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Math Help - Parametric equations

  1. #1
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    Feb 2010
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    Parametric equations

    I am hoping someone can help me to solve this problem.

    In the xy-plane, a particle moves along the parabola y = x^2 - x with a constant speed of 2 times the square root of 10, units per second. If dx/dt is greater than 0, what is the value of dy/dt when the particle is at the point (2,2)? The answer is 6, but I do not know how to get there. Thanks in advance for your help.
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  2. #2
    Junior Member
    Joined
    Nov 2009
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    Speed equals  \sqrt{\frac{dx}{dt}^2 + \frac{dy}{dt}^2} = 2\sqrt{10}.

     y = x^2 - x implies that  \frac{dy}{dt} = (2x-1) \frac{dx}{dt}.

    So \sqrt{\frac{dx}{dt}^2 +\frac{dy}{dt}^2} = \sqrt{\frac{dx}{dt}^2 + (2x-1)^2 \frac{dx}{dt}^2} = \frac{dx}{dt} \sqrt{4x^2 - 4x + 2} = 2\sqrt{10}.

    Letting x = 2, you should get \frac{dx}{dt} = 2, which means \frac{dy}{dt} = (2*2-1)*2 = 6.
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