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March 30, 2015

March 30, 2015

Posted by **bryan** on Sunday, February 3, 2013 at 12:28pm.

- Calculus -
**Steve**, Sunday, February 3, 2013 at 2:30pmIf the rope is of length h, and the boat is x away from the dock,

h^2 = 1+x^2

When x=8, h = √65

2h dh/dt = 3x dx/dt

Since dh/dt = -1,

2√65 (-1) = 3*8 dx/dt

dx/dt = -√65/12

Since dx/dt < 0, the boat is approaching the dock at about 0.67 m/s

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