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Math Help - Tangent planes to a parametric surface

  1. #1
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    Tangent planes to a parametric surface

    How would you do the following problem???

    For the parametric surface r(u,v) = <u + v, u^2, v^2>, find an equation of the tangent plane at r(1,2).
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  2. #2
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    Quote Originally Posted by intrepidy View Post
    How would you do the following problem???

    For the parametric surface r(u,v) = <u + v, u^2, v^2>, find an equation of the tangent plane at r(1,2).
    to find the normal vector we need

    r_u \times r_v

    evaluated at (1,2)

    taking the dervitvies we get

    r_u(u,v)=<1,2u,0> \mbox{ eval at (1,2) } r_u(1,2)=<1,2,0>

    r_v(u,v)=<1,0,2v> \mbox{ eval at (1,2) } r_v(1,2)=<1,0,4>

    Now taking the cross product we get

    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
1 && 2 && 0 \\<br />
1 && 0 && 4 \\<br />
\end{vmatrix}= (8-0) \vec i -(4-0) \vec j +(0-2) \vec k =<br />
 8 \vec i -4 \vec j -2 \vec k = < 8,-4,- 2><br />

    so now we need to know the point for the tangent plane

    r(1,2)=(1+2,1^2,2^2)=(3,1,4)

    now we need a vector in the plane containing (x,y,z) so

    v_1=<x-3,y-1,z-4>

    Now if we dot the normal vector with the above generic vector they will equal zero. (why?)

    <(x-3),(y-1),(z-4)> \cdot <8,-4,-2>=0

    8(x-3)-4(y-1)-2(z-4)=0 \iff 8x-4y-2z-12 \iff 8x-4y-2z=12

    Yeah!!
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