3.8 Tangent Plane

If a surface $S$ representing the equation $z = f (x, y)$ is “smooth” near the point $P = (x_{0}, y_{0}, z_{0})$ , where $z_{0} = f (x_{0}, y_{0})$ , then it will have a tangent plane (see Fig. 1). In this section, we will derive the equation of this plane.

Figure 1

Later in this chapter, we will learn under which conditions the surface is smooth.

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We recall that the slope of the curve formed by the intersection of $S$ and the plane $y = y_{0}$ at $P$ is $f_{x} (x_{0}, y_{0})$ . Let’s call the tangent line to the curve of of $f (x, y_{0})$ at the point $P$ , $L_{1}$ .
The equation of $L_{1}$ is
$z - z_{0} = f_{x} (x_{0}, y_{0}) \times (x - x_{0}) and y = y_{0} .$

- Recall from single variable calculus that the equation of the line tangent to the curve $\small {\color{Gray} z=f(x)}$ at $\small {\color{Gray}(x_0,y_0) }$ is $\small {\color{Gray} z-z_0=f'(x_0)(x-x_0). }$ Here we just need to replace $\small {\color{Gray} f'(x_0)}$ by $\small {\color{Gray} f_x(x_0,y_0) }$ .

If $f_{x} (x_{0}, y_{0}) \neq 0$ , we can rewrite the decription of $L_{1}$ in the regular format as:
$L_{1} : \frac{x - x_{0}}{1} = \frac{z - z_{0}}{f_{x} (x_{0}, y_{0})}, and y = y_{0} .$
Similarly $L_{2}$ that is the tangent line to the intersection of $S$ and the plane $x = x_{0}$ at the point $P$ is described by:
$L_{2} : \frac{y - y_{0}}{1} = \frac{z - z_{0}}{f_{y} (x_{0}, y_{0})} and x = x_{0} .$
The direction vectors of the lines $L_{1}$ and $L_{2}$ are
$\begin{array}{r} {\vec{v}}_{1} = (1, 0, f_{x} (x_{0}, y_{0})), and {\vec{v}}_{2} = (0, 1, f_{y} (x_{0}, y_{0})), \end{array}$
respectively.

- Recall that the line that is parallel to $\small {\color{Gray}\vec{v}=(a,b,c) }$ and passing through $\small {\color{Gray}(x_0,y_0,z_0) }$ is described by $\small {\color{Gray}\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c} }$ . $\small {\color{Gray}\vec{v} }$ is called the direction vector of this line.

The tangent plane to the surface $S$ at the point $P$ is defined to be the plane that contains both lines $L_{1}$ and $L_{2}$ . Equivalently, we can say the tangent plane contains both direction vectors ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ . Therefore, the cross product of ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ is normal to the tangent plane and to the surface $S$ at the point $P$ :
$\vec{n} = {\vec{v}}_{2} \times {\vec{v}}_{1} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & f_{y} (x_{0}, y_{0}) \\ 1 & 0 & f_{x} (x_{0}, y_{0}) \end{matrix} | = f_{x} (x_{0}, y_{0}) \hat{i} + f_{y} (x_{0}, y_{0}) \hat{j} - \hat{k}$

- Recall that if two non-parallel vectors $\small {\color{Gray} \vec{v}_1 }$ and $\small {\color{Gray} \vec{v}_2 }$ are in the plane $\small {\color{Gray} \Omega }$ , then $\small {\color{Gray} \vec{v}_1\times\vec{v}_2 }$ is perpendicular to the plane $\small {\color{Gray} \Omega }$ and to any vector that is in $\small {\color{Gray} \Omega }$ .

Because the equation of a plane passing through $(x_{0}, y_{0}, z_{0})$ with normal $\vec{n} = (n_{1}, n_{2}, n_{3})$ is:
$n_{1} (x - x_{0}) + n_{2} (y - y_{0}) + n_{3} (z - z_{0}) = 0,$
the equation of the tangent plane reads:
$f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0}) - (z - z_{0}) = 0$
or it can be rewritten as:
$z = z_{0} + f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0})$

We will see later in this chapter that not only does the tangent plane contain $L_{1}$ and $L_{2}$ but also it contains the tangent line to the intersection of $S$ and any plane which is perpendicular to the $x y$ -plane at $(x_{0}, y_{0}, z_{0})$ . That’s why it is called the “tangent plane.”

The equation of the tangent plane to the surface

z = f (x, y)

at the point

P = (x_{0}, y_{0}, z_{0})

is given by:

\begin{matrix} (*) & z = z_{0} + f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0}), \end{matrix}

where

z_{0} = f (x_{0}, y_{0})

If $f_{x} (x_{0}, y_{0}) = f_{y} (x_{0}, y_{0}) = 0$ , the tangent plane is perpendicular to the $z$ -axis and is given by
$z = z_{0} .$

Because $\vec{n} = (f_{x} (x_{0}, y_{0}), f_{y} (x_{0}, y_{0}), - 1)$ is perpendicular to $S$ at $(x_{0}, y_{0}, z_{0})$ , the equation of the line normal to $S$ at the point $P$ is:
$\frac{x - x_{0}}{f_{x} (x_{0}, y_{0})} = \frac{y - y_{0}}{f_{y} (x_{0}, y_{0})} = \frac{z - z_{0}}{- 1} .$
If either $f_{x} (x_{0}, y_{0}) = 0$ or $f_{y} (x_{0}, y_{0}) = 0$ , the above equation needs to be modified.

Example

Find the tangent plane and the normal line to the surface

S

given by

z = x^{2} y - x + 2

at the point

(1, 0, 1)

Solution

First we calculate

\frac{\partial z}{\partial x} (1, 0)

and

\frac{\partial z}{\partial y} (1, 0)

\frac{\partial z}{\partial x} = 2 x y - 1 \Rightarrow \frac{\partial z}{\partial x} (1, 0) = - 1

\frac{\partial z}{\partial y} = x^{2} \Rightarrow \frac{\partial z}{\partial y} (1, 0) = 1

Therefore the equation of the tangent plane at

(1, 0, 1)

is:

z = \underset{= 1}{\underset{⏟}{z_{0}}} + \underset{= - 1}{\underset{⏟}{\frac{\partial z}{\partial x} (x_{0}, y_{0})}} (x - \underset{= 1}{\underset{⏟}{x_{0}}}) + \underset{= 1}{\underset{⏟}{\frac{\partial z}{\partial y} (x_{0}, y_{0})}} (y - \underset{= 0}{\underset{⏟}{y_{0}}})

\Rightarrow z = 1 - (x - 1) + y

The equation of the line normal to $S$ at $(1, 0, 1)$ is:
$\frac{x - 1}{- 1} = \frac{y - 0}{1} = \frac{z - 1}{- 1}$

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