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Related rate

  1. Feb 22, 2007 #1
    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

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    3. The attempt at a solution

    I simply don't have enough information..

    I set up an equation relating the three variables..

    [tex]s^{2}=x^{2}+(12-y)^{2}[/tex]

    and differentiate with respect to time and end up with

    [tex]s\frac{ds}{dt}=x\frac{dx}{dt}-(12-y)\frac{dy}{dt}[/tex]

    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. jcsd
  3. Feb 22, 2007 #2

    arildno

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    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.
     
  4. Feb 22, 2007 #3
    Thankyou arildno :smile:

    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 [tex]\theta[/tex]...

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

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

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

    which eventually leads to

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

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

    and s=12 by similiar reasoning

    Then [tex]tan\theta=\frac{6\sqrt{3}}{6}[/tex]

    [tex]\theta=\frac{\pi}{3}[/tex]

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

    I get [tex]\frac{dx}{dt}=\frac{\sqrt{3}}{15}[/tex]

    Which is the correct magnitude, but the incorrect sign... :/
     
    Last edited: Feb 22, 2007
  5. Feb 22, 2007 #4

    arildno

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    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.
     
  6. Feb 22, 2007 #5
    Ok, let me try that and see if I can get it to work.
     
  7. Feb 22, 2007 #6
    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.
     
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