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Point on a Curve

  1. Jan 20, 2009 #1

    jgens

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

    A particle moves along a path described by y = 4 - x^2. At what point along the curve are x and y changing at the same rate

    2. Relevant equations

    Simple equations regarding derivatives.

    3. The attempt at a solution

    It's been a while before I've done any related rates problems, could someone please let me know if this is correct:

    Since, x and y must be changing at the same rate (presumably with respect to time) x' = y' and y' = -2xx'. Therefore, -2x = 1 and x = -1/2. Placing my x value into the original equation yields 15/4. Hence, the point is (-1/2, 15/4).

    Thanks.
     
  2. jcsd
  3. Jan 20, 2009 #2
    SEEMS correct...
     
  4. Jan 20, 2009 #3

    Dick

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    Of course, it's right. What could go wrong?
     
  5. Jan 20, 2009 #4

    jgens

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    Plenty, I could have made an incorrect assumption ultimately leading to false conclusions.
     
  6. Jan 20, 2009 #5

    quasar987

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    Good answer but you do not need to assume that x and y are varying wrt an external parameter. The derivative y'(x) = dy/dx of y wrt x expresses the instantaneous rate of change of y wrt a change in x.

    The points where y and x are changing at the same rate are those where y'(x)=1.
     
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