## Constructing a Tangent Plane to a Surface Given in Parametric Coordinates

The equation of the plane to a surface at a point with position vector
$\mathbf{r_0}$
is
$(\mathbf{r} - \mathbf{r_0}) \cdot \mathbf{n} -=0$
.
For a surface
$S$
given in parametric coordinates
$(u,v)$
$(u,v)$
returns distinct points, apoint
$P$
in the surface has position vector
$\mathbf{r} = x(u,v) \mathbf{i}+ y(u,v) \mathbf{j} +z(u,v) \mathbf{j}$
, we can construct a normal by taking the cross product of the partial derivatives with respect to
$u,v$
.
Taking the partial derivatives with respect to
$u$
and
$v$
respectively,
$\frac{\partial \mathbf{r}}{\partial u} = \frac{\partial x}{\partial u} \mathbf{i} + \frac{\partial y}{\partial u} \mathbf{j} + \frac{\partial z}{\partial u} \mathbf{k}$

$\frac{\partial \mathbf{r}}{\partial v} = \frac{\partial x}{\partial v} \mathbf{i} + \frac{\partial y}{\partial v} \mathbf{j} + \frac{\partial z}{\partial v} \mathbf{k}$

A normal is then
\begin{aligned} \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} &=( \frac{\partial x}{\partial u} \mathbf{i} + \frac{\partial y}{\partial u} \mathbf{j} + \frac{\partial z}{\partial u} \mathbf{k}) \times (\frac{\partial x}{\partial v} \mathbf{i} + \frac{\partial y}{\partial v} \mathbf{j} + \frac{\partial z}{\partial v} \mathbf{k}) \\ &=(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial z}{\partial u} \frac{\partial y}{\partial v}) \mathbf{i} + (\frac{\partial z}{\partial u} \frac{\partial x}{\partial v} - \frac{\partial x}{\partial u} \frac{\partial z}{\partial v}) \mathbf{j} + (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v}) \mathbf{k} \end{aligned}

Example: If
$\mathbf{r} = (u+v)^2 \mathbf{i} + uv \mathbf{j} + v^3 \mathbf{j}$
,
$\frac{\partial \mathbf{r}}{\partial u}= 2(u+v) \mathbf{i} + v \mathbf{j}$

$\frac{\partial \mathbf{r}}{\partial v}= 2(u+v) \mathbf{i} + u \mathbf{j} + 3v^2 \mathbf{k}$

Then
\begin{aligned} \mathbf{n} &= \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \\ &=(2(u+v) \mathbf{i} + v \mathbf{j}) \times ( 2(u+v) \mathbf{i} + u \mathbf{j} + 3v^2 \mathbf{k}) \\ &=(v \times 3v^2 - u \times 0) \mathbf{i} + (0 \times 2(u+v) -2(u+v) \times 3v^2) \mathbf{j} + (2(u+v) \times u - 2(u+v) \times v) \mathbf{k} \\ &= 3v^3 \mathbf{i} 6v^2 (u+v) \mathbf{j} +2(u^2 - v^2 ) \mathbf{k} \end{aligned}

At the point where
$u=2,v=1$
, we have
$\mathbf{r_0} = 9 \mathbf{i} +2 \mathbf{j} + \mathbf{k}$
and
$\mathbf{n}= 3 \mathbf{i} - 18 \mathbf{j} + 6 \mathbf{k}$
The equation of the plane is then
$(\mathbf{r} -(9 \mathbf{i} + 2 \mathbf{j} + \mathbf{k}) \cdot ((3 \mathbf{i}- 18 \mathbf{j} + 6 \mathbf{k}) ($

Put
$\mathbf{r_0} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}$
we have
$(x-9) \times 3 + (y-2) \times -18 + (z-1) \times 6=0 \rightarrow 9x-18y+6z=-3$