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Math Help - Parametric Surfaces

  1. #1
    No one in Particular VonNemo19's Avatar
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    Parametric Surfaces

    Find the rectangular equation for the surface given by r(u,v)=\frac{v}{2}i+uj+vk by eliminating the paranmeter.

    I'm having a bad day. I can see that 2x=z. But, that's as far as I get. Any Help?
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  2. #2
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    Quote Originally Posted by VonNemo19 View Post
    Find the rectangular equation for the surface given by r(u,v)=\frac{v}{2}i+uj+vk by eliminating the parameter.

    I'm having a bad day. I can see that 2x=z. But, that's as far as I get. Any Help?

    As it turns out, 2x=z is the equation of the plane.

    The position vector, \displaystyle \vec r(u,v)=\frac{v}{2}\,\hat i+u\hat j+v\hat k, gives the coordinates of a point at \displaystyle (x,\ y,\ z) \displaystyle =\left(\frac{v}{2},\ u,\ v\right)\,.

    \displaystyle \vec r(0,0)=0\hat i+0\hat j+0\hat k, so the surface passes through the origin, (0, 0, 0).

    Define vector, \displaystyle \vec A \displaystyle =\vec{r}(2,\ 0)-\vec{r}(0,\ 0)=1\hat i+0\hat j+2\hat k.

    Define vector, \displaystyle \vec B \displaystyle =\vec{r}(0,\ 1)-\vec{r}(0,\ 0)=0\hat i+1\hat j+0\hat k.

    Notice that any vector of the form \displaystyle \vec r(u,v)=\frac{v}{2}\,\hat i+u\hat j+v\hat k, may be written as a linear

    combination of the constant vectors, \displaystyle \vec A and \displaystyle \vec B.

    Therefore, the surface specified by \displaystyle \vec r(u,v) is a plane.

    Vectors, \displaystyle \vec A and \displaystyle \vec B lie in the plane, so their vector product (cross product) is perpendicular to the plane.

    \displaystyle \vec A\times\vec B = \begin{vmatrix}<br />
\ \hat i&\hat j&\hat k\ \\<br />
\ 1&0&2\ \\ <br />
\ 0&1&0\ \end{vmatrix}=-2\,\hat i+0\,\hat j+1\,\hat k

    So, the equation of the plane is \displaystyle -2(x-0)+0(y-0)+1(z-0)=0,

    which may be written as \displaystyle 2x-z=0\,.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Hey man, thanks a lot.
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