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.?

- anonymous

- schrodinger

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- anonymous

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

- anonymous

@ChmE |dw:1348980587824:dw|

- anonymous

@ChmE ??

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## More answers

- anonymous

So let \(t\) be our variable, and it will represent hours.

- anonymous

what happen next? @wio

- anonymous

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

- anonymous

f(0) = g(2)

- anonymous

then?

- anonymous

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

- anonymous

I thought it would be one hour

- anonymous

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

- 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.

- anonymous

ChmE, what is your solution?

- anonymous

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

- 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

- anonymous

can you show it step by step ?? @wio

- anonymous

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

- anonymous

anti derivative??? @wio

- 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

- anonymous

Yes, what does \(\int5dx\) equal?

- anonymous

we havent discussed it in our class.. no idea

- 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?

- 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

- anonymous

Since distance = speed * time.

- 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.

- anonymous

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

- anonymous

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

- anonymous

@wio yes..

- anonymous

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

- anonymous

|dw:1348982375609:dw|

- anonymous

See where I'm headed ChmE?

- anonymous

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

- 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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

No!

- anonymous

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

- 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!

- anonymous

what should we do next?

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