Question

One end of a rope is fastened to a boat and the other end is wound around a windlass located on a dock at a point 4 meters above the level of the boat. (see picture in the book) If the boat is drifting away from the dock at the rate of 2 meters/min , how fast is the rope unwinding at the instant when the length of the rope is 5 meters . (Let D(t) denote the length of rope at time t .)

Answer 1

Can you post the picture?

Answer 2

Yes, hold on

Answer 3

Actually never mind, there's no picture in the book like it said there was

Answer 4

I'm not understanding the problem. Bump your question as soon as the option becomes available, maybe someone else can help.

Answer 5

I don't understand

Answer 6

you want
y′
and you know that
y2=42+x2
and therefore
2yy′=2xx′
you are given that
x′=2

Answer 7

I know that y' is 1.56, but that's not the answer

Answer 8

Now that you know x^ prime

yy′=xx′
x′=2,x=5

and by pythagoras we find
\[y=\sqrt{4^2+5^2}= \sqrt{41}\]

Answer 9

I saw this on the other problem like this that was posted. That doesn't help.

Answer 10

given the height and hypot. of the right triangle, we can use pythag. thm. to find the base is 3.
The height of the dock is not changing so, dy/dt = 0.

What we know:
x = 3 y = 4 z = 5 dx/dt = 2 dy/dt = 0 and x^2 + y^2 = z^2

What we want:
dz/dt = ??

Sol:
take derivative wrt time,
2x dx/dt + 2y dy/dt = 2z dz/dt
substitute known values and simplify,
2(3)(2) + 2(4)(0) = 2(5) dz/dt
6 = 10 dz/dt
6/10 = 3/5 or 0.6 = dz/dt

so the rope is moving at a rate of 0.6 m/min.