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Math Help - Finding tangency points of a polar equation

  1. #1
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    Finding tangency points of a polar equation

    Question: Find all the points where the tangent line is horizontal and vertical for the polar equation r=1-2cos{\theta}

    I'm trying to find the tangencies by passing through the cartesian.

    So far I have this:
    x=cos\theta-2cos^2\theta
    y=sin\theta-2cos\theta sin\theta

    \frac{dx}{d\theta}=-sin{\theta}(1-4cos{\theta})

    \frac{dy}{d\theta}= cos -2cos2{\theta}

    I'm currently on the horizontal tangency but got stuck/wondering if there is another way:

     cos-2cos2{\theta}=0


    Thanks in advance
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  2. #2
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    That looks good to me. \frac{dy}{dx}= \frac{cos(\theta)- 2cos^2(\theta)}{-sin(\theta)(1- cos(\theta))}.

    The tangent will be horizontal when the numerator is 0:
    [tex]cos(\theta)- 2cos^2(\theta)= cos(\theta)(1- 2cos(
    \theta))= 0. That will be true wnen [tex]cos(\theta)= 0[tex] or when cos(\theta)= \frac{1}{2}.

    The tangent will be vertical when the denominator is 0:
    -sin(\theta)(1- cos(\theta))= 0 when sin(\theta)= 0 or cos(\theta)= 1.

    Since both numerator and denominator are 0 when cos(\theta)= 0, that is not necessarily a vertical or horizontal tangent. You will need to look more closely at that.
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