Normal Lines to Level Surfaces Examples 1

# Normal Lines to Level Surfaces Examples 1

Recall from the Normal Lines to Level Surfaces page that if we have a level surface $f(x, y, z) = k$ then we can determine the normal line that passes through the point $P(x_0, y_0, z_0)$ with the following parametric equations:

(1)
\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}

Let's look at some examples of computing normal lines to level surfaces.

## Example 1

**Find the equation of the normal line for the surface $y = \sin (x) \cos (z)$ that passes through the point $\left ( \frac{\pi}{6}, -\frac{1}{2}, \pi \right )$.**

We can rewrite the equation above as a function $f(x, y, z) = \sin (x) \cos (z) - y$. We first compute all first partial derivatives of the surface above as follows:

(2)
\begin{align} \quad \frac{\partial f}{\partial x} = \cos (x) \cos (z) \\ \quad \frac{\partial f}{\partial y} = -1 \\ \quad \frac{\partial f}{\partial z} = - \sin (x) \sin (z) \end{align}

Evaluating these partial derivatives at the point $\left ( \frac{\pi}{6}, -\frac{1}{2}, \pi \right )$ and we get that:

(3)
\begin{align} \quad \frac{\partial}{\partial x} f \left ( \frac{\pi}{6}, -\frac{1}{2}, \pi \right ) = -\frac{\sqrt{3}}{2} \\ \quad \frac{\partial}{\partial y} f \left ( \frac{\pi}{6}, -\frac{1}{2}, \pi \right ) = -1 \\ \quad \frac{\partial}{\partial z} f \left ( \frac{\pi}{6}, -\frac{1}{2}, \pi \right ) = 0 \end{align}

Thus applying the formula above and we get that the normal line that passes through the point $\left ( \frac{\pi}{6}, -\frac{1}{2}, \pi \right )$ is:

(4)
\begin{align} \left\{\begin{matrix} x = \frac{\pi}{6} - \frac{\sqrt{3}}{2} t \\ y = -\frac{1}{2} -t \\ z = \pi \end{matrix}\right. \quad -\infty < t < \infty \end{align}

## Example 1

**Find the equation of the normal line for the surface $x = y \cos (yz)$ that passes through the origin.**

We can first rewrite the equation of the surface above as $f(x, y, z) = y \cos (yz) - x$. We now find all of the first partial derivatives of $f$:

(5)
\begin{align} \quad \frac{\partial f}{\partial x} = -1 \\ \quad \frac{\partial f}{\partial y} = -yz \sin (yz) + \cos (yz) \\ \quad \frac{\partial f}{\partial z} = -y^2 \sin (yz) \end{align}

We now evaluate our partial derivatives at the origin to get that:

(6)
\begin{align} \quad \frac{\partial}{\partial x} f(0, 0, 0) = -1 \\ \quad \frac{\partial}{\partial y} f(0, 0, 0)= 1 \\ \quad \frac{\partial}{\partial z} f(0, 0, 0) = 0 \end{align}

Applying the formula above and we have that:

(7)
\begin{align} \left\{\begin{matrix} x = - t \\ y = t \\ z = 0 \end{matrix}\right. \quad -\infty < t < \infty \end{align}