Differentiation involving Related Rates

[Haianel]

Okay. Here is the problem.

A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat.

(a) The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock?

(b) Suppose the boat is moving at a constant rate of 4 feet per second. Determine the speed at which the winch pulls in rope when there is a total of 13 of rope feet out. What happens to the speed at which the winch pulls in rope as the boat gets closer to the dock?

So, first off... for the ends of both "a" and "b", my (personal) logic says, for "a", that the closer the boat gets, the slower it will go. For "b", I think that the winch should pull faster.

Secondly, using the pythagorean theorem, I found that the missing "x" axis side was 5 feet. I know that 13 feet is my hypotenuse, and that 12 feet, as the height, will remain constant. Now, using "h" to denote the hypotenuse, I can say that "dh/dt" is equal to -4 feet per second. I know that in "a" I am looking for "dx/dt" and that in "b" I am looking for "dh/dt". That is about as far as I have gotten. Any help at all is appreciated.

 

[galactus]

Yes, use the Pythagorean theorem.

\(\displaystyle D^{2}=x^{2}+y^{2}\)

\(\displaystyle D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

Enter in the knowns and solve for dx/dt.

You already know that x=5 and dy/dt=0 (because it remains constant).

You can use the same equation to find dD/dt in part b.