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X- and y-coordinate's rate of change

  1. Nov 25, 2006 #1
    Hi, I have a problem with an exercise in the book "A First Course in Calculus, fifth edition". The problem is stated as follows:
    "A particle moves differentiably on the parabola [tex]y=x^2[/tex]. At what point on the curve are its x- and y-coordinate moving at the same rate? (You may assume dx/dt and dy/dt is not equal to 0 for all t)"
    According to the book the answers should be 1/2 and 1/4 however I only get 1/2.

    I solve it, as follows:
    Their rate is the same when
    [tex]\frac{dx}{dy}=\frac{dy}{dx}[/tex]
    so I differentiate the equation with respect to x.
    [tex]\frac{dy}{dx}=2x[/tex]
    And then with respect to y.
    [tex]1=2x\frac{dx}{dy}[/tex]
    [tex]\frac{dx}{dy}=\frac{1}{2x}[/tex]

    Then to find when their rate of change is the same I write:
    [tex]\frac{1}{2x}=2x[/tex]

    [tex]4x^2=1[/tex]

    [tex]x=\sqrt{1/4}[/tex]

    [tex]x=\frac{1}{2}[/tex]

    So, what have I done wrong? I'm not all that familiar with this stuff yet so I may have misunderstood the exercise.
     
    Last edited: Nov 25, 2006
  2. jcsd
  3. Nov 25, 2006 #2

    dextercioby

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    Maybe that 1/4 is the "y" value for the 1/2 "x" value ?

    Daniel.
     
  4. Nov 25, 2006 #3
    Thanks, that must be it. I can't believe I failed to notice that.
     
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