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Stuck on a related rates problem

  1. Apr 28, 2005 #1
    Im going through odd number related rate problems in preparation for an exam tmmrw. The correct answer is [tex]\frac{10}{\sqrt{133}}[/tex]. There is something wrong with the relation I construct; first the problem:

    The diagram they give is similar to an isocelles triangle with PQ running 12 ft down the middle. Point A is on the left, point B is on the right, representing carts A and B respectively.

    I figure they want me to find [tex]\frac{dB}{dt} = \frac{dB}{dA} \frac{dA}{dt}[/tex] So i expressed point QB in terms of QA and got [tex]\sqrt{ (39-\sqrt{QA^2+144})^2-144}[/tex]. However this taking the deriviative of this times 2ft/s does not yeild the correct answer. What would be the correct way to relate QB in terms of QA?
  2. jcsd
  3. Apr 28, 2005 #2


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    You have two right triangles sharing a common leg, with the sum of their hypotenuses constant at 39 feet. You need to work out the relationship between the lengths of the remaining two legs. It appears you know that, but have lost track of something somewhere. You need not solve for B in terms of A. Just take advantage of the fact that AP + BP is constant and find d(BQ)/dt in terms of d(BP)/dt and d(AQ)/dt in terms of d(AP)/dt. How are d(AP)/dt and d(BP)/dt related?
    Last edited: Apr 29, 2005
  4. Apr 28, 2005 #3
    well, then, I could solve for B in terms of A? Does that just mean that my work above is incorrect? Or that [tex]\frac{dB}{dt} = \frac{dB}{dA} \frac{dA}{dt}[/tex] is incorrect? I'm trying to find my conceptual flaw.
  5. Apr 28, 2005 #4


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    Your expression for QB appears to be correct, and the chain rule is correct. Perhaps taking a derivative that is a bit complicated is where you are going wrong, or maybe its finding the length of QB. I have some numbers worked out. What are you getting for QB and for the answer?
  6. Apr 28, 2005 #5


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    Your expression is right. I get the answer. Just be extra careful when taking the derivative.
  7. Apr 29, 2005 #6
    ahhh... I made the silly mistake of taking [tex]\frac{dA}{dB} \frac{dA}{dt}[/tex], not [tex]\frac{dB}{dt} = \frac{dB}{dA} \frac{dA}{dt}[/tex] thank you both for your assistence.
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