Still Need Help Please
A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat.
Thats a 90 degree right triangle, the boat is at B, the winch at A, and line AC is the 12 feet.
The winch pulls in rope at a rate of 4 feet per second. there is 13 feet of rope out (line AB), find the acceleration of the boat.
okay, so i solved for line CB using the pythagorean theorem, which is 5 feet. i know i have to implicitely differentiate, then a second time because it is acceleration. could anyone help me.
tell us the eqution you got and we'll take it from there
Originally Posted by doctorgk
x^2 + y^2 =s^2
13^2 = x^2 + 12^2
2x(dx/dt) + 2y(dy/dt) = 2s(ds/dt)
so... 5(dx/dt) = s(ds/dt)
ds/dt = (13*4)/5
how would i get the second derivative to get acceleration??
Let us call the length of rope out (AB in diagram) u, and the distance of the boat from the dock (CB in diagram) x. We want .
We can get the velocity by the chain rule and that we know is -4 ft/s (from the problem). This gives velocity in terms of u. Applying the chain rule similarly to v gives us acceleration a in terms of u, and so you plug in u=13 ft for the moment in question to get the result.
so u would differentiate again leaving only the constant y/x, which is just going to be 5/144 ?