Normal Lines to Level Surfaces

Recall from the Normal Lines on a Surface page that if $z = f(x, y)$ is a two variable real-valued function that generates the smooth surface $S$, then the normal line at point $P(x_0, y_0, z_0)$ can be obtained by the following set of parametric equations:

(1)

\begin{align} \left\{\begin{matrix} x = x_0 + \frac{\partial}{\partial x} f(x_0, y_0) t\\ y = y_0 + \frac{\partial}{\partial y} f(x_0, y_0) t\\ z = z_0 - t \end{matrix}\right. \quad -\infty < t < \infty \end{align}

This formula works great if the variable $z$ is isolated, however, often times it is not, and using the formula above can get complicated in such cases.

Recall from the Finding a Tangent Plane on a Surface page that we developed a more general method for determining the equation of a tangent plane by considering the level surface of a three variable function $f(x, y, z) = k$, and we found that the gradient vector at point $P(x_0, y_0, z_0)$ was perpendicular to the tangent vector at $P$:

(2)

\begin{align} \quad \nabla f(x_0, y_0, z_0) \cdot (x - x_0, y - y_0, z - z_0) = 0 \end{align}

Since the gradient vector $\nabla f(x_0, y_0, z_0)$ at $P(x_0, y_0, z_0)$ is perpendicular to the tangent vector (the tangent vector being parallel to the tangent line) at $P$, then the gradient vector $\nabla f(x_0, y_0, z_0)$ is parallel to the normal line at $P$. Therefore, the equation of the normal line at $P(x_0, y_0, z_0)$ is given by the following set of parametric equations:

(3)

\begin{align} \left\{\begin{matrix} x = x_0 + \frac{\partial}{\partial x} f(x_0, y_0, z_0)t \\ y = y_0 + \frac{\partial}{\partial y} f(x_0, y_0, z_0)t \\ z = z_0 + \frac{\partial}{\partial z} f(x_0, y_0, z_0)t \end{matrix}\right. \quad -\infty < t < \infty \end{align}

Example 1

Find the equation of the normal line that passes through the point $\left (0, \frac{\pi}{2}, \pi \right)$ on the surface $x \sin (y) + z \cos (y) = 0$.

We note that $x \sin (y) + z \cos (y) = 0$ is the level surface to $w= f(x,y,z) = x \sin (y) + z \cos (y)$ at $k = 0$.

We need to find the partial derivatives of $f$. We have that $\frac{\partial w}{\partial x} = \sin y$, $\frac{\partial w}{\partial y} = x \cos y - z \sin y$, and $\frac{\partial w}{\partial z} = \cos y$. Therefore $\frac{\partial}{\partial x} f \left (0, \frac{\pi}{2}, \pi \right) = 1$, $\frac{\partial}{\partial y} f \left (0, \frac{\pi}{2}, \pi \right) - \pi$, and $\frac{\partial}{\partial z} f \left (0, \frac{\pi}{2}, \pi \right) = 0$. Therefore we have that the normal line at $\left (0, \frac{\pi}{2}, \pi \right)$ is given by:

(4)

\begin{align} \left\{\begin{matrix} x = 0 + t \\ y = \frac{\pi}{2} -\pi t \\ z = \pi + 0t \end{matrix}\right. \quad -\infty < t < \infty \end{align}