# Thread: Tangent planes to a parametric surface

1. ## 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).

2. Originally Posted by intrepidy
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

$\displaystyle r_u \times r_v$

evaluated at (1,2)

taking the dervitvies we get

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

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

Now taking the cross product we get

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

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

$\displaystyle r(1,2)=(1+2,1^2,2^2)=(3,1,4)$

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

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

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

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

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

Yeah!!