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 $$\phi(t)=F(x(t),y(t),z(t))$$, the chain rule states (also see ($$\dagger$$) on page ) that:
$\phi'(t)=\overrightarrow{\nabla} F(\mathbf{r}(t))\bullet\mathbf{r}'(t).$
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:
$\overrightarrow{\nabla} F(\mathbf{x}_0)\bullet (\mathbf{x}-\mathbf{x}_0)=0,$
$\text{or}\quad \frac{\partial F}{\partial x}(x_0,y_0,z_0)(x-x_0)+\frac{\partial F}{\partial y}(x_0,y_0,z_0)(y-y_0)+\frac{\partial F}{\partial z}(x_0,y_0,z_0)(z-z_0)=0.$

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

The equations of the normal line to the level surface $$F(x,y,z)=0$$ at $$(x_0,y_0,z_0)$$ are:

$\frac{x-x_0}{\frac{\partial F}{\partial x}(x_0,y_0,z_0)}=\frac{y-y_0}{\frac{\partial F}{\partial y}(x_0,y_0,z_0)}=\frac{z-z_0}{\frac{\partial F}{\partial z}(x_0,y_0,z_0)}$

(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.);

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 $$\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{8}}\right)$$.

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:
$\left(\sqrt{2},0,\sqrt{8}\right)\bullet\left(x-\frac{1}{\sqrt{2}},y,z-\frac{1}{\sqrt{8}}\right)=0,$
$\text{or}\quad \sqrt{2}x+\sqrt{8}z=2.$ 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 $$F(x,y,z)=x^2 y+zy^2+x z^2$$. $$S$$ 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:
$\frac{\partial F}{\partial x}=\frac{\partial f}{\partial x},\quad \frac{\partial F}{\partial y}=\frac{\partial f}{\partial y},\quad \frac{\partial F}{\partial z}=-1,$
according to Theorem 3.15.1, the equation of the tangent plane to $$S$$ at $$(x_0,y_0,z_0)$$ is
$\frac{\partial f}{\partial x}(x_0,y_0,z_0)(x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0,z_0)(y-y_0)-(z-z_0)=0,$
which is the same as Eq. (*) on 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 3.15.1. 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.)

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 $F_x(x_0,y_0)(x-x_0)+F_y(x_0,y_0)(y-y_0)=0. \tag{*}$ 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: $y-y_0=f'(x_0)(x-x_0)\Rightarrow y-2=0(x-0) \Rightarrow y=2$

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.