Tangent Planes to Level Surfaces

# Tangent Planes to Level Surfaces

We have already looked at computing tangent planes to surfaces described by a two variable function $z = f(x, y)$ on the Finding a Tangent Plane on a Surface page. This method is convenient when the variable $z$ is isolated from the variables $x$ and $y$ as we can apply the following formula to obtain the tangent plane at a point $P(x_0, y_0, z_0)$:

(1)
\begin{align} \quad z - z_0 = f_x (x_0, y_0) (x - x_0) + f_y (x_0, y_0) (y - y_0) \quad \mathrm{or} \quad z - z_0 = \frac{\partial}{\partial x} f(x_0, y_0) (x - x_0) + \frac{\partial}{\partial y} f(x_0, y_0) (y - y_0) \end{align}

If $z$ is not isolated from the variables $x$ and $y$, then finding the tangent plane at a point can be messy with this method, so instead, we will view these surfaces as "level surfaces" to a three variable function.

Let $w = f(x, y, z)$ be a three variable real-valued function, and consider the equation $f(x, y, z) = k$ where $k \in \mathbb{R}^2$. The equation $f(x, y, z) = k$ represents a level surface corresponding to the real number $k$ (this is analogous to obtaining level curves for functions of two variables).

Let $P(x_0, y_0, z_0)$ be a point on this level surface, and let $C$ be any curve that passes through $P$ and that is on the surface $S$. This curve $C$ can be parameterized as a vector-valued function $\vec{r}(t) = (x(t), y(t), z(t))$. Let $P$ correspond to $t_0$, that is $\vec{r}(t_0) = (x_0, y_0, z_0)$.

Now since the curve $C$ is on the surface $S$, we must have that $f(x(t), y(t), z(t)) = k$ for any defined $t$ for the curve $C$. Suppose that $x = x(t)$, $y = y(t)$, and $z = z(t)$ are differentiable functions. If we apply the chain rule to both sides of the equation $f(x(t), y(t), z(t)) = k$ we get that:

(2)
\begin{align} \quad \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} + \frac{\partial w}{\partial z} \frac{dz}{dt} = 0 \\ \quad \left ( \frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z} \right ) \cdot \left (\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right ) = 0 \\ \quad \nabla f(x, y, z) \cdot \vec{r'}(t) = 0 \end{align}

Therefore, the gradient vector $\nabla f(x, y, z)$ and the derivative (tangent) vector $\vec{r}(t)$ are perpendicular since their dot product is equal to $0$. Therefore, at any $t_0$, we have that:

(3)
\begin{align} \quad \nabla f(x_0, y_0, z_0) \cdot \vec{r'}(t_0) = 0 \end{align}

Since $\nabla f(x_0, y_0, z_0)$ is perpendicular to the tangent vector at point $P$ (corresponding to $t_0$) and passes through the point $P(x_0, y_0, z_0)$, then we can obtain the equation of the tangent plane at $P$ as:

(4)
\begin{align} \quad \quad \nabla f(x_0, y_0, z_0) \cdot (x - x_0, y - y_0, z - z_0) = 0 \\ \quad \quad \left ( \frac{\partial}{\partial x} f(x_0, y_0, z_0), \frac{\partial}{\partial y} f(x_0, y_0, z_0), \frac{\partial}{\partial z} f(x_0, y_0, z_0) \right ) \cdot (x - x_0, y - y_0, z - z_0) = 0 \\ \quad \quad \frac{\partial}{\partial x} f(x_0, y_0, z_0) (x - x_0) + \frac{\partial}{\partial y} f(x_0, y_0, z_0) (y - y_0) + \frac{\partial}{\partial z} f(x_0, y_0, z_0) (z - z_0) = 0 \end{align}

## Example 1

Find the equation of the tangent plane to the sphere $x^2 + y^2 + z^2 = 16$ at $(1, 2, \sqrt{11})$.

We note that the sphere $x^2 + y^2 + z^2 = 16$ is the level surface of the function $w = f(x,y,z) = x^2 + y^2 + z^2$ when $k = 3$.

The partial derivatives of $f$ are $\frac{\partial w}{\partial x} = 2x$, $\frac{\partial w}{\partial y} = 2y$, and $\frac{\partial w}{\partial z} = 2z$. Therefore we have that $\frac{\partial}{\partial x} f(1, 2, \sqrt{11}) = 2$, $\frac{\partial}{\partial y} f(1, 2, \sqrt{11}) = 4$, and $\frac{\partial}{\partial z} f(1, 2, \sqrt{11}) = 2 \sqrt{11}$. Applying these values to the formula above and we get that the equation of the tangent plane is:

(5)
\begin{align} \quad 2(x - 1) + 4(y - 2) + 2 \sqrt{11}(z - \sqrt{11}) = 0 \end{align}