# 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)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)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)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)## 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)