have three points that form a right triangle. Ship A is on the left. Ship B's starting point is on the right, at the right angle of the triangle, and B's current position is north of where it was at noon. Draw this picture.

Write an expression that represents the distance D between ship A and ship B as a function of t. The rate of change of distance between the ships is dD/dt. You should end up with an equation for dD/dt in terms of the lengths of the sides of this triangle, and the rates of change (the speeds) of the ships. Evaluate dD/dt at 4pm (i.e., when t = 4 hrs).

Is that enough to get you going?

8. Mar 13, 2009

### MeTh0Dz

Hmm well I did notice that I screwed up one line. Although it doesn't change the answer much, perhaps tell us the answer if you know it?

Below is what I actually meant for the one line.

$$\frac{dh}{dt}$$ = $$\frac{1234t + 320}{2\sqrt{(617t^2 + 320t + 100)}}$$

9. Mar 13, 2009

### HallsofIvy

I think you will find it simpler to deal with h2.

$$h^2= 19^2t^2+ (10+ 16t)^2$$
so
$$2h\frac{dh}{dt}= 2(19^2)t+ 16(10+ 16t)^2$$

10. Mar 13, 2009

### MeTh0Dz

True, but it still leaves you with the ugly task of plugging h in at the end.

11. Mar 13, 2009

### HallsofIvy

But that's just arithmetic. h is approximately 106 nm at 4:00