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Math Help - Calculus with Parametric Curves

  1. #1
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    Calculus with Parametric Curves

    x=2cos(theta)
    y=sin(2theta)

    I have to find the coordinates of the horizontal and vertical tangents to this problem. I know that the top of the derivative must be 0 for horizontal and 0 on the bottom for the vertical but how do I go about finding the cartesian coordinates?
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  2. #2
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    Quote Originally Posted by aaronb View Post
    x=2cos(theta)
    y=sin(2theta)

    I have to find the coordinates of the horizontal and vertical tangents to this problem. I know that the top of the derivative must be 0 for horizontal and 0 on the bottom for the vertical but how do I go about finding the cartesian coordinates?
    Here is how you would find the horizontal tangents...

    \frac{d y}{d \theta}=2\cos(2\theta)

    So we set it equal to zero to get

    \cos(2\theta)=0 \iff 2\theta=\frac{\pi}{2}+n \pi where n is an integer

    \theta=\frac{\pi+2n\pi}{4}=\frac{(2n+1)\pi}{4}

    Now to get the cartesian coordinates plug the values of theta into your parametric equations so when n=0 we get pi/4 so

    x=2\cos(\frac{\pi}{4})=\sqrt{2}

    y=\sin(\frac{\pi}{2})=1

    so ONE of the ordered pairs is (\sqrt{2},1)

    This should get you started you will need to use more values of n to find the rest.

    Hint there are 4 horizontal tangents and 2 vertical good luck.
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  3. #3
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    I think I got this.

    I keep on plugging in integers for n until it exceeds the period of 2pi then I would be done.
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  4. #4
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    x = 2cosu
    y = sin2u = 2sinu cosu = sinu (2cosu)=x sinu
    sinu = y/x
    cosu = | sqrt[ 1- (sinu)^2 ] | = | sqrt[ 1 - y^2 / x^2 ] |

    x = 2| sqrt[ 1 - y^2 / x^2 ] | => x^2 = 2| sqrt(x^2 - y^2) |
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