Althea

High School Teacher - Tutor for 3 years

MARCH 16, 2024

Linearization of a function is a mathematical process used to approximate a function near a given point using a linear function. This technique is particularly useful in calculus and applied mathematics for simplifying complex functions at specific points to make them easier to analyze or work with, especially when dealing with non-linear behavior in functions.

The linear approximation or linearization of a function $f(x)$ at a point $a$ is given by the tangent line to the function at that point. The equation of this tangent line provides the linear approximation of $f(x)$ near $a$.

The linearization $L(x)$ of the function $f(x)$ at a point $a$ is given by:

$L(x) = f(a) + f'(a)(x - a)$

where:

- $L(x)$ is the linear approximation of $f(x)$ near $a$,
- $f(a)$ is the value of the function at point $a$,
- $f'(a)$ is the derivative of $f$ at $a$, representing the slope of the tangent line at $a$,
- $x$ is the point at which the linear approximation is evaluated, and
- $a$ is the point of tangency where the linear approximation is based.

Finding the linearization of a function around a point involves creating a linear approximation that mimics the function's behavior near that point. The linear approximation is essentially the equation of the tangent line at that point. Here's how you can find it:

**Identify the Function and Point of Interest**: Determine the function $f(x)$ and the specific point $a$ at which you want to find the linearization.**Calculate $f(a)$**: Evaluate the function at the point $a$ to get $f(a)$. This value is the y-coordinate of the point where the tangent meets the curve.The Formula: $f(a)=a^2$

**Find the Derivative $f'(x)$**: Calculate the first derivative of $f(x)$, which represents the slope of the tangent line at any point along the curve.The Formula: $f'(x)=2x$

**Evaluate $f'(a)$**: Find the derivative value at point $a$, i.e., $f'(a)$. This value gives the slope of the tangent line at $x = a$.**Write the Linearization Equation**: Use the formula $L(x) = f(a) + f'(a)(x - a)$ to write the linear equation. This equation represents the tangent line at $x = a$, which is the linear approximation of $f(x)$ near this point.

Let's find the linearization of the function $f(x) = \sqrt{x}$ at $a = 4$.

**Function and Point**: $f(x) = \sqrt{x}$, $a = 4$.**Calculate $f(a)$**: $f(4) = \sqrt{4} = 2$.**Find the Derivative $f'(x)$**: $f'(x) = \frac{1}{2\sqrt{x}}$**Evaluate $f'(a)$**: $f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}$**Construct the Linearization Formula**: $L(x) = f(4) + f'(4)(x - 4)$ $L(x) = 2 + \frac{1}{4}(x - 4) = 2 + \frac{1}{4}x - 1 = \frac{1}{4}x + 1$

**Conclusion for the Example:**

The linearization $L(x)$ of the function $f(x) = \sqrt{x}$ at $a = 4$ is $L(x) = \frac{1}{4}x + 1$. This linear function approximates $f(x)$ near $x = 4$, providing a simpler way to estimate the function's values around this point.

Linearization of a function of two variables at a point involves approximating the function near that point using the plane tangent to the function's surface at the given point. The linear approximation, in this case, is akin to a two-variable function's "first-order Taylor expansion" around the point.

**Identify the Function and Point**: Let $f(x, y)$ be your function and $(a, b)$ the point where you want to linearize $f$.**Find Partial Derivatives**: Calculate the partial derivatives of $f$ with respect to $x$ and $y$, denoted as $f_x$ and $f_y$, respectively. These represent the slopes of the tangent plane in the directions of the $x$ and $y$ axes.**Evaluate Partial Derivatives at the Point**: Compute $f_x(a, b)$ and $f_y(a, b)$, the values of the partial derivatives at the point $(a, b)$.**Calculate $f(a, b)$**: Evaluate the function at point $(a, b)$ to find $f(a, b)$, which is the height of the function at that point.**Write the Linearization Formula**: Use the values found to construct the linear approximation formula: $L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)$

Consider the function $f(x, y) = x^2 + xy + y^2$, and we want to find its linearization around the point $(1, 1)$.

**Given Function**: $f(x, y) = x^2 + xy + y^2$**Find Partial Derivatives**:- $f_x = 2x + y$ (derivative of $f$ with respect to $x$)
- $f_y = x + 2y$ (derivative of $f$ with respect to $y$)

**Evaluate Partial Derivatives at (1, 1)**:- $f_x(1, 1) = 2(1) + 1 = 3$
- $f_y(1, 1) = 1 + 2(1) = 3$

**Calculate $f(1, 1)$**:- $f(1, 1) = 1^2 + 1 \cdot 1 + 1^2 = 3$

**Write the Linearization Formula**:- Starting with the general formula: $L(x, y) = f(1, 1) + f_x(1, 1)(x - 1) + f_y(1, 1)(y - 1)$
- Substitute the values: $L(x, y) = 3 + 3(x - 1) + 3(y - 1)$
- Simplify: $L(x, y) = 3 + 3x - 3 + 3y - 3 = 3x + 3y - 3$

**Conclusion for the Example:**

The linear approximation of the function $f(x, y) = x^2 + xy + y^2$ near the point $(1, 1)$ is given by the equation $L(x, y) = 3x + 3y - 3$. This linear function approximates $f(x, y)$ close to $(1, 1)$, providing a plane that tangentially touches the surface of $f(x, y)$ at that point.

Linearization of a function $f(x, y, z)$ of three variables at a point involves finding a plane that best approximates the function near that point. This plane is tangent to the surface defined by $f(x, y, z)$ at the given point. This process extends the concept of linearization for functions of one or two variables, incorporating partial derivatives with respect to all three variables.

**Identify the Function and Point**: Consider a function $f(x, y, z)$ and a point $(a, b, c)$ where you want to approximate the function.**Compute Partial Derivatives**: Calculate the first-order partial derivatives of the function with respect to each variable:- $f_x$ (partial derivative with respect to $x$),
- $f_y$ (partial derivative with respect to $y$), and
- $f_z$ (partial derivative with respect to $z$).

**Evaluate Partial Derivatives at the Point**: Find the values $f_x(a, b, c)$, $f_y(a, b, c)$, and $f_z(a, b, c)$ by substituting the coordinates of the point into each partial derivative.**Calculate $f(a, b, c)$**: Evaluate the original function at the point to find the value of $f$ at $(a, b, c)$.**Construct the Linearization**: Use the linearization formula for three variables: $L(x, y, z) = f(a, b, c) + f_x(a, b, c)(x - a) + f_y(a, b, c)(y - b) + f_z(a, b, c)(z - c)$

**Function**: $f(x, y, z) = x^2 + y^2 + z^2 + xyz$

**Point for Linearization**: $(1, 1, 1)$

**Step 1**: Our function and point of interest are already defined.

**Step 2**: Compute Partial Derivatives:

- $f_x = 2x + yz$,
- $f_y = 2y + xz$,
- $f_z = 2z + xy$.

**Step 3**: Evaluate Partial Derivatives at $(1, 1, 1)$:

- $f_x(1, 1, 1) = 2(1) + (1)(1) = 3$,
- $f_y(1, 1, 1) = 2(1) + (1)(1) = 3$,
- $f_z(1, 1, 1) = 2(1) + (1)(1) = 3$.

**Step 4**: Calculate $f(1, 1, 1)$:

- $f(1, 1, 1) = 1^2 + 1^2 + 1^2 + 1 \cdot 1 \cdot 1 = 4$.

**Step 5**: Construct the Linearization:

- $L(x, y, z) = 4 + 3(x - 1) + 3(y - 1) + 3(z - 1)$,
- Simplify: $L(x, y, z) = 3x + 3y + 3z - 5$.

**Conclusion for the Example:**

The linear approximation (linearization) of the function $f(x, y, z) = x^2 + y^2 + z^2 + xyz$ near the point $(1, 1, 1)$ is $L(x, y, z) = 3x + 3y + 3z - 5$. This approximation gives us a linear function that approximates $f$ close to $(1, 1, 1)$, essentially representing the tangent plane to the surface at that point.

Linearization is a powerful tool for approximating and understanding the local behavior of functions, especially when direct analysis is complicated due to the function's non-linear nature. It's widely used in various fields, including engineering, physics, and economics, to deal with complex systems by approximating them linearly near points of interest.