Normal Lines on a Surface

Normal Lines on a Surface

We have just looked at Tangent Planes to Surfaces to a point on a surface of a two variable real-valued function $z = f(x, y)$. We will now look at what a normal line on a surface $S$ at point $P$ is.

First, let $z = f(x, y)$ be a two variable real-valued function that generates the smooth surface $S$, and let $(x_0, y_0) \in D(f)$. Let $P$ be the point that lies on the surface $S$ with coordinates $P(x_0, y_0, z_0)$, and let $\Pi$ be the tangent plane to $S$ at $P$.

Recall that the vertical plane $y = y_0$ intersects $S$ forming a curve of intersection $C_1$. The tangent line $T_1$, of this curve at point $P$ has slope $\frac{\partial}{\partial x} f(x_0, y_0)$. Also recall that the tangent plane $\Pi$ contains $T_1$ by definition, and more precisely, the tangent plane $\Pi$ intersects the vertical plane $y = y_0$ to form $T_1$. Therefore $T_1$ is parallel to the vector we define as $\vec{T_1} = \vec{i} + 0 \vec{j} + \frac{\partial}{\partial x} f(x_0, y_0) \vec{k}$.

Similarly, the vertical plane $x = x_0$ intersects $S$ forming a curve of intersection $C_2$. The tangent line $T_2$, of this curve at point $P$ has sloe $\frac{\partial}{\partial y} f(x_0, y_0)$. Similarly, the tangent plane $\Pi$ contains $T_2$ by definition, and more precisely, the tangent plane $\Pi$ intersects the vertical plane $x = x_0$ to form $T_2$. Therefore $T_2$ is parallel to the vector we define as $\vec{T_2} = 0\vec{i} + \vec{j} + \frac{\partial}{\partial y} f(x_0, y_0) \vec{k}$.

Since $\vec{T_1}$ and $\vec{T_2}$ are parallel to both $T_1$ and $T_2$ respectively, then the cross product $\vec{T_2} \times \vec{T_1}$ produces a vector that is perpendicular to both of these tangent lines. Since $\Pi$ contains both of these tangent lines, it follows that the cross product $\vec{T_2} \times \vec{T_1}$ produces a vector that is perpendicular to the tangent plane $\Pi$, or rather, produces a normal vector for $\Pi$.

(1)
\begin{align} \quad \quad \vec{T_2} \times \vec{T_1} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & 1 & \frac{\partial}{\partial y} f(x_0, y_0)\\ 1 & 0 & \frac{\partial}{\partial x} f(x_0, y_0) \end{vmatrix} = \frac{\partial}{\partial x} f(x_0, y_0) \vec{i} + \frac{\partial}{\partial y} f(x_0, y_0) \vec{j} - \vec{k} \end{align}

Notice that we can once again obtain the equation of the tangent plane of $P$ by expanding the dot product $(\vec{T_2} \times \vec{T_1} ) \cdot (x - x_0, y - y_0, z - z_0) = 0$.

However, now that we have this normal vector $\vec{T_2} \times \vec{T_1}$ to the tangent plane at $P$, we note that $\vec{T_2} \times \vec{T_1}$ will be parallel to the normal line corresponding to $P$. We have all we need to construct the equation of a normal line according to the Equations of Lines in Three-Dimensional Space page. Since we have a point ($P(x_0, y_0, z_0)$) that passes through this line and a vector ($\vec{T_2} \times \vec{T_1}$) that is parallel to this line, then we obtain the following parametric equations for this normal line.

(2)
\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}
 Definition: Let $z = f(x, y)$ be a two variable real-valued function, and let $P(x_0, y_0, z_0)$ be a point on the surface generated by $f$. The Normal Line at $P$ is the line that passes through $P$ and is perpendicular to the tangent plane at $P$ and perpendicular to the surface $S$ at $P$.

Let's look at some examples of finding normal lines on a surface.

Example 1

Let $z = f(x, y) = -2xe^xy^2 - 2y$. Find the equation of the normal line at $(-1, -1) \in D(f)$.

We first find the partial derivatives of $f$ as $\frac{\partial z}{\partial x} = [-2e^x - 2xe^x]y^2$, and $\frac{\partial z}{\partial y} = -4xe^xy - 2$. If we evaluate both of these partial derivatives at $x = -1$ and $y = -1$, we get that:

(3)
\begin{align} \quad \quad \frac{\partial}{\partial x} f(-1, -1) = \frac{\partial z}{\partial x} \biggr \rvert_{x = -1, y = -1} = \left [ -2xe^xy^2 - 2y \right ] \biggr \rvert_{x = -1, y = -1} = 0 \end{align}
(4)
\begin{align} \quad \quad \frac{\partial}{\partial y} f(-1, -1) = \frac{\partial z}{\partial y} \biggr \rvert_{x = -1, y = -1} = \left [ -4xe^xy - 2 \right ] \biggr \rvert_{x = -1, y = -1} = \frac{-4}{e} - 2 \end{align}

We also note that $f(-1, -1) = \frac{2}{e} + 2$. Therefore we have that the equation of the normal line has the following parametric equations:

(5)
\begin{align} \left\{\begin{matrix} x = 1 + 0t\\ y = 1 + \left (-\frac{4}{e} - 2 \right)t\\ z = \left (\frac{2}{e} + 2 \right) - t \end{matrix}\right. \quad -\infty < t < \infty \end{align}

The graph of $f$ as well as the tangent plane and normal line are depicted below: 