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

Figure 2.
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.)

Figure 3.

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.

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