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Math Help - Polar coordinates.

  1. #1
    Super Member Showcase_22's Avatar
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    Polar coordinates.

    Let C be the curve in the plane defined by r=\sin \theta, 0 \leq \theta \leq 2 \pi.
    a). Draw a sketch of C.
    My sketch looks like a figure of 8.

    b). Find the co-ordinates of the four points at which the tangent to C is vertical.

    Hint: The formula \frac{ dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d \theta}} may be helpful.
    I know that:

    x=r \cos \theta \Rightarrow \ x=\sin^2 \theta \cos \theta

    y=r \sin \theta \Rightarrow \ y=\sin^3 \theta

    Therefore:

    \frac{dx}{d  \theta}=2 \sin \theta \ cos^2 \theta -\sin^3 \theta

    \frac{dy}{d \theta}=3 \sin^2 \theta

    Hence:

    \frac{dy}{dx}=\frac{ 3 \sin^2 \theta}{ 2 \sin \theta \cos^2 \theta- \sin^3 \theta}=\frac{3 \sin \theta}{2 \cos^2 \theta-\sin^2 \theta}

    I want when \frac{dy}{dx}= \infty. This will be when the denominator of my expression is equal to 0.

    2 \cos^2 \theta=1- \cos^2 \theta

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

    3 \cos ^2 \theta=1 \Rightarrow \ \boxed{\cos \theta= \pm \frac{1}{\sqrt{3}}}

    Since I have an expression for \cos \theta, it is useful to write x and y in terms of \cos \theta.

    x=r \cos \theta \Rightarrow \ x=\sin^2 \theta \cos \theta \Rightarrow \ x=(1-\cos^2 \theta) \cos \theta

    y=r \sin \theta \Rightarrow \ y=\sin^3 \theta=(1-\cos^2 \theta)\sqrt{1-\cos^2 \theta}

    Therefore when \cos \theta=\frac{1}{\sqrt{3}} we have:

    x=\left( 1- \frac{1}{3} \right) \frac{1}{\sqrt{3}}=\left( \frac{2}{3} \right) \frac{1}{\sqrt{3}}=\frac{2}{3 \sqrt{3}}

    y=\left( 1-\frac{1}{3} \right) \sqrt{1-\frac{1}{3}}=\frac{2}{3}\sqrt{\frac{2}{3}}=\pm \left( \frac{2}{3} \right)^{\frac{3}{2}}

    The first two points are \left( \frac{2}{3 \sqrt{3}}, \left( \frac{2}{3} \right)^{\frac{3}{2}} \right) and \left( \frac{2}{3 \sqrt{3}}, -\left( \frac{2}{3} \right)^{\frac{3}{2}} \right).

    It's also apparent that the next x coordinate is -\frac{2}{3 \sqrt{3}} with the same y coordinates as before.

    The other two points are \left( -\frac{2}{3 \sqrt{3}}, \left( \frac{2}{3} \right)^{\frac{3}{2}} \right) and \left( -\frac{2}{3 \sqrt{3}}, -\left( \frac{2}{3} \right)^{\frac{3}{2}} \right).

    The four points are (listed together for clarity) \left( -\frac{2}{3 \sqrt{3}}, \left( \frac{2}{3} \right)^{\frac{3}{2}} \right), \left( -\frac{2}{3 \sqrt{3}}, -\left( \frac{2}{3} \right)^{\frac{3}{2}} \right), \left( \frac{2}{3 \sqrt{3}}, \left( \frac{2}{3} \right)^{\frac{3}{2}} \right) and \left( \frac{2}{3 \sqrt{3}}, -\left( \frac{2}{3} \right)^{\frac{3}{2}} \right).

    I'm pretty sure this is right. I would just like someone to confirm my answer.
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    y=r \sin \theta \Rightarrow \ y=\sin^3 \theta

    Therefore:

    \frac{dy}{d \theta}=3 \sin^2 \theta
    Unfortunately not

    Lucky you are : it does not change anything and your results are correct !
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  3. #3
    Super Member Showcase_22's Avatar
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    I mean \frac{dy}{d \theta}=3\sin^2 \theta cos \theta!
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  4. #4
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    Your results are correct anyway !
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  5. #5
    Super Member Showcase_22's Avatar
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    yay!

    MATH WIN!

    Thankyou kindly running gag.
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