Linear Approximations of Functions of Several Variables

Linear Approximations of Functions of Several Variables

Recall from the Linear Approximation of Single Variable Functions page that for a single variable differentiable function $y = f(x)$, the tangent line at the point $(a, f(a))$ with equation $y - f(a) = f'(a)(x - a)$ approximates the graph of $f$ near $x = a$. For example, consider the function $f(x) = x^2$, and suppose that we wanted to compute $f(2.95)$. The equation of the tangent line at $(3, 9)$ is $y - 9 = 6(x - 3)$ or rather $y = 6x - 9$. Since $2.95$ is relatively close to $3$, then we can approximately $f(2.95) \approx 6(2.95) - 9 = 8.7$, which is relatively good considering $f(2.95) = 8.7025$.

We will now looking at applying the same technique to approximating variables of functions of several variables.

Linear Approximations of Two Variable Functions

Consider the two variable real-valued function $f(x, y)$, the point $P(a, b, c)$, and the tangent plane to $P$. If we zoom in closer and closer to $P$, the surface $S$ generated by $f$ resembles the tangent plane of $P$ more and more accurately. Another way to view this is that the closer we get to the point $P$, the small region around $P$ resembles the tangent plane of $P$ more and more accurately as the diagram below depicts:    Recall that the tangent plane of $P$ can be obtained by the following formula: $z - c = f_x (a, b) (x - a) + f_y (a, b) (y - b)$ which can be rearranged as:

(1)
\begin{align} \quad z = c + f_x (a, b) (x - a) + f_y (a, b) (y - b) \\ \quad z = f(a, b) + f_x (a, b) (x - a) + f_y (a, b) (y - b) \end{align}
 Definition: If $f(x, y)$ is a two variable real-valued function whose partial derivatives $f_x (x, y)$ and $f_y (x, y)$ exist, then for $(x, y)$ close to $(a, b)$ in the domain of $f$, $f(x, y) \approx f(a, b) + f_x (a, b) (x - a) + f_y (a, b) (y - b)$ is the Linear Approximation of $f$ at $(a, b)$, and the function $L(x, y) = f(a, b) + f_x (a, b) (x - a) + f_y (a, b) (y - b)$ is known as the Linearization of $f$ at $(a, b)$.

We should note that the linear approximation of $f$ at $(a, b)$ is more and more accurate the closer $(x, y) \in D(f)$ is to $(a,b)$.

Example 1

Find the linearization of the function $f(x, y) = x^3y^2$ at $(3, 1)$ in the domain and use it to estimate the value of $f(3.01, 0.99)$.

We first note that $f_x (x, y) = 3x^2y^2$ and so $f_x (3, 1) = 27$. Also $f_y (x, y) = 2x^3y$ and so $f_y ( 3, 1) = 54$. We also note that $f(3, 1) = 27$, and so the linearization of $f$ at $(3, 1)$ is:

(2)
\begin{align} \quad L(x, y) = f(a, b) + f_x (a, b) (x - a) + f_y (a, b) (y - b) \\ \quad L(x, y) = f(3, 1) + f_x (3, 1) (x - 3) + f_y (3, 1) (y - 1) \\ \quad L(x, y) = 27 + 27(x - 3) + 54(y - 1) \end{align}

Thus we have that $L(3.01, 0.99) = 27 + 27(3.01 - 3) + 54(0.99 - 1) = 26.73 \approx f(3.01, 0.99)$. We note that the true value of $f(3.01, 0.99) = 26.728...$, so this linear approximation is pretty close.

Example 2

Find the linearization of the function $f(x, y) = 2x^2e^{3y}$ at $(1, 1)$ in the domain and use it to estimate the value of $f(0.99, 0.99)$.

We first note that $f_x (x, y) = 4xe^{3y}$ and so $f_x (1, 1) = 4e^3$. Also $f_y (x, y) = 6x^2e^{3y}$ and so $f_y (1, 1) = 6e^3$. We also note that $f(1, 1) = 2e^3$, and so the linearization of $f$ at $(1, 1)$ is:

(3)
\begin{align} \quad L (x, y) = f(a, b) + f_x (a, b) (x - a) + f_y (a, b) (y - b) \\ \quad L (x, y) = f(1, 1) + f_x (1, 1) (x - 1) + f_y (1, 1) (y - 1) \\ \quad L (x, y) = 2e^3 + 4e^3 (x - 1) + 6e^3 ( y - 1) \end{align}

Thus we have that $L(0.99, 0.99) = 2e^3 + 4e^3 (0.99 - 1) + 6e^3 (0.99 - 1) = 2e^3 + 4e^3(-0.01) + 6e^3(-0.01) = 38.16 \approx f(0.99, 0.99) = 38.208...$, so this linear approximation is pretty close.