## anonymous 3 years ago One ship is sailing south at a rate of 5 knots, and another is sailing east at a rate of 10 knots. At 2 P.M. the second ship was at the place occupied by the first ship one hour before. at what time was the distance between the ships not changing.?

1. anonymous

Did you draw a picture yet. i find this really helps.

2. anonymous

@ChmE |dw:1348980587824:dw|

3. anonymous

@ChmE ??

4. anonymous

So let $$t$$ be our variable, and it will represent hours.

5. anonymous

what happen next? @wio

6. anonymous

Let f(t) be the position of ship sailing south, and g(t) be the position ship sailing east.

7. anonymous

f(0) = g(2)

8. anonymous

then?

9. anonymous

I'm interested in the sol'n as well. How did you come up with f(0) = g(2)

10. anonymous

I thought it would be one hour

11. anonymous

You're right, it's f(0) = g(1)

12. anonymous

@bii17 sry for not responding earlier. I wasn't confident in my sol'n and didn't want to possibly steer you down a wrong path at the risk of confusing you.

13. anonymous

14. anonymous

@ChmE that's ok.. @wio can i see your solution.. i really dont get the problem..

15. anonymous

My solution is to find the position functions, then use the Pythagorean theorem on them to get a 'change in distance' function. Then take the derivative. Then find when that is 0

16. anonymous

can you show it step by step ?? @wio

17. anonymous

Okay bii17. What is the anti-derivative of 5?

18. anonymous

anti derivative??? @wio

19. anonymous

I don't understand what the question is asking. Is it looking for at what time is the distance the same as the initial change in distance? Thats how I read it

20. anonymous

Yes, what does $$\int5dx$$ equal?

21. anonymous

we havent discussed it in our class.. no idea

22. anonymous

Hmmmm, okay, well let's say that the ship started as position 0, it is moving 5 knots, how many knot-hours has it traveled in t hours?

23. anonymous

Let a knot-hour be the distance you travel in an hour at the speed of a knot. Our ship will have traveled 5t

24. anonymous

Since distance = speed * time.

25. anonymous

So f(t) = 5t. Now g(t) = 2t + c. Where c is the initial position. We don't know it yet, but we can use f(0) = g(1) to find out c.

26. anonymous

5(0) = 2(1) + c => c = -2

27. anonymous

So g(t) = 2t-2 Are you following?

28. anonymous

@wio yes..

29. anonymous

Now, since they are traveling perpendicular to each other... how can we find the distance between them with respect to time?

30. anonymous

|dw:1348982375609:dw|

31. anonymous

32. anonymous

The reason is because we can give them any initial position we want, so long as they are consistent. Right?

33. anonymous

I made it so the initial position would be the point that they both eventually reach. Then I made it so that at time 0, the first boat was at that position.

34. anonymous

No problem, it's a bit complicated but I can try to explain it better if you'd like.

35. anonymous

Basically I made the origin the point where they both reach at some point.

36. anonymous

I made t=0 when the first boat is at that position.

37. anonymous

so the second boat doesn't get into that position until 1 hour later

38. anonymous

so since it is moving at 2 knots, it is 2knot-hours behind, giving us that -2 in 2t-2

39. anonymous

No!

40. anonymous

It is the time at which the distance between the ships isn't changing.

41. anonymous

So first we find the position of the ships with respect to the origin. Then.... well look at that triangle I drew up top!

42. anonymous

what should we do next?