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January 30, 2015

January 30, 2015

Posted by **Zoe** on Wednesday, April 3, 2013 at 12:28am.

I'm totally stuck...

- Math-Calculus -
**Steve**, Wednesday, April 3, 2013 at 12:07pmDraw a diagram. Label as P the point on the ocean directly above the pulley. If the ship being pulled is x meters from P, and the pulling ship is at distance y,

√(x^2+45^2) + √(y^2+45^2) = 200

3 km/hr = 50m/min, so we want

dx/dt when t=1 and dy/dt=50.

at t=0, x=y=89.30

after 1 minute, y=139.30, so x=29.14

x/√(x^2+45^2) dx/dt + y/√(y^2+45^2) dy/dt = 0

29.14/53.61 dx/dt + 139.30/146.39 * 50 = 0

dx/dt = -87.53 m/s = 5.25 km/hr

Note that this cannot continue for long, as the ship will be pulled under by the cable.

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