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April 24, 2014

April 24, 2014

Posted by **Anonymous** on Monday, March 30, 2009 at 9:14am.

A boat is pulled into a dock by a rope attatched to the bow of the boat and passing through a pulley on the dock that is 1m higher than the bow of the boat. If the rope is pulled in at a rate of 1m/s, how fast is the boat approaching the dock when it is 8m from the dock?

You would set up a right triangle and use the pythagoreom therom to solve I think. I need to figure out dx/dt right? How would I do that?

- calculus -
**Reiny**, Monday, March 30, 2009 at 9:37amyes, label your right-angled triangle this way:

hypotenuse = h , (the rope)

vertical = 1 , (height of dock above bow)

horizontal = x , (path of the boat)

you are given dh/dt = 1 m/s

when x = 8, h^2 = 1^2 + 8^

h = √65

from x^2 + 1^2 = h^2

2x dx/dt = 2h dh/dt

then dx/dt = 2(8)(1)/(2√65)

= 8/√65 m/s or appr. 0.992 m/s

- oops typocalculus -
**Reiny**, Monday, March 30, 2009 at 9:41amgot my substitution backwards in the last two lines

should be:

dx/dt = 2√65(1)/16

= √65/8 or 1.0079 m/s

- calculus -
**Anonymous**, Monday, March 30, 2009 at 9:48amthanx, that is what i thought i just had trouble sustituting all the numbers back in.

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