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Math Help - Finding Vertical Tangent of Parametric Curve

  1. #1
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    Finding Vertical Tangent of Parametric Curve

    I know from graphing this that there are 6 vertical tangents to the following curve, but I can only determine two of them:

    x = \cos(3\theta)    y = 3 sin \theta

    Clearly, \frac{dx}{d\theta} = -3\sin(3\theta), so I have to find the points at which this value is equal to zero. As I see it, this happens at each multiple of \pi.

    As I plug multiples of \pi into the original equation, I get back either (1,0) or (-1,0), so I know that these points are have vertical tangents. But, I can see from the graph that there are vertical tangents at each multiple of \frac {\pi}{3} rather than simply \pi.

    Obviously, I am doing something wrong, but I don't know what. Isn't the idea to simply look at where \frac{dx}{d\theta} = 0? What am I missing?

    Thanks!
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  2. #2
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    Re: Finding Vertical Tangent of Parametric Curve

    Hello, joatmon!

    Your algebra/trig is off . . .


    We have: . \begin{Bmatrix}x &= &\cos(3\theta) \\ y &=& 3\sin\theta \end{Bmatrix}

    We want: . \frac{dx}{d\theta} \,=\,0

    Hence: . -3\sin(3\theta) \:=\:0

    . . . . . . . . \sin(3\theta) \:=\:0

    . . . . . . . . . . . 3\theta \:=\:\pi n

    . . . . . . . . . . . . \theta \:=\:\frac{\pi}{3}n

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  3. #3
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    Re: Finding Vertical Tangent of Parametric Curve

    Thanks. I had a feeling that I was missing something obvious.
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