# Homework Help: Related Rates Ship Distance Problem

1. Sep 10, 2010

### bobjefri

Problem:
At 12 noon, ship A travels due west at 100 knotsshiop B. After 4hours, ship B leaves the same port on the direction N30'W at 80 knots. How fast is the distance between two ships changing at 8PM in the same day?

Thanks.

2. Sep 11, 2010

### HallsofIvy

Ships traveling at 80 and 100 knots? I can see hydroplane boats going that fast but ships?

Any way: ship A travels at 100 knots starting at noon. In t hours it will have gone a distance 100t nautical miles. Ship B travels at 80 knots, starting 4 hours later. In t hours after noon, it will have gone 80(t- 4) nautical miles. Since A goes due west and B goes "N30W" (you have ' after the 30 and normally that means "minutes" (1/60 of a degree) but that would be a very unusual way of giving a direction- I am going to assume it is "North 30 degrees West". That makes the angle between the two paths 90- 30= 60 degrees. The third side of the triangle, the distance between the two ships is given by the "cosine rule"- if triangle ABC sides AC of length a and BC of length b then the third side, c, is given by $c^2= a^2+ b^2- 2abcos(C)$.

Here, that is $c^2= 10000t^2+ 6400(t- 4)^2- 2(100t)(80(t- 4))cos(60)$. Of course, cos(60)= 1/2 so that is just $c^2= 10000t^2+ 6400(t- 4)^2- 4000t(t-4)$. Differentiate with respect to t to find the speed with which that is changing.

Last edited by a moderator: Sep 14, 2010
3. Sep 13, 2010

### bobjefri

HallsofIvy, thank you so much.I think it's not complete or was cutted,idk..
About the 'knots' covered by the ships,hehe,well,that's what the reference book gives..
thank you so much.
oh, sorry for the ' that i used in N30'W, that's actually degrees.i did not found the degree sign.hehe.im so sorry.

4. Sep 14, 2010

### HallsofIvy

No, there was nothing cut off. What happened was that I had suggested a solution using coordinates, then said "not using coordinates" and started showing how to use the cosine law, but then went back and changed the original to "cosine law" solution. I have now taken off that last paragraph.