No idea how to start this related rates problem!

[voidflux]

Hello everyone!
I'm doing related rates, usually doing with volume or area of a shape, but now we are getting into different types of problems, problems dealing with a right triangle and i'm getting lost! The problem is as fallows:
A boat is pulled into a dock by means of a winch 12 ft aove the deck of the boat
(a) The winch pulls in rope at a rate of 4 ft per second. Determine the speed of the boat when there is 13 feet of rope out.
(b) Suppose the boat is moving at a cosntant rate of 4 feet per second. Determine the speed at which the winch pulls in rope when there is a total of 13 feet of rope out.

OKay the picture looks like this

x[winch]
|-.
|   -.
|      -.
|         -.
|            -.  <|
--------------  [boat]

The height of this is 12 feet
the hypotenus is 13 ft
So the winch is that x up there, and the hypotenus is the rope, so the witch is pulling at a horizontal direction.
Thats what confuses me....
I know how to solve latter problems like:
A 13' ladder is leaning against a wall if we pull the bottom at 3ft per second, how fast is the top going down when the ladder is 5 feet from the wall.
I know how to solve problems like these, But I don't know what the top problem wants me to solve for.
For instance, in question (a) if the winch is pulling the boat, does that mean the hypotenus, we'll call r for rope, dr/dt = 4ft/sec?
Then using c^2=b^2+a^2; we can find the x, which would come out to be 5. So do they want me to solve for dx/dt? meaning dx/dt is the speed of the boat?
I'm also lost at the fact that it seems like I have too much information, because from what they gave me, i end up with
(13)^2 = 12^2 + x^2;
d/dx[13^2] = d/dx [12^2] + d/dx[x^2];
0 = 0 + 2x dx/dt
0 = dx/dt?
which is wrong in so many ways, any help?
thanks for listening!

[sql_lall]

Split into horizontal and vertical:

Find F_h and F_v in terms of t.
I.e. find the components of the pull in terms of the time.

=> you have d F_h / dt and d F_v /dt
now you can use d F_h / dt to find the pull on the boat, and the overall speed at a certain time.

[voidflux]

I figured it out alittle differnetly, I don't nkow if i'm getting the rgiht answer though because this is an even question.
I did the following:
c^2 = x^2 + y^2;
2c dc/dt = 2x dx/dt + 2y dy/dt
u know dy/dt is going to be 0 because the height is never changing, so
u now have
2c dc/dt = 2x dx/dt
U want to find dx/dt because thats the speed of the boat
dx/dt = c/x dc/dt
dx/dt = (13)/5 (4)
dx/dt = 10.4 ft/sec

for the 2nd part, part b
i got
dc/dt = x/c dx/dt
dc/dt = 5/13(4)
dc/dt = 6.15 ft/sec