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Related Rates

  1. Jan 20, 2009 #1

    jgens

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

    Mr. Wilson is standing near the top of a ladder 24 feet long which is leaning against a vertical wall of his house. Dennis the little boy next door, ties a rope from his tricycle to the bottom of the ladder and starts to pull the foot of the ladder away from the house wall. The bottom end of the ladder begins to slide away from the wall at the rate of 1 foot per second. How fast is the angle between the top of the ladder and the wall changing when the foot of the ladder is 8 feet from the wall?

    2. Relevant equations

    Again, simple derivatives.

    3. The attempt at a solution

    Well, here's my best guess:

    Listing the knowns: z = 24, x' = 1, x = 8.

    One of my relating equations is x^2 + y^2 = z^2; therefore, xx' + yy' = 0. While the second relating equation is tan(theta) = x/y, and consequently sec^2(theta)(theta)' = (x'y - y'x)/y^2. Or similarly: (theta)' = (x'y - y'x)/(sec^2(theta)y^2). A few intermediate calculations give y' = -8/sqrt(512), y = sqrt(512), and theta = arctan(8/sqrt(512)). Plugging everything in produces (theta)' = 0.0405.
     
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  3. Jan 20, 2009 #2

    Dick

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    Now, I'm not sure I agree with that one. The angle between the wall and the ladder is theta=arcsin((8+x)/24), right? Find theta' and use x'=1. I seem to be getting a different answer from yours. But it's late, and I could be wrong.
     
  4. Jan 20, 2009 #3

    jgens

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    I suppose your approach if considerably simpler. We let sin(theta) = x/z = 8/24. Therefore, cos(theta)(theta)' = x'/z; a little algebra yields (theta)' = x'/(zcos(theta)). Plugging everything in should yield, theta' = 1/sqrt(512) which is approximately 0.0442. Is this what you got Dick?
     
  5. Jan 20, 2009 #4

    jgens

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    I'll try the method you presented directly. theta = arcsin(x/z); therefore, theta' = x'/(zsqrt(1 - (x/z)^2)) and plugging everything in I get the same result.
     
  6. Jan 20, 2009 #5

    Dick

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    Yes, I got 0.0442. But it's late enough that I'm starting to get a different answer everytime I do it. So have faith in yourself.
     
  7. Jan 20, 2009 #6

    jgens

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    That's pretty funny. Thanks.
     
  8. Jan 21, 2009 #7

    Dick

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    Then vote for me for funniest HW helper of the year for next year. I just got 0.0442 again. So the odds are 2/3 that it's right. 3/4 if we count your post.
     
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