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Math Help - Parametric Equation Integration

  1. #1
    Del
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    Parametric Equation Integration

    Notice that the curve given by the parametric equations

    x = 81 - t^2

    y = t^3 - 25t

    is symetric about the x-axis. (If t gives us the point (x,y), then -t will give (x,-y) ).

    At which x value is the tangent to this curve horizontal?

    The curve makes a loop around the x-axis. What is the total area of this loop?

    -----------
    Please help!
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  2. #2
    Super Member wingless's Avatar
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    x = 81 - t^2

    \frac{dx}{dt} = -2t

    y = t^3 - 25t

    \frac{dy}{dt} = 3t^2 - 25


    \frac{dy}{dx} = \frac{dy}{dt}\cdot \frac{dt}{dx} = \frac{3t^2-25}{-2t}

    y' = \frac{25-3t^2}{2t}

    Now set y'=0 (cause its slope is 0) and find its corresponding x coordinate.
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  3. #3
    Del
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    **post deleted** Thanks for the help.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Del View Post
    Notice that the curve given by the parametric equations

    x = 81 - t^2

    y = t^3 - 25t

    is symetric about the x-axis. (If t gives us the point (x,y), then -t will give (x,-y) ).

    At which x value is the tangent to this curve horizontal?

    The curve makes a loop around the x-axis. What is the total area of this loop?

    -----------
    Please help!

    The curve is horizontal when:

    \frac{dy}{dx}=0,

    differentiating the equation for y in terms of t by x give:

    \frac{dy}{dx}=3t^2\frac{dt}{dx}-25\frac{dt}{dx}

    so at the point of interest either \frac{dt}{dx}=0 or t=\sqrt{\frac{25}{3}}.

    I will leave it to you to show that the first of these does not occur, and to
    find the value of x corresponding to the latter.

    RonL
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  5. #5
    Del
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    Thanks guys, but what about the area of the loop about the x-axis?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Del View Post
    x = 81 - t^2

    y = t^3 - 25t

    is symetric about the x-axis. (If t gives us the point (x,y), then -t will give (x,-y) ).
    Quote Originally Posted by Del View Post
    Thanks guys, but what about the area of the loop about the x-axis?
    The relation is symmetric about the x-axis. So we need to know for what t values the loop is traced. So we know that the curve must meet itself on the x-axis, so we look for where that happens:
    t^3 - 25t = 0

    t(t + 5)(t - 5) = 0

    So t = -5, 0, and 5. This corresponds to x values:
    x = 56, 81, and 56 respectively.

    So if we find the area between the curve and the x-axis over the [0, 5] t interval and double it, we will have the area of the loop.

    Does this help?

    -Dan
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  7. #7
    Del
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    This is helpful, but I keep getting a negative value when I integrate. I got -2333.333, but this is incorrect.
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  8. #8
    Behold, the power of SARDINES!
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    counter clock wise :)

    Quote Originally Posted by Del View Post
    This is helpful, but I keep getting a negative value when I integrate. I got -2333.333, but this is incorrect.

    The curve is being traversed in a clockwise manner NOT counter clockwise.

    Green's Theorem states that the area is

    A=\int_{c}xdy=-\int_{c}ydx if c is positively oriented

    so we also know that

    \int_{c}xdy=-\int_{-c}xdy where -c is going around the curve clockwise.

    I hope this helps.
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