text: # Related rate
  
  ## Homework Statement
  
  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.
  
  ## Homework Equations
  
  http://www.imagehosting.com/out.php/i249658_ahhh.JPG
  
  ## 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:
  
  ## Answers and Replies
  
  arildno
  Science Advisor
  Homework Helper
  Gold Member
  Dearly Missed
  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.
  
  orangeflover
  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:
  arildno
  Science Advisor
  Homework Helper
  Gold Member
  Dearly Missed
  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.
  
  Ok, let me try that and see if I can get it to work.
  
  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.