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1. Feb 22, 2007

dontdisturbmycircles

1. The problem statement, all variables and given/known data

A winch at the top of 12 meter high building pulls a pipe of the same length to a vertical position. The winch pulls in rope at a rate of -0.2 meters per second. Find the rate of the vertical change and the rate of the horizontal change at the end of the pipe when y=6.

2. Relevant equations

3. The attempt at a solution

I simply don't have enough information..

I set up an equation relating the three variables..

$$s^{2}=x^{2}+(12-y)^{2}$$

and differentiate with respect to time and end up with

$$s\frac{ds}{dt}=x\frac{dx}{dt}-(12-y)\frac{dy}{dt}$$

But I simply don't have all of the variables needed. I can solve for x (with pythagorean theorem), but not dx/dt...

Any help is appreciated. :yuck:

2. Feb 22, 2007

arildno

Hint:
First find the ANGULAR rate of change by a smart law of cosines.
THEN, find the rates of change of (x,y) by using a right-angled triangle.

3. Feb 22, 2007

dontdisturbmycircles

Thankyou arildno

I was able to get the correct numerical answer but something is bothering me...

Take the angle made at the top of the wall to be $$\theta$$...

We know that $$\frac{d\theta}{dt}$$ must be positive since the angle is tending towards 90deg right?

I got $$\frac{d\theta}{dt}=\frac{/sqrt{3}}{80}$$... but this leads to something that is incorrect. Because when I set up the following equation to relate x and s, I get this..

$$sin\theta=\frac{x}{s} = x(s)^{-1}$$
$$cos\theta\frac{d\theta}{dt}=x(-1)s^{-2}\frac{ds}{dt}+s^{-1}\frac{dx}{dt}$$

which eventually leads to

$$\frac{dx}{dt}=s(cos\theta\frac{d\theta}{dt}+x*s^{-2}\frac{ds}{dt})$$

Now, solving for x using the right triangle at the base, $$x=\sqrt{12^{2}-6^{2}} = 6*\sqrt{3}$$

and s=12 by similiar reasoning

Then $$tan\theta=\frac{6\sqrt{3}}{6}$$

$$\theta=\frac{\pi}{3}$$

So solving $$\frac{dx}{dt}=s(cos\theta\frac{d\theta}{dt}+x*s^{-2}\frac{ds}{dt})$$

I get $$\frac{dx}{dt}=\frac{\sqrt{3}}{15}$$

Which is the correct magnitude, but the incorrect sign... :/

Last edited: Feb 22, 2007
4. Feb 22, 2007

arildno

Well, I used the lowest angle, with the variable length s(t) as the opposite side. I didn't calculate it through, but my signs looked good.

5. Feb 22, 2007

dontdisturbmycircles

Ok, let me try that and see if I can get it to work.

6. Feb 22, 2007

dontdisturbmycircles

Wow it worked very smoothly using the lower angle... Correct sign and all. I'll have to think about this problem for a bit. Thanks again.