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\centerline
{
\bf Dr. Z's Math251 Handout \#16.4  (2nd ed.)
[Parametrized Surfaces]
}

By Doron Zeilberger

{\bf Problem Type 16.4a}: 
Find an equation of the tangent plane to the given
parametric surface at the specified point.
$$
x=x(u,v) \quad , \quad y=y(u,v) \quad , \quad z=z(u,v) \quad ; 
\quad u=1, v=1 \quad .
$$

{\bf Example Problem 16.4a}: 
Find an equation of the tangent plane to the given
parametric surface at the specified point.
$$
x=u^2 \quad , \quad y=v^2 \quad , \quad z=uv \quad ; 
\quad u=1, v=1 \quad .
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Set-up 
$$
{\bf r}=x \, {\bf i}+ y \, {\bf j}+ z \, {\bf k} \quad ,
$$
and compute the partial derivatives w.r.t. $u$ and
w.r.t. $v$ :
$$
{\bf r}_u=x_u \, {\bf i}+ y_u \, {\bf j}+ z_u \, {\bf k} \quad ,
$$
$$
{\bf r}_v=x_v \, {\bf i}+ y_v \, {\bf j}+ z_v \, {\bf k} \quad .
$$
Then {\bf plug-in} the given values of $u$ and $v$.
&
{\bf 1.} In this problem
$$
{\bf r}=u^2 \, {\bf i}+ v^2 \, {\bf j}+ uv \, {\bf k} \quad ,
$$
We have
$$
{\bf r}_u=2u \, {\bf i}+ 0 \, {\bf j}+ v \, {\bf k} \quad ,
$$
$$
{\bf r}_v=0\, {\bf i}+ 2v \, {\bf j}+ u \, {\bf k} \quad .
$$
Now {\bf plug-in} $u=1, v=1$ to get {\bf numerical vectors}.
$$
{\bf r}_u(1,1)=2 \, {\bf i}+ 0 \, {\bf j}+ 1 \, {\bf k} \quad ,
$$
$$
{\bf r}_v(1,1)=0\, {\bf i}+ 2 \, {\bf j}+ 1 \, {\bf k} \quad .
$$
So at this point, $r_u=\langle\, 2,0,1 \,\rangle$,
$r_v=\langle\, 0,2,1 \,\rangle$.
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{\bf 2.}  Find the cross-product ${\bf r}_u \times {\bf r}_v$.
This is a vector {\bf normal} to the tangent plane.
&
{\bf 2.} 
$$
{\bf r}_u \times {\bf r}_v =
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
2 & 0& 1\cr
0 & 2 & 1
} \right \vert \quad
$$
$$
=-2 \, {\bf i} -2 \, {\bf j}+ 4 \, {\bf k} \quad .
$$
Or in $\langle \rangle$ notation
$$
{\bf N}= \langle -2 , -2 ,4 \rangle \quad .
$$

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{\bf 3.} Find the {\bf point} $(x_0,y_0,z_0)$
by plugging into $x,y,z$ the specific values of $u$ and $v$
given in the problem
The desired equation of the tangent plane is
$$
a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \quad .
$$
where $N=\langle a,b,c \rangle$ and the point is $(x_0,y_0,z_0)$.

&
{\bf 3.} The point is $(1^2,1^2, 1 \cdot 1 )=(1,1,1)$.
The desired equation of the tangent plane is
$$
(-2)(x-1)-2(y-1)+4(z-1)=0 \quad .
$$
Or, in expanded form
$$
-2x-2y+4z=0 \quad .
$$
Dividing by $-2$ (to make it nicer), we get:

{\bf Ans.:} $x+y-2z=0$.


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{\bf A Problem from a previous Final}:
Find an equation for the tangent plane to the parametric surface
$$
x=u^2 \quad , \quad y=u+v \quad, \quad z=v^2 \quad ,
$$
at the point $(1,2,1)$. Simplify as much as you can!

{\bf Ans.}: $x-2y+z=-2$.

\end
