1. ## Still Need Help Please

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

A-._
-..... _
- ........_
- ..........._
- ............._
- ................_
C-................ _B

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.

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

A-._
-..... _
- ........_
- ..........._
- ............._
- ................_
C-................ _B

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

3. ## equation

x^2 + y^2 =s^2
13^2 = x^2 + 12^2
x=5

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??

4. Originally Posted by doctorgk
x^2 + y^2 =s^2
13^2 = x^2 + 12^2
x=5

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??
first of all, you do not want 3 unknowns, that just makes the problem harder. we use variables to represent quantities that are changing. so your equation should be:

If we let $AC = 12$, $BC = x$ and $AB = y$, then:

$12^2 + x^2 = y^2$ .......and our aim is to find $\frac {d^2 x}{dt^2}$

$\Rightarrow 2x ~\frac {dx}{dt} = 2y~\frac {dy}{dt}$

$\Rightarrow \frac {dx}{dt} = \frac yx~\frac {dy}{dt}$

Now what?

5. ## Chain rule

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 $a = \frac{d^{2}x}{dt^2}$.
We can get the velocity $v=\frac{dx}{dt}$ by the chain rule $\frac{dx}{dt}=\frac{dx}{du}\frac{du}{dt}$ and that we know $\frac{du}{dt}$ 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.

--Kevin C.

6. ## equation 2

so u would differentiate again leaving only the constant y/x, which is just going to be 5/144 ?