If a surface $S$ representing the equation $z=f(x,y)$ is “smooth” near the point $P=\left(x_0,y_0,z_0\right)$, 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.

#### Read more about the derivation of the equation of the tangent plane

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
$$z-z_0=f_x(x_0,y_0)\times(x-x_0)\quad\text{and}\quad y=y_0.$$

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

If $f_x(x_0,y_0)\neq 0$, we can rewrite the decription of $L_1$ in the regular format as:
$$L_1:\quad\frac{x-x_0}{1}= \frac{z-z_0}{f_x(x_0,y_0)},\quad\text{and}\quad y=y_0.$$
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:
$$L_2:\quad \frac{y-y_0}{1}=\frac{z-z_0}{f_y(x_0,y_0)}\quad\text{and}\quad x=x_0.$$
The direction vectors of the lines $L_1$ and $L_2$ are
\begin{align*}
\end{align*}
respectively.

• Recall that the line that is parallel to $\small&space;{\color{Gray}\vec{v}=(a,b,c)&space;}$ and passing through $\small&space;{\color{Gray}(x_0,y_0,z_0)&space;}$ is described by $\small&space;{\color{Gray}\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}&space;}$ . $\small&space;{\color{Gray}\vec{v}&space;}$ 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 $P$:
$$\vec{n}=\vec{v}_2\times\vec{v}_1=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & f_y(x_0,y_0)\\ 1 & 0 & f_x(x_0,y_0) \end{vmatrix}=f_x(x_0,y_0)\hat{i}+f_y(x_0,y_0)\hat{j}-\hat{k}$$

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

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 $xy$-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:
$z=z_0+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0),\label{TangentPlane-0} \tag{*}$ 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 $\vec{n}=\left(f_x(x_0,y_0),f_y(x_0,y_0),-1\right)$ is perpendicular to $S$ at $(x_0,y_0,z_0)$, the equation of the line normal to $S$ at the point $P$ is:
$\bbox[#F2F2F2,5px,border:2px solid black]{\large \frac{x-x_0}{f_x(x_0,y_0)}=\frac{y-y_0}{f_y(x_0,y_0)}=\frac{z-z_0}{-1}.}$
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 $z=x^2y-x+2$ at the point $(1,0,1)$.
Solution
First we calculate $\frac{\partial z}{\partial x}(1,0)$ and $\frac{\partial z}{\partial y}(1,0)$:
$$\frac{\partial z}{\partial x}=2xy-1 \Rightarrow \frac{\partial z}{\partial x}(1,0)=-1$$
$$\frac{\partial z}{\partial y}=x^2 \Rightarrow \frac{\partial z}{\partial y}(1,0)=1$$
Therefore the equation of the tangent plane at $(1,0,1)$ is:
$$z=\underbrace{z_0}_{=1}+\underbrace{\frac{\partial z}{\partial x}(x_0,y_0)}_{=-1} (x-\underbrace{x_0}_{=1})+\underbrace{\frac{\partial z}{\partial y}(x_0,y_0)}_{=1} (y-\underbrace{y_0}_{=0})$$
$$\Rightarrow z=1-(x-1)+y$$

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