Tangent Planes and Linear Approximation

Just as we can visualize the line tangent to a curve at a point in 2-space, in 3-space 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 $$f_x (x_0 , y_0 )(x - x_0 ) + f_y (x_0 , y_0 )(y - y_0 ) - (z - z_0 ) = 0.$$

Notes
• Recall that the equation of the plane containing a point $(x_0 , y_0 , z_0 )$ and normal to the vector ${\bf n} = (a, b, c)$ is $$a(x - x_0 ) + b(y - y_0 ) + c(z - z_0 ) = 0.$$ The derivation of the equation for the tangent plane just involves showing that the tangent plane is normal to the vector ${\bf n} = (f_x (x_0 , y_0 ), f_y (x_0 , y_0 ), -1)$.

• For surfaces $F (x, y, z) = 0$ that are not easily solved for $z$, the equation of the tangent plane at $(x_0 , y_0 , z_0 )$ is $$F_x (x_0 , y_0 , z_0 )(x - x_0 ) + F_y (x_0 , y_0 , z_0 )(y - y_0 ) + F_z (x_0 , y_0 , z_0 )(z - z_0 ) = 0$$ provided that $\nabla F (x_0 , y_0 , z_0 ) \neq 0$. Note that if we let $F (x, y, z) = f (x, y) - z$, we obtain the equation given for the tangent plane to $z = f (x, y)$ at $(x_0 , y_0 , z_0 )$.

#### Example

Let'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 Approximation

The 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 $(x_0 , y_0 )$.

Notes
• The linear approximation is really just the multivariable Taylor polynomial of degree 1 for $f (x, y)$ about $(x_0 , y_0 )$. It is only accurate near $(x_0 , y_0 )$. Better approximations can be obtained by using higher-order Taylor polynomials.

• These concepts can be extended to functions of more than two variables: $$f (x, y, z) \approx f (x_0 , y_0 , z_0 )+f_x (x_0 , y_0 , z_0 )(x-x_0 )+f_y (x_0 , y_0 , z_0 )(y-y_0 )+f_z (x_0 , y_0 , z_0 )(z-z_0 )$$ where $f (x, y, z)$ is differentiable at $(x_0 , y_0 , z_0 )$.

#### Example

From 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

• Tangent Plane to a Surface

Let $(x_0 , y_0 , z_0 )$ be any point on the surface $z = f (x, y)$. If $f (x, y)$ is differentiable at $(x_0 , y_0 )$, then the surface has a tangent plane at $(x_0 , y_0 , z_0 )$ given by $$f_x (x_0 , y_0 )(x - x_0 ) + f_y (x_0 , y_0 )(y - y_0 ) - (z - z_0 ) = 0.$$

• Linear Approximation to a Surface

If $f (x, y)$ is differentiable at $(x_0 , y_0 )$, then near $(x_0 , y_0 )$ $$f (x, y) \approx f (x_0 , y_0 ) + f_x (x_0 , y_0 )(x - x_0 ) + f_y (x_0 , y_0 )(y - y_0 ).$$