Normal Lines to Level Surfaces Examples 2

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 parametric equations for the normal line to the sideways paraboloid $y = x^2 + z^2$ at the point $(1, 2, 1)$.

Let $f(x, y, z) = y - x^2 - z^2 = 0$. The partial derivatives of this function are:

(2)

\begin{align} \quad \frac{\partial f}{\partial x} = -2x \quad , \quad \frac{\partial f}{\partial y} = 1 \quad , \quad \frac{\partial f}{\partial z} = -2z \end{align}

The partial derivatives of $f$ evaluated at the point $(1, 2, 1)$ are:

(3)

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

Therefore the normal line to the paraboloid $y = x^2 + z^2$ at the point $(1, 2, 1)$ is given parametrically by:

(4)

\begin{align} \left\{\begin{matrix} x = 1 -2t \\ y = 2 + t \\ z = 1 -2t \end{matrix}\right. \quad -\infty < t < \infty \end{align}

Example 2

Find parametric equations for the normal line to the surface $y^3 = (x - z)^2$ at the point $(2, 1, 1)$.

Let $f(x, y, z) = y^3 - (x - z)^2 = y^3 - (x^2 - 2xz + z^2) = 0$. The partial derivatives of this function are:

(5)

\begin{align} \quad \frac{\partial f}{\partial x} = -2x + 2z \quad , \quad \frac{\partial f}{\partial y} = 3y^2 \quad , \quad \frac{\partial f}{\partial z} = 2x - 2z \end{align}

The partial derivatives of $f$ evaluated at the point $(2, 1, 1)$ are:

(6)

\begin{align} \quad \frac{\partial}{\partial x} f(2, 1, 1) = -4 + 2 = -2 \quad , \quad \frac{\partial}{\partial y} f(2, 1, 1) = 3 \quad , \quad \frac{\partial}{\partial z} f(2, 1, 1) = 4 - 2 = 2 \end{align}

Therefore the normal line to the surface $y^3 = (x - z)^2$ at the point $(2, 1, 1)$ is given parametrically by:

(7)

\begin{align} \left\{\begin{matrix} x = 2 + -2t \\ y = 1 + 3t \\ z = 1 + 2t \end{matrix}\right. \quad -\infty < t < \infty \end{align}