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Math Help - Rate of Change, Circle

  1. #1
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    Rate of Change, Circle

    Point moves anticlockwise around circle x^2 + y^2 =1 at uniform speed of 2 m/s

    a) Find expression for rates of change of its x - coordinate in terms of x, when the point is above the x -axis

    My attempt:
    Since v = 2,
    Then dx/dt^2 + dy/dt^2 = v = 2

    then dx/dt^2 = 2- dy/dt

    Is that correct, cause the book has -2sqrt(1-x^2)
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    Point moves anticlockwise around circle x^2 + y^2 =1 at uniform speed of 2 m/s

    a) Find expression for rates of change of its x - coordinate in terms of x, when the point is above the x -axis

    My attempt:
    Since v = 2,
    Then dx/dt^2 + dy/dt^2 = v = 2

    then dx/dt^2 = 2- dy/dt

    Is that correct, cause the book has -2sqrt(1-x^2)
    The easy way is to work in polar coordinates:

    x = r \cos (\theta) = \cos \theta

    y = r \sin (\theta) = \sin \theta

    \displaystyle \frac{dx}{dt} = \frac{dx}{d \theta} \cdot \frac{d \theta}{dt} = -sin (\theta) \cdot \frac{d \theta}{dt}.

    But since r = 1, \frac{d \theta}{dt} = 2.

    Therefore \displaystyle \frac{dx}{dt} = - 2 sin (\theta) = -2y = -2 \sqrt{1 - y^2}.

    ------------------------------------------------------------------------------------------------------------

    The longer way is to note that:

    \displaystyle 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = - \frac{x}{y} \frac{dx}{dt} .... (1)

    \displaystyle \left( \frac{dx}{dt}\right)^2 + \left( \frac{dy}{dt}\right)^2 = 4 .... (2)

    Substitute (1) into (2) and solve for \frac{dx}{dt}. Reject the positive solution since x is decreasing for points above the x-axis if the point moves anticlockwise.
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    Thanks so much.
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