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  
\[(u,v)\]
  so that adistinct  
\[(u,v)\]
  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
\[\begin{equation} \begin{aligned} \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} &=( \frac{\partial x}{\partial u} \mathbf{i} + \frac{\partial y}{\partial u} \mathbf{j} + \frac{\partial z}{\partial u} \mathbf{k}) \times (\frac{\partial x}{\partial v} \mathbf{i} + \frac{\partial y}{\partial v} \mathbf{j} + \frac{\partial z}{\partial v} \mathbf{k}) \\ &=(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial z}{\partial u} \frac{\partial y}{\partial v}) \mathbf{i} + (\frac{\partial z}{\partial u} \frac{\partial x}{\partial v} - \frac{\partial x}{\partial u} \frac{\partial z}{\partial v}) \mathbf{j} + (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v}) \mathbf{k} \end{aligned} \end{equation}\]

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

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
\[(x-9) \times 3 + (y-2) \times -18 + (z-1) \times 6=0 \rightarrow 9x-18y+6z=-3\]

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