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Math Help - Points of intersection of 2 curves

  1. #1
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    Points of intersection of 2 curves

    Find the coordinates of the points of intersection of the curve with parametric equations: x=8sin^3\theta, y=8cos^3\theta, where 0 \leq \theta < \pi with the line y=\sqrt3x-8

    What I have tried is to substitute x and y in y=\sqrt3x-8 with those in the parametric equations, but this forms cos^3\theta=\sqrt3sin^3\theta-1 which seems hard to solve, unless a graphical calculator is used.
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Points of intersection of 2 curves

    Quote Originally Posted by Punch View Post
    Find the coordinates of the points of intersection of the curve with parametric equations: x=8sin^3\theta, y=8cos^3\theta, where 0 \leq \theta < \pi with the line y=\sqrt3x-8

    What I have tried is to substitute x and y in y=\sqrt3x-8 with those in the parametric equations, but this forms cos^3\theta=\sqrt3sin^3\theta-1 which seems hard to solve, unless a graphical calculator is used.
    cos^3\theta=\sqrt3sin^3\theta-1

     1 = \sqrt{3} \sin ^3 \theta - \cos ^3 \theta

    since i see \sqrt{3} I think about an angle which has a value include \sqrt{3}
    we have two angles  \frac{\pi}{3} and  \frac{\pi}{6}
    and I find that \frac{\pi}{3} fit for that
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  3. #3
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    Re: Points of intersection of 2 curves

    Quote Originally Posted by Amer View Post
    cos^3\theta=\sqrt3sin^3\theta-1

     1 = \sqrt{3} \sin ^3 \theta - \cos ^3 \theta

    since i see \sqrt{3} I think about an angle which has a value include \sqrt{3}
    we have two angles  \frac{\pi}{3} and  \frac{\pi}{6}
    and I find that \frac{\pi}{3} fit for that
    But when I sub them in x, the x-coordinates i get are 5.196 and 1 whereas the answer is 3.18. In addition, this is not a mathematical approach.
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  4. #4
    MHF Contributor Amer's Avatar
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    Re: Points of intersection of 2 curves

    Ok
    x = 8 \sin ^3 \theta \Rightarrow \sin \theta = \frac{\sqrt[3]{x}}{2}

    y = 8 \cos ^3 \theta \Rightarrow \cos \theta = \frac{\sqrt[3]{y}}{2}

     \sin ^2 \theta + \cos ^2 \theta = 1 = \frac{\sqrt[3]{x^2}}{4} + \frac{\sqrt[3]{y^2}}{4}

     0 \leq x \leq 8 \;\;\; and \;\;\; -8 < y \leq 8

    y  =\sqrt{3} x - 8 \Rightarrow y+8 = \sqrt{3} x\Rightarrow (y+8)^2 = 3x^2

    so now we have to curves see this link which solve these two curves and give x = 3\sqrt{3} which is equal to 5.196
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  5. #5
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    Re: Points of intersection of 2 curves

    Quote Originally Posted by Amer View Post
    Ok
    x = 8 \sin ^3 \theta \Rightarrow \sin \theta = \frac{\sqrt[3]{x}}{2}

    y = 8 \cos ^3 \theta \Rightarrow \cos \theta = \frac{\sqrt[3]{y}}{2}

     \sin ^2 \theta + \cos ^2 \theta = 1 = \frac{\sqrt[3]{x^2}}{4} + \frac{\sqrt[3]{y^2}}{4}

     0 \leq x \leq 8 \;\;\; and \;\;\; -8 < y \leq 8

    y  =\sqrt{3} x - 8 \Rightarrow y+8 = \sqrt{3} x\Rightarrow (y+8)^2 = 3x^2

    so now we have to curves see this link which solve these two curves and give x = 3\sqrt{3} which is equal to 5.196
    Thank you! However, the answer states x=3.18
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  6. #6
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    Re: Points of intersection of 2 curves

    I too get x=3\sqrt{3}.

    I began by eliminating the parameter θ.

    \frac{x}{8}=\sin^3\theta\:\therefore\:\theta=\sin^  {\small{-1}}\left(\frac{x^{\small{\frac{1}{3}}}}{2}\right)

    y=8\cos^3\left(\sin^{\small{-1}}\left(\frac{x^{\small{\frac{1}{3}}}}{2}\right) \right)

    y=8\left(1-\left(\frac{x^{\small{\frac{1}{3}}}}{2}\right)^2 \right)^{\small{\frac{3}{2}}}

    y=8\left(\frac{4-x^{\small{\frac{2}{3}}}}{4}\right)^{\small{\frac{3  }{2}}}=\left(4-x^{\small{\frac{2}{3}}}\right)^{\small{\frac{3}{2}  }}

    Equating this to the given line, we find:

    (1) \left(4-x^{\small{\frac{2}{3}}}\right)^{\small{\frac{3}{2}  }}=\sqrt{3}x-8

    Squaring both sides (and being aware that extraneous solutions may be introduced):

    \left(4-x^{\small{\frac{2}{3}}}\right)^3=\left(\sqrt{3}x-8\right)^2

    Expanding gives:

    64-48x^{\small{\frac{2}{3}}}+12x^{\small{\frac{4}{3}}  }-x^2=3x^2-16\sqrt{3}x+64

    Collecting like terms, rearranging and factoring yields:

    4x^{\small{\frac{2}{3}}}\left(x^{\small{\frac{4}{3  }}}-3x^{\small{\frac{2}{3}}}-4\sqrt{3}x^{\small{\frac{1}{3}}}+12\right)=0

    By substitution into (1), we see x = 0 is an extraneous solution, so we need only consider:

    x^{\small{\frac{4}{3}}}-3x^{\small{\frac{2}{3}}}-4\sqrt{3}x^{\small{\frac{1}{3}}}+12=0

    This equation has two real roots, one of which is extraneous. By substitution into (1), we find the only valid root is:

    x^{\small{\frac{1}{3}}}=3^{\small{\frac{1}{2}}}

    x=3^{\small{\frac{3}{2}}}=3\sqrt{3}\approx5.196
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  7. #7
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    Re: Points of intersection of 2 curves

    the graph shows three solutions ...

    x = 0 , x = 3.18 , x = 5.196
    Attached Thumbnails Attached Thumbnails Points of intersection of 2 curves-parametricintersection.jpg  
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Points of intersection of 2 curves

    I see now where I erred...

    I failed to account for:

    \theta=\pi-\sin^{\small{-1}}\left(\frac{x^{\small{\frac{1}{3}}}}{2}\right)

    which would lead to:

    (1) \pm\left(4-x^{\small{\frac{2}{3}}}\right)^{\small{\frac{3}{2}  }}=\sqrt{3}x-8

    where in addition, we also have the solutions:

    x=0

    x=\left(1-\frac{7}{\sqrt[3]{13+16\sqrt{2}}}+\sqrt[3]{13+16\sqrt{2}}\right)^{\small{\frac{3}{2}}} \approx3.1815

    My apologies for bungling my first post here!
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