
Just as we can visualize the line tangent to a curve at a point in 2space, in 3space we can picture the plane tangent to a surface at a point.
Consider the surface given by $z = f (x, y)$. Let $(x_0 , y_0 , z_0 )$ be any point on this surface. If
$f (x, y)$ is differentiable at $(x_0 , y_0 )$, then the surface has a tangent plane at $(x_0 , y_0 , z_0 )$.
The equation of the tangent plane at $(x_0 , y_0 , z_0 )$ is given by
$$f_x (x_0 , y_0 )(x  x_0 ) + f_y (x_0 , y_0 )(y  y_0 )  (z  z_0 ) = 0.$$
ExampleLet's find the equation of the plane tangent to the surface $z = 4x^3 y^2 + 2y$ at the point $(1, 2, 12)$. Since $f (x, y) = 4x^3 y^2 + 2y$, $$ f_x (x, y) = 12x^2 y^2 \textrm{ and } f_y (x, y) = 8x^3 y + 2.$$ With $x = 1$ and $y = 2$, \begin{eqnarray*} f_x (1, 2) &=& 12(1)^2 (2)^2 = 48 \\ f_y (1, 2) &=& 8(3)^3(2) + 2 = 14. \end{eqnarray*} Thus, the tangent plane has normal vector $ {\bf n} = (48, 14, 1) $ at $(1, 2, 12)$ and the equation of the tangent plane is given by $$ 48(x  1)  14 (y  (2))  (z  12) = 0.$$ Simplifying, $$ 48x  14y  z = 64. $$
Linear ApproximationThe tangent plane to a surface at a point stays close to the surface near the point. In fact, if $f (x, y)$ is differentiable at the point $(x_0 , y_0 )$, the tangent plane to the surface $z = f (x, y)$ at $(x_0 , y_0 )$ provides a good approximation to $f (x, y)$ near $(x_0 , y_0 )$:
$\qquad$ Solving $f_x (x_0 , y_0 )(x  x_0 ) + f_y (x_0 , y_0 )(y  y_0 )  (z  z_0 ) = 0$ for $z$,
$$ z = z_0 + f_x (x_0 , y_0 )(x  x_0 ) + f_y (x_0 , y_0 )(y  y_0 ). $$
$\qquad$ Since $z_0 = f (x_0 , y_0 )$, we have that
$$ z = f (x_0 , y_0 ) + f_x (x_0 , y_0 )(x  x_0 ) + f_y (x_0 , y_0 )(y  y_0 ). $$
$\qquad$ Near $(x_0 , y_0 )$, the surface is close to the tangent plane. Thus,
$$ f (x, y) \approx f (x_0 , y_0 ) + f_x (x_0 , y_0 )(x  x_0 ) + f_y (x_0 , y_0 )(y  y_0 ). $$
We call this the linear approximation or local linearization of $f (x, y)$ near $(x_0 , y_0 )$.
ExampleFrom our work in the previous example, the linear approximation to $f (x, y) = 4x^3 y^2 + 2y$ near $x = 1,\quad y = 2$ is $$ f (x, y) \approx 48x  14y  64. $$ This is, of course, exact at $x = 1, \quad y = 2$: $$ f (1, 2) = 12 = 48(1)  14(2)  64. $$ At $x = 1.1$ and $y = 1.9$, according to the linear approximation, $$ f (1.1, 1.9) \approx 48(1.1)  14(1.9)  64 = 15.4, $$ which is indeed very close to the exact value $f (1.1, 1.9) = 15.41964$.
Key Concepts
