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Math Help - Vector Function that represents the Curve

  1. #1
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    Vector Function that represents the Curve

    I just need to check if my work has been done accordingly.

    Q: Find the vector function that represents the curve of the intersection of the 2 surfaces.
    The cylinder x^2 + y^2 = 4 and z= xy

    Solution:

    x^2 + y^2 = 4 \rightarrow x+y = 2

    x = cos(t) y=sin(t) 0 \lesseq t \lesseq \frac{pi}{2}
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  2. #2
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    Quote Originally Posted by lllll View Post
    I just need to check if my work has been done accordingly.

    Q: Find the vector function that represents the curve of the intersection of the 2 surfaces.
    The cylinder x^2 + y^2 = 4 and z= xy

    Solution:

    x^2 + y^2 = 4 \rightarrow x+y = 2

    x = \cos (t) y= \sin (t) 0 \leq t \leq \frac{pi}{2}
    I think you want

    x = {\color{red}2} \cos (t)

     y= {\color{red}2} \sin (t)

    {\color{red}z = 4 \sin (t) \, \cos (t) = 2 \sin (2t)}

    for  0 \leq t \leq {\color{red}2 \pi}
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    x^2 + y^2 = 4 \rightarrow x+y = 2

    x = {\color{red}2}cos(t) \ \ y={\color{red}2}sin(t) 0 \leq t \leq {\color{red}2\pi}

    z = xy = {\color{red}4}cos(t)sin(t)

    \therefore x = 2cos(t) \ \ y=2sin(t) \ \ z = 4cos(t)sin(t) 0 \leq t \leq {\color{red}2\pi}

    r(t) = 2cos(t) i + 2sin(t) j + 4cos(t)sin(t) k

    r(t) = 2cos(t) i + 2sin(t) j + 2sin(2t) k

    So here it is completed with the changes is this correct?. And sorry for the double post.
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    Quote Originally Posted by lllll View Post
    x^2 + y^2 = 4 \rightarrow x+y = 2

    x = {\color{red}2}cos(t) \ \ y={\color{red}2}sin(t) 0 \leq t \leq \frac{\pi}{2}

    z = xy = {\color{red}4}cos(t)sin(t)

    \therefore x = 2cos(t) \ \ y=2sin(t) \ \ z = 4cos(t)sin(t) 0 \leq t \leq \frac{\pi}{2}

    r(t) = 2cos(t) i + 2sin(t) j + 4cos(t)sin(t) k

    r(t) = 2cos(t) i + 2sin(t) j + 2sin(2t) k

    So here it is completed with the changes is this correct?. And sorry for the double post.
    Yes. By the way, I made an edit on the value of the parameter t .....

    If you consider the level curves of the surface z = xy you'll quickly see that it cuts the cylinder in all quadrants .....
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    Quote Originally Posted by mr fantastic View Post
    Yes. By the way, I made an edit on the value of the parameter t .....

    If you consider the level curves of the surface z = xy you'll quickly see that it cuts the cylinder in all quadrants .....
    *Ahem* ....But not on the axes .... The values {\color{red}t = 0, ~ \frac{\pi}{2}, ~ \frac{3\pi}{2}, ~ 2 \pi} \, need to be excluded.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    ... you'll quickly see that it cuts the cylinder in all quadrants .....
    It looks like that:
    Attached Thumbnails Attached Thumbnails Vector Function that represents the Curve-flaechendurchdringung.gif  
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  7. #7
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    Quote Originally Posted by earboth View Post
    It looks like that:
    Holy cow!
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