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Tricky Related Rates Problem

  1. Jul 22, 2011 #1
    1. The problem statement, all variables and given/known data

    A winch at the top of a 12 metre building pulls a pipe of the same length to a vertical positon, as shown in the figure below. The winch pulls the rope at a rate of -0.2 m/s. Find the rate of the vertical change and horiziontal change at the end of the pipe when y=6.


    Sorry for no diagram I couldn't find a picture of it. It's question 27 in section 2.6 of Larson and Edwards Calculus 9th editon if that helps. bacially the diagram looks like a buliding on the left that's 12 metres tall with the winch on the top of the buliding on the right edge. The pipe is 12 metres long and is being pulled with a rope (s). Well I guess I can't draw it using characters on this site, sorry I hope the discribtion was enough.

    3. The attempt at a solution

    I know I need come up with an equation that uses all sides so I can use the ds/dt, I just can't seem to think of one. I don't think I can use Pythagorean Theorem because it's not a right triangle. And the only other equation for a triangle that I know is the Law of Cosines and I'm not given an angle. I also tried using Area formulas but I don't know dA/dt or s.

    Please help me with this problem, it's driving me nuts that I can't think of what to do.
     
  2. jcsd
  3. Jul 22, 2011 #2

    LCKurtz

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    The pipe leans against the building. Call the distance from the foot of the building to the foot of the pipe x, the distance from the bottom of the building to the top of the pipe y, and the length of rope hanging over the edge s. You are given ds/dt. Look for equations linking s, x, and y to get started. And, yes, I see a right triangle there.
     
  4. Jul 22, 2011 #3
    No see, the rope is attached to the part of the pipe that is in the air. So the triangle becomes two sides 12 metres (the buliding and the pipe) and the third side is the rope attached to the top of the building to the top of the pipe.

    Sorry I really wish I had a diagram here.
     
  5. Jul 22, 2011 #4

    LCKurtz

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    OK, so the pipe is butted up against the building at the bottom I guess. Draw a picture with the pipe partially raised and call the coordinates of the rising end (x,y). Draw a line from (x,y) perpendicular to the building. That forms a right triangle with hypotenuse s. Work with that triangle.
     
  6. Jul 22, 2011 #5
    But when I take d/dt of it I'll get a dy/dt and a dx/dt how will I be able to solve for both if both are unknown?
     
  7. Jul 22, 2011 #6
    So I tried it and got:

    x^2 + y^2 = s^2

    2x(dx/dt) + 2y(dy/dt) = 2s(ds/dt)

    Now I know what y equals and what ds/dt is, but I don't know x or s. So what did I overlook?
     
  8. Jul 22, 2011 #7

    LCKurtz

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    You have two right triangles. x and y are in the triangle that has the pipe as its hypotenuse; they are not the legs of the triangle having s as its hypotenuse. You don't have x2+y2=s2. Look again.
     
  9. Jul 22, 2011 #8
    Oh okay so I will get

    x^2 + y^2 = 12^2
    Then I could find x in terms of y and s, maybe? Because I do need that ds/dt in there somewhere.
     
  10. Jul 22, 2011 #9

    LCKurtz

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    Look again at the above. Use the other triangle too.
     
  11. Jul 23, 2011 #10
    Have you considered drawing the picture with MS Paint, uploading it to imagehost, flickr, imageshack, etc. & then posting the link?
     
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