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Math Help - Extrema polar equation

  1. #1
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    Extrema polar equation

    Given r = 4cos(2Ɵ) , 0 < Ɵ < π/4

    Calculate the point on this curve with the highest y-coordinate.

    How would you calculate this? I do not even know how to start this

    Thanks already!
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  2. #2
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    Re: Extrema polar equation

    Hey jones123.

    Hint: How do you represent the (x,y) co-ordinates given the polar representation?
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  3. #3
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    Re: Extrema polar equation

    you're going to have to determine where \frac{dy}{d\theta} = 0 in the given interval ... this will require doing the grunt work of finding both \frac{dy}{d\theta} and \frac{dr}{d\theta}.

    Derivatives of polar equations are a P-I-T-A.
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    Re: Extrema polar equation

    Quote Originally Posted by chiro View Post
    Hey jones123.

    Hint: How do you represent the (x,y) co-ordinates given the polar representation?
    So I know that the conversion to cartesian form gives: x^4 + 2(x^2)(y^2) + y^4 - 4x^2 + 4y^2 = 0
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  5. #5
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    Re: Extrema polar equation

    Good. Now, differentiate that with respect to x and set y' equal to 0.

    Another way to do this, without converting to Cartesian coordinates, is to maximise y= rsin(\theta) subject to the constraint r^2- 4cos(2\theta)= 0, perhaps using Lagrange multipliers.
    Last edited by HallsofIvy; December 22nd 2012 at 04:58 AM.
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  6. #6
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    Re: Extrema polar equation

    y = r\sin{\theta}

    \frac{dy}{d\theta} = r\cos{\theta} + \sin{\theta} \cdot \frac{dr}{d\theta}


    r^2 = 4\cos(2\theta)

    \frac{dr}{d\theta} = -\frac{4\sin(2\theta)}{r}


    \frac{dy}{d\theta} = r\cos{\theta} - \frac{8\sin^2{\theta}\cos{\theta}}{r}

    0 = r\cos{\theta} - \frac{8\sin^2{\theta}\cos{\theta}}{r}

    r\cos{\theta} = \frac{8\sin^2{\theta}\cos{\theta}}{r}

    r^2\cos{\theta} = 8\sin^2{\theta}\cos{\theta}

    4\cos(2\theta) = 8\sin^2{\theta}

    4(1-2\sin^2{\theta}) = 8\sin^2{\theta}

    \sin^2{\theta} = \frac{1}{4}

    \sin{\theta} = \frac{1}{2}

    \theta = \frac{\pi}{6}
    Attached Thumbnails Attached Thumbnails Extrema polar equation-polarmax.png  
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  7. #7
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    Re: Extrema polar equation

    Quote Originally Posted by skeeter View Post
    r^2 = 4\cos(2\theta)

    \frac{dr}{d\theta} = -\frac{4\sin(2\theta)}{r}


    \frac{dy}{d\theta} = r\cos{\theta} - \frac{8\sin^2{\theta}\cos{\theta}}{r}
    Thanks a lot! Just one more question: I do not understand the part of your calculation i have quoted...

    If r = 4cos(2Ɵ) than r = sqrt(4cos(2Ɵ)), shouldn't the derivative dr/dƟ be - 2sin(2Ɵ) / sqrt(cos(2Ɵ)) ?
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    Re: Extrema polar equation

    \frac{d}{d\theta}\left[r^2 = 4\cos(2\theta)\right]

    2r \cdot \frac{dr}{d\theta} = -8\sin(2\theta)

    \frac{dr}{d\theta} = \frac{-8\sin(2\theta)}{2r} = - \frac{-4\sin(2\theta)}{r}


    If r = 4cos(2Ɵ) than r = sqrt(4cos(2Ɵ)), shouldn't the derivative dr/dƟ be - 8sin(2Ɵ) / 2sqrt(cos(2Ɵ)) ?
    note that sqrt(cos(2Ɵ)) = r
    Last edited by skeeter; December 23rd 2012 at 06:18 AM.
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