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.
explanation: The document is a discussion from a homework help forum, where a student is seeking assistance with a related rates problem in calculus. The problem involves a winch pulling a pipe to a vertical position, and the student is asked to find the rates of vertical and horizontal change at a specific point. The student provides their attempt at solving the problem and expresses difficulty in finding the necessary variables. Another user, arildno, provides a hint to solve the problem using the law of cosines and angular rates of change. The student then follows the hint and successfully solves the problem, but initially encounters an issue with the sign of the answer. After further guidance, the student resolves the issue.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion from a homework help forum, where a student is seeking assistance with a related rates problem in calculus. The problem involves a winch pulling a pipe to a vertical position, and the student is asked to find the rates of vertical and horizontal change at a specific point. The student provides their attempt at solving the problem and expresses difficulty in finding the necessary variables. Another user, arildno, provides a hint to solve the problem using the law of cosines and angular rates of change. The student then follows the hint and successfully solves the problem, but initially encounters an issue with the sign of the answer. After further guidance, the student resolves the issue.
  
  Label: question-and-answer