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Thread: [SOLVED] Horizontal tangent and parametric equation?

  1. #1
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    [SOLVED] Horizontal tangent and parametric equation?

    x=25-t^2
    y=t^3-25t

    I can't figure out why the answer is not the square root of 25/3.

    I found dy/dx and used the denominator to find the vertical tangent but it isn't working with the numerator.

    It says to find the "x" value (for the vertical tangent it asked for the "t" value) that makes the tangent horizontal.
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  2. #2
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    Horizontal tangent occurs when \frac{dy}{dx} = 0.
    So: \frac{dy}{dx} = \frac{\displaystyle \ \frac{dy}{dt} \ }{\displaystyle \ \frac{dx}{dt} \ } = \frac{\left(t^3 - 25t\right)'}{\left(25-t^2\right)'}
    Once you differentiate and set the expression equal to 0, you should get: {\color{red}t} = \pm \sqrt{\frac{25}{3}}
    Now substitute it into: x = 25-{\color{red}t}^2
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  3. #3
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    Ok I knew I was close to the answer, I just needed that last step. Thanks.

    The problem also says that the curve makes a loop along the x-axis, and asks for the area inside the loop. How do I go about finding this? This seems slightly different than the other area problems we have been given.
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