Tangent Planes to Parametric Surfaces

Recall from the Parametric Surfaces page that we can parameterize surfaces (much like parameterizing curves) as a two variable vector-function $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $a ≤ u ≤ b$ and $c ≤ v ≤ d$.

Of course, it would be nice to be able to find the equations of tangent planes to specific points on a surface generated parametrically.

Consider a generic surface $\delta$ given parametrically by $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v)$, and let $P_0$ be a point on $\delta$ whose positive vector is $\vec{r}(u_0, v_0)$.

By holding $u = u_0$ constant then $\vec{r}(u_0, v)$ is a parameterization of a grid curve, call it $C_1$, that lies on $\delta$.

The tangent vector on $C_1$ is going to be the partial derivative of $\vec{r}(u, v)$ with respect to $v$:

(1)

\begin{align} \quad \frac{\partial \vec{r}}{\partial v} = \frac{\partial x}{\partial v} \vec{i} + \frac{\partial y}{\partial v} \vec{j} + \frac{\partial z}{\partial v} \vec{k} \end{align}

Therefore the tangent vector at $P_0$ on $C_1$ is going to be $\frac{\partial \vec{r}}{\partial v}$ evaluated at $(u_0, v_0)$, that is:

(2)

\begin{align} \quad \frac{\partial}{\partial v} \vec{r}(u_0, v_0) = \frac{\partial}{\partial v} x(u_0, v_0) \vec{i} + \frac{\partial}{\partial v} y(u_0, v_0) \vec{j} + \frac{\partial}{\partial v} z(u_0, v_0) \vec{k} \end{align}

Similarly, by holding $v = v_0$ constant the $\vec{r}(u, v_0)$ is a parameterization of a gird curve, call it $C_2$, that lies on $\delta$.

The tangent vector on $C_2$ is going to be the partial derivative of $\vec{r}(u, v)$ with respect to $u$:

(3)

\begin{align} \quad \frac{\partial \vec{r}}{\partial u} = \frac{\partial x}{\partial u} \vec{i} + \frac{\partial y}{\partial u} \vec{j} + \frac{\partial z}{\partial u} \vec{k} \end{align}

Therefore the tangent vector at $P_0$ on $C_2$ is going to be $\frac{\partial \vec{r}}{\partial u}$ evaluated at $(u_0, v_0)$, that is:

(4)

\begin{align} \quad \frac{\partial}{\partial u} \vec{r}(u_0, v_0) = \frac{\partial}{\partial u} x(u_0, v_0) \vec{i} + \frac{\partial}{\partial u} y(u_0, v_0) \vec{j} + \frac{\partial}{\partial u} z(u_0, v_0) \vec{k} \end{align}

Now the tangent plane at $P_0$ will be spanned by $\frac{\partial \vec{r}}{\partial u}$ and $\frac{\partial \vec{r}}{\partial v}$. To find a formula for this tangent plane, we need a vector that is perpendicular to both $\frac{\partial \vec{r}}{\partial u}$ and $\frac{\partial \vec{r}}{\partial v}$. We can obtain one with the vector cross product, that is:

(5)

\begin{align} \quad \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u}\\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v}\\ \end{bmatrix} = \frac{\partial (y, z)}{\partial (u, v)} \vec{i} + \frac{\partial (z, x)}{\partial (u, v)} \vec{j} + \frac{\partial (x, y)}{\partial (u, v)} \vec{k} \end{align}

Therefore the equation of the tangent plane at a point $P_0$ on the parametrically defined surface $\delta$ is:

(6)

\begin{align} \quad \frac{\partial (y, z)}{\partial (u, v)} \biggr \rvert_{(u_0, v_0)} (x - x(u_0, v_0)) + \frac{\partial (z, x)}{\partial (u, v)} \biggr \rvert_{(u_0, v_0)} (y - y(u_0, v_0)) + \frac{\partial (x, y)}{\partial (u, v)} \biggr \rvert_{(u_0, v_0)} (z - z(u_0, y_0)) = 0 \end{align}

Example 1

Find the equation of the tangent plane that passes through the point $(2, 1, 2)$ and lies on the surface $\delta$ given parametrically by $\vec{r}(u, v) = 2u^3 \vec{i} + uv^2 \vec{j} + 2v \vec{k}$.

The surface $\delta$ above is graphed below:

We first note that the point $(2, 1, 2)$ corresponds to $(u, v) = (1, 1)$ as you should verify. Now let's compute $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}$:

(7)

\begin{align} \quad \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{bmatrix} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ 6u^2 & v^2 & 0 \\ 0 & 2uv & 2 \end{bmatrix} \end{align}

From this we see that $\frac{\partial (y, z}{\partial (u, v)} = 2v^2$, $\frac{\partial (z, x)}{\partial (u, v)} = -12u^2$ and $\frac{\partial (x, y)}{\partial (u, v)} = 12u^3v$ and so:

(8)

\begin{align} \quad \frac{\partial (y, z)}{\partial (u, v)} \biggr \rvert_{(u_0, v_0)} (x - x(u_0, v_0)) + \frac{\partial (z, x)}{\partial (u, v)} \biggr \rvert_{(u_0, v_0)} (y - y(u_0, v_0)) + \frac{\partial (x, y)}{\partial (u, v)} \biggr \rvert_{(u_0, v_0)} (z - z(u_0, y_0)) = 0 \\ \quad 2v^2 \biggr \rvert_{(u, v) = (1, 1)} (x - 2) + -12u^2 \biggr \rvert_{(u, v) = (1, 1)} (y - 1) + 12u^3v \biggr \rvert_{(u, v) = (1, 1)} (z - 2) = 0 \\ \quad 2(x - 2) - 12(y - 1) + 12(z - 2) = 0 \end{align}