Forums / Public / General discussion / Calculus!!...Related Rates!!....Please I need alot of HELP!!!?
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Gothicbunny
28 posts |
#10509 2007-10-25 21:18 GMT |
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A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher then the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 m from the dock?
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PrehistoricSwimmer
40 posts |
#10510 2007-10-25 21:26 GMT |
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1 m.s no acceleration!
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ShoeLover
45 posts |
#10511 2007-10-25 21:28 GMT |
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The rope, the dock, and the boat form a right triangle. When the boat is 8m from the dock, the rope is 8/(squareroot(65)) m long. So the ratio is:
8/(squareroot(65)) = x/1 x = 8/(squareroot(65)) |
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Stare
39 posts |
#10512 2007-10-25 21:33 GMT |
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OK, so when the boat is x m away from the dock the length of the rope is √(x^2 + 1) m. So we have
y = √(x^2 + 1) dy/dt = -1 dx/dt = ? dy/dt = dy/dx . dx/dt So -1 = (1/2) (x^2 + 1)^(-1/2) . (2x) dx/dt so dx/dt = -1 / [x . (x^2+1)^(-1/2)] = -√(x^2+1) / x So when x is 8 we have dx/dt = -√(65)/8 ≈ -1.008 m/s. So the boat is approaching the dock at approximately 1.008 m/s. |
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JetSki
35 posts |
#10513 2007-10-25 21:35 GMT |
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The closer it get, the faster it get there.
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FlowerPower
37 posts |
#10514 2007-10-29 19:39 GMT |
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i don't know that
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