3.15 Tangent Planes – Revisited

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).

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 $\varphi (t)=F(x(t),y(t),z(t))$, the chain rule states (also see ($†$) on this page ) that:
$${\varphi }^{\prime }(t)=\overrightarrow{\mathrm{\nabla }}F(\mathbf{r}(t))\mathbf{\cdot }{\mathbf{r}}^{\prime }(t).$$
Because $\varphi (t)=c$ is constant, we have ${\varphi }^{\prime }(t)=0$. In particular, ${\varphi }^{\prime }(t_0)=0$ and therefore:
$$\overrightarrow{\mathrm{\nabla }}F(\underset{=\mathbf{r}(t_0)}{\underbrace{x_0,y_0,z_0}})\cdot {\mathbf{r}}^{\prime }(t_0)=0.$$
This means the gradient of $f$ at $(x_0,y_0,z_0)$ is normal to the tangent vector ${\mathbf{r}}^{\prime }(t)$. 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 $\overrightarrow{\mathrm{\nabla }}F(x_0,y_0,z_0)\ne (0,0,0)$, because $\overrightarrow{\mathrm{\nabla }}F(x_0,y_0,z_0)$ 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:
$$\overrightarrow{\mathrm{\nabla }}F({\mathbf{x}}_0)\mathbf{\cdot }(\mathbf{x}-{\mathbf{x}}_0)=0,$$ or
$$F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0.$$

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

$$\frac{x-x_0}{F_x({\mathbf{x}}_0)}=\frac{y-y_0}{F_y({\mathbf{x}}_0)}=\frac{z-z_0}{F_z({\mathbf{x}}_0)}$$

• 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

Let $F(x,y,z)=z^2-x^2-y^2$.
$$\overrightarrow{\mathrm{\nabla }}F(x,y,z)=(-2x,-2y,2z)\Rightarrow \overrightarrow{\mathrm{\nabla }}F(0,0,0)=(0,0,0).$$
Because $\overrightarrow{\mathrm{\nabla }}F(0,0,0)=(0,0,0)$, 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.)

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 $\overrightarrow{\mathrm{\nabla }}F$ is normal to its level curves. The equation of the line tangent to the level curve at $(x_0,y_0)$ becomes: $$\overrightarrow{\mathrm{\nabla }}F(x_0,y_0)\mathbf{\cdot }(x-x_0,y-y_0)=0$$ or $$\begin{array}{}\text{(*)}& F_x(x_0,y_0)(x-x_0)+F_y(x_0,y_0)(y-y_0)=0.\end{array}$$ Note that (*) can also be written as: $$y-y_0=-\frac{F_x(x_0,y_0)}{F_y(x_0,y_0)}(x-x_0).$$

¹ In 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.