3.15 Tangent Planes - Revisited

Table of Contents

Tangent Plane and Normal Line of a Level Surface

Let $S$ be a level surface of a differentiable function $F$ having the equation: \[F(x,y,z)=c.\] Consider a curve $C$ on $S$ that passes through $\mathbf{x}_0=(x_0,y_0,z_0)$ (Fig. 1).

Figure 1

Assume $C$ is given parametrically by a differentiable vector-valued function $\mathbf{r}(t)=(x(t),y(t),z(t))$ and $\mathbf{r}(t_0)=(x_0,y_0,z_0)$. Because $C$ is on $S$, \[F(\mathbf{r}(t))=F(x(t),y(t),z(t))=c.\] If $\phi(t)=F(x(t),y(t),z(t))$, the chain rule states (also see ($\dagger$) on this page ) that:
\[\phi'(t)=\overrightarrow{\nabla} F(\mathbf{r}(t))\boldsymbol{\cdot}\mathbf{r}'(t).\]
Because $\phi(t)=c$ is constant, we have $\phi'(t)=0$. In particular, $\phi'(t_0)=0$ and therefore:
\[\overrightarrow{\nabla} F(\underbrace{x_0,y_0,z_0}_{=\mathbf{r}(t_0)})\cdot\mathbf{r}'(t_0)=0.\]
This means the gradient of $f$ at $(x_0,y_0,z_0)$ is normal to the tangent vector $\mathbf{r}'(t)$. Consider all curves on $S$ passing through $(x_0,y_0,z_0)$. A plane that contains the tangent vectors of all these curves is called the tangent plane.¹ If $\overrightarrow{\nabla} F(x_0,y_0,z_0)\neq (0,0,0)$, because $\overrightarrow{\nabla} F(x_0,y_0,z_0)$ is normal to the tangent vectors at $(x_0,y_0,z_0)$, the (Recall that a plane through $\mathbf{x}_0=(x_0,y_0,z_0)$ with normal $\mathbf{n}=(n_1,n_2,n_3)$ consists of all points $\mathbf{x}=(x,y,z)$ that satisfy: $\mathbf{n}\cdot (\mathbf{x}-\mathbf{x}_0)=0$ )equation of tangent plane; is:
\[\overrightarrow{\nabla} F(\mathbf{x}_0)\boldsymbol{\cdot} (\mathbf{x}-\mathbf{x}_0)=0,\] or
\[F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0.\]

Theorem 1. Let $F:U\subseteq\mathbb{R}^3\to \mathbb{R}$ be a differentiable function. If $\mathbf{x}_0=(x_0,y_0,z_0)$ lies on the level surface $S$ defined by $F(x,y,z)=c$ and $\overrightarrow{\nabla} F(\mathbf{x}_0)\neq \mathbf{0}$, then the equation of the tangent plane at $(x_0,y_0,z_0)$ is:
\[F_x(\mathbf{x}_0)(x-x_0)+F_y(\mathbf{x}_0)(y-y_0)+F_z(\mathbf{x}_0)(z-z_0)=0.\] or
\[\overrightarrow{\nabla} F(\mathbf{x}_0)\boldsymbol{\cdot} (\mathbf{x}-\mathbf{x}_0)=0,\]

The equations of the normal line to the level surface $F(x,y,z)=c$ at $(x_0,y_0,z_0)$ are:

\[\bbox[#F2F2F2,5px,border:2px solid black]{\frac{x-x_0}{F_x(\mathbf{x}_0)}=\frac{y-y_0}{F_y(\mathbf{x}_0)}=\frac{z-z_0}{F_z(\mathbf{x}_0)}}\]

Example 1

Consider the surface $S$ defined by the equation $x^2+y^2+4z^2=1$. Find the tangent plane to $S$ at the point $\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{8}}\right)$.

The equation of the ellipsoid has the standard form$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.$ The points $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$ lie on the surface of the ellipsoid.

Solution

Here $F(x,y,z)=x^2+y^2+4z^2$ and $S$ is the level surface $F(x,y,z)=1$. To find the equation of the tangent plane at $\mathbf{x}_0=\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{8}}\right)$, we need to find $\overrightarrow{\nabla} F(\mathbf{x}_0)$.

\[F(x,y,z)=x^2+y^2+4z^2\Rightarrow \overrightarrow{\nabla} F=(2x,2y,8z)\]
\[\Rightarrow \overrightarrow{\nabla} F\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{8}}\right)=\left(\sqrt{2},0,\sqrt{8}\right).\]
Thus the equation of the tangent plane is:
\[\left(\sqrt{2},0,\sqrt{8}\right)\boldsymbol{\cdot}\left(x-\frac{1}{\sqrt{2}},y,z-\frac{1}{\sqrt{8}}\right)=0,\]
\[\text{or}\quad \sqrt{2}x+\sqrt{8}z=2.\] Fig. 2 shows the level surface and its tangent plane at $\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{8}}\right)$.

Example 2

Consider the surface $S$ defined by $x^2 y+zy^2+x z^2=2$. Find the equation of the tangent plane to $S$ at $(-2,2,3)$.

Solution

Let $F(x,y,z)=x^2 y+zy^2+x z^2$. Then $S$ is the level surface specified by $F(x,y,z)=2$.
\[\overrightarrow{\nabla} F(x,y,z)=(2xy+z^2,x^2+2zy,y^2+2xz)\]\[\Rightarrow \overrightarrow{\nabla} F(-2,2,3)=(1,16,-8)\]
Therefore the equation of the tangent plane is
\[1\times (x+2)+16 (y-2)-8 (z-3)=0\]
\[\text{or}\quad \quad x+16y-8z=6\]

Example 3

Determine the tangent plane to the surface $S$ specified by the explicit equation \[z=f(x,y)\]

Solution

Let $F(x,y,z)=f(x,y)-z$. The graph of $z=f(x,y)$ is the same as the level surface $F(x,y,z)=f(x,y)-z=0$. Because:
\begin{align}
\frac{\partial F}{\partial x}(x_0,y_0,z_0)&=\frac{\partial f}{\partial x}(x_0,y_0),\\
\frac{\partial F}{\partial y}(x_0,y_0,z_0)&=\frac{\partial f}{\partial y}(x_0,y_0),\\
\frac{\partial F}{\partial z}(x_0,y_0,z_0)=-1,
\end{align}
where $z_0=f(x_0,y_0)$, according to Theorem 1, the equation of the tangent plane to $S$ at $(x_0,y_0,z_0)$ is
\[f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)-(z-z_0)=0,\]
which is the same as Eq. (*) on this page.

Example 4

Determine the tangent plane — if it exists — to the surface specified by
$z^2=x^2+y^2$ at $(0,0,0)$.

Solution

Let $F(x,y,z)=z^2-x^2-y^2$.
\[\overrightarrow{\nabla} F(x,y,z)=(-2x,-2y,2z) \Rightarrow \overrightarrow{\nabla} F(0,0,0)=(0,0,0).\]
Because $\overrightarrow{\nabla} F(0,0,0)=(0,0,0)$, we cannot use Theorem 1. In fact, if we plot the surface level, we realize that the surface does not have a tangent plane. If we solve $F(x,y,z)=0$ for $z$ and write the result as two functions $z=f(x,y)=\sqrt{x^2+y^2}$ and $z=g(x,y)=-\sqrt{x^2+y^2}$, we can show $f$ and $g$ are not differentiable at $(0,0)$ and therefore they do not have a tangent plane at the origin. (You should try to show the first partial derivatives at the origin do not exist; therefore, $f$ and $g$ are not differentiable.)

Figure 3: The level surface of $z^2-x2-y^2=0$.

Tangent Line of a Level Curve

The arguments are the same if we consider the level curves of $F(x,y)=c$. In the previous section we saw that $\overrightarrow{\nabla} F$ is normal to its level curves. The equation of the line tangent to the level curve at $(x_0,y_0)$ becomes: \[\overrightarrow{\nabla} F(x_0,y_0)\boldsymbol{\cdot} (x-x_0,y-y_0)=0\] or \[\bbox[#F2F2F2,5px,border:2px solid black]{F_x(x_0,y_0)(x-x_0)+F_y(x_0,y_0)(y-y_0)=0.} \tag{*}\] Note that (*) can also be written as: \[{y-y_0=-\frac{F_x(x_0,y_0)}{F_y(x_0,y_0)}(x-x_0).}\]

Example 5

Find the equation of the tangent line to $y e^{x^2}=2$ at $x=0$ and $y=2$.

Solution

Method (a): Letting $F(x,y)=y e^{x^2}$, we find $\overrightarrow{\nabla} F(x,y)=\left(2x e^{x^2},y\right)$. So $\overrightarrow{\nabla} F(0,2)=(0,2)$. Using Eq (*), the equation of the tangent line at $(0,2)$ is:
\[0\times(x-0)+2 (y-2)=0\quad \text{or} \quad y=2.\]

Method (b): We can solve for $y$ and write $y=2/e^{x^2}=2 e^{-x^2}$. We know the slope of the tangent line to the curve $y=f(x)$ is $f'(x)$. Thus, we first find $dy/dx$: $\frac{dy}{dx}=-2xe^{-x^2}\Rightarrow \frac{dy}{dx}\Big|_{x=0}=0$. The tangent line at $x=0$ is: \[y-y_0=f'(x_0)(x-x_0)\Rightarrow y-2=0(x-0) \Rightarrow y=2\]

¹In Section 3.8 we defined the tangent plane as a plane that contains the tangent vectors in the x and y directions. When F is differentiable, the surface is smooth enough and the tangent plane contains the tangent vectors in all directions.↩