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Calculus/an angle problem
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Question
Hi,
I'm hoping you can help me!
Here's the question:
A boat is being pulled by rope towards a dock that is 6 feet above where it is tied. The rope is being pulled in at a rate of 2 feet per second. How fast is the boat approaching the dock when there is 10 feet of rope left? At what rate is the angle from the dock changing at that moment?
It is known that 6²+y²=x² where y is the diagonal and x is the base of the right triangle with the dock at the top corner and the boat at the far end on the bottom. The right angle is below the dock and over from the ship.
Taking the derivative, 0 +2y(dy/dt) = 2x(dx/dt).
It is known that x=8, y=10, and dy/dt=-2. The variable dx/dt can be found using the preceding three values.
Take the angle Θ up at the dock between straight down and the cord.
cos(Θ)=y/6. Take the derivative of both sides with Θ and y as functions of t. You get -sin(Θ)dΘ/dt = 6(dy/dt).
Θ is the angle whose sin is 8/10. dy/dt is -2. Θ can be found by moving the values to the right side and then taking arcsin.
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Scotto
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