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Math Help - Maximum Y-Coordinate Values Of A Cardioid

  1. #1
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    Maximum Y-Coordinate Values Of A Cardioid

    Algebraically determine the maximum value of the y-coordinate of points on the cardioid r = 1 + cos \theta for 0 θ π.

    I used the formula:

    \frac{dy}{dx}=\frac{\frac{dr}{d \theta} \sin \theta +r \cos \theta}{\frac{dr}{d \theta} \cos \theta-r \sin \theta} , which yielded \frac{-sin\theta(sin\theta+cos\theta+cos^{2}\theta)}{-sin\theta(cos\theta-sin\theta-sin\theta cos\theta)} .

    If I solve -sin\theta(sin\theta+cos\theta+cos^{2}\theta), setting it to zero, it will give the value of theta at which horizontal tangents exist. The problem is that I don't know how to solve it.

    After graphing it, I know that the max value occurs when \theta = \frac{\pi}{3} and r=1.5, but I'm not sure how to get to those values.
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  2. #2
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    Re: Maximum Y-Coordinate Values Of A Cardioid

    All you need is that

    y=r\cos(\theta)=(1+\cos(\theta))\sin(\theta)=\sin(  \theta)+\sin(\theta)\cos(\theta)

    Now we have

    \frac{dy}{d\theta}=\cos(\theta)+\cos^2(\theta)-\sin^2(\theta)=2\cos^2(\theta)+\cos(\theta)-1

    Setting equal to zero and factoring gives

    (2\cos(\theta)-1)(\cos(\theta)+1)=0

    This will give you both the max and min values.
    Thanks from Algebrah
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  3. #3
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    Re: Maximum Y-Coordinate Values Of A Cardioid

    Wow, that's so much better than my inchoate solution.

    Thanks a lot!
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  4. #4
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    Re: Maximum Y-Coordinate Values Of A Cardioid

    Hello, Algebrah!

    Determine the maximum value of the y-coordinate of points on the cardioid: r \:=\: 1 + \cos \theta\;\text{ for }0\,\le\,\theta\,\le\,\pi

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    We have: . r' \:=\:\text{-}\sin\theta

    \text{Formula: }\:\frac{dy}{dx} \;=\;\frac{r\cos\theta + r'\sin\theta}{\text{-}r\sin\theta + r'\cos\theta}

    \text{Then: }\:\frac{dy}{dx} \;=\;\frac{(1+\cos\theta)\cos\theta + (\text{-}\sin\theta)\sin\theta}{\text{-}(1+\cos\theta)\sin\theta + (\text{-}\sin\theta)\cos\theta} \;=\;\frac{\cos\theta + \cos^2\theta - \sin^2\theta}{\text{-}\sin\theta - 2\sin\theta\cos\theta}


    We have: . \cos\theta + \cos^2\theta - \sin^2\theta \;=\;0

    . . . . \cos\theta + \cos^2\theta - (1 - \cos^2\theta) \;=\;0

    . . . . . \cos\theta + \cos^2\theta - 1 + \cos^2\theta \;=\;0

    . . . . . . . . . . 2\cos^2\theta + \cos\theta - 1 \;=\;0

    n . . . . . . (\cos\theta + 1)(2\cos\theta - 1) \;=\;0

    . . \begin{array}{cccccccccccc}\cos\theta + 1 \:=\: 0 &\Rightarrow& \cos\theta \:=\: \text{-}1 & \Rightarrow & \theta \:=\: \pi & {\color{red}**}\\ \\[-3mm] 2\cos\theta - 1 \:=\: 0 & \Rightarrow & \cos\theta \:=\: \frac{1}{2} &\Rightarrow & \theta \:=\: \frac{\pi}{3} \end{array}

    ** There is a horizontal tangent at \theta = \pi
    . . .but it gives a minimum value for y.

    Hence, the maximum y occurs at \theta \,=\,\tfrac{\pi}{3}

    \text{Therefore: }\:y \;=\;r\sin\theta \;=\;\left(1 + \cos\tfrac{\pi}{3}\right)\sin\tfrac{\pi}{3}\;=\; \left(\frac{3}{2}\right)\left(\frac{\sqrt{3}}{2} \right)\;=\;\frac{3\sqrt{3}}{4}
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