Any idea in rate of change?

Question

A winch at the top of a 12 meter building pulls a pipe of the same
length to a vertical position. the winch pulls in rope at a rate of -
.2 meter per second. Find the rate of vertical change and the rate
of horizontal change at the end of the pipe when y=6.

[Difficulty]
can't get the amswer the book says.

[Thoughts]
S^2=y^2+X^2
2S(ds/dt)=2y(dy/dt)+0
2S(-.2)/2y= dy/dt
2/5= dy/dt
but the Answer is 1/4 and -.1154700

Answer 1

I assume that the winch is pulling the far end of the pipe toward the top of the building.

The distance between the end of the pipe, (x,y), and the top of the building, (0,12), is
D=sqrt(x^2 + (12-y)^2)
We know that dD/dt = -0.2 m/s.

Differentiate D with respect to t, and we get
dD/dt = (1/2)1/sqrt(x^2+(12-y)^2)*[2x dx/dt + 2(12-y)(-dy/dt)].

We have x^2 + y^2 = 12^2, so when y=6,
x = sqrt(12^2 - 6^2) = sqrt(144-36) = sqrt(108) = 6*sqrt(3).

Plug x and y into dD/dt, and fill in dD/dt = 0.2:
-0.2 = 1/2*[sqrt(3) dx/dt - dy/dt].

however, because x^2 + y^2 = 12^2, you know that
2x dx/dt + 2y dy/dt = 0, so when x=6sqrt(3) and y=6, you have
12 sqrt(3) dx/dt + 12 dy/dt = 0, or
dx/dt = -(1/sqrt(3))dy/dt.

You then get
-0.2 = -dy/dt, so dy/dt = 0.2 and
dx/dt = -(1/sqrt(3))dy/dt = -0.1154700.

Interesting, I got the same dx/dt as the book, but not the dy/dt.

Answer 2

Related rate problems are lots of fun. You didn't describe how the winch and pipe are attached, so I'm assuming the rope runs straight down the side of the building to the ground and is attached to the end of the pipe at that point. It then pulls the pipe up the side of the building.

Your equation is correct, but your derivative is missing a term:

2S ds/dt = 2x dx/dt + 2y dy/dt
or
S ds/dt = x dx/dt + y dy/dt

Since the pipe doesn't change length, ds/dt = 0. The vertical change, dy/dt, is always .2 m/sec (the pipe climbs up the building at the same rate that the winch draws in the rope. (Are you sure the book says dy/dy=1/4??) Finally, from the pythagorean theorem, when y=6, x=6 sqrt 3

The only thing left to change is dx/dt, the horizontal rate of change.

0 = 6 sqrt 3 dx/dt + 6 (.2)
dx/dt = -1.2 / 6 sqrt 3
dx/dt = -1.2 sqrt 3 / 18

dx/dt = -0.115 m/sec.

Answer 3

I think James L has the right answer, even though it doesn't match the book answer that you quote.

You can demonstrate that dy/dt = -sqrt(3) dx/dt, so the ratio of the two answers should be sqrt(3). This is true of James L's answer (0.2 and -.11547), and is not true of the book answers in your question.

Answer 4

Your error comes in your third line, where you have "+0". x is also a function of t. Fix this, and you should come out alright.