Problem 18 - Calculus for Biology and Medicine

Question

Find the linearization of \(f(x, y)\) at the indicated point \(\left(x_{0}, y_{0}\right) .\) \(f(x, y)=2 x y ;(1,-1)\)

Step-by-Step Solution

Answer

The linearization of the function is \(L(x, y) = -2x + 2y + 2.\)

Step 1: Understand the Linearization Formula

The linearization of a function \(f(x, y)\) at a point \((x_0, y_0)\) is given by the formula: \[ L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0). \] Here, \(f_x\) and \(f_y\) represent the partial derivatives of \(f\) with respect to \(x\) and \(y\) respectively.

Step 2: Calculate \(f(x_0, y_0)\)

Substitute \(x_0 = 1\) and \(y_0 = -1\) into the function. \[ f(1, -1) = 2 \times 1 \times (-1) = -2. \] This value is used as the constant term in the linear approximation.

Step 3: Find the Partial Derivative with Respect to \(x\)

Calculate \(f_x(x, y)\), the partial derivative of \(f\) with respect to \(x\): \[ f_x(x, y) = \frac{\partial}{\partial x}(2xy) = 2y. \] Evaluate this at \((x_0, y_0)\): \[ f_x(1, -1) = 2(-1) = -2. \]

Step 4: Find the Partial Derivative with Respect to \(y\)

Calculate \(f_y(x, y)\), the partial derivative of \(f\) with respect to \(y\): \[ f_y(x, y) = \frac{\partial}{\partial y}(2xy) = 2x. \] Evaluate this at \((x_0, y_0)\): \[ f_y(1, -1) = 2(1) = 2. \]

Step 5: Form the Linearization Function

Substitute \(f(x_0, y_0)\), \(f_x(x_0, y_0)\), and \(f_y(x_0, y_0)\) into the linearization formula: \[ L(x, y) = -2 + (-2)(x - 1) + 2(y + 1). \] Simplify the expression: \[ L(x, y) = -2 - 2x + 2 + 2y + 2 = -2x + 2y + 2. \]

Key Concepts

Partial Derivatives

Partial derivatives are an essential part of analyzing multivariable functions. They represent how a function changes as each individual variable changes, treating all other variables as constant. For a given function like \(f(x, y)\), partial derivatives examine the function's sensitivity to changes in just one variable, either \(x\) or \(y\).

• The partial derivative of \(f\) with respect to \(x\) is denoted as \(f_x\). To find it, you differentiate the function while treating \(y\) as a constant. In mathematical terms: \[ f_x(x, y) = \frac{\partial}{\partial x}(f) \]
• Similarly, the partial derivative with respect to \(y\) is \(f_y\). Here, \(x\) is kept constant: \[ f_y(x, y) = \frac{\partial}{\partial y}(f) \]

In our example, after partial differentiation, we found \(f_x\) equals \(2y\) and \(f_y\) equals \(2x\). These derivatives are crucial for building the linear approximation, or linearization, of a function.

Multivariable Functions

Multivariable functions are an extension of single-variable functions, incorporating two or more independent variables. These functions, such as \(f(x, y)\), require techniques from multivariable calculus to analyze.

• They can describe surfaces in three-dimensional space, with \(x\) and \(y\) specifying coordinates on a plane and \(f(x, y)\) providing the height or depth from that plane.
• When analyzing these functions, tools like partial derivatives become important to understand how the function changes in different directions.
• In the exercise, \(f(x, y) = 2xy\) is a simple bilinear example where the product of the two variables determines the value.

The goal of linearization in multivariable functions is to create a linear function that closely approximates \(f(x, y)\) near a specific point. This helps simplify complex functions into manageable, linear forms that are easier to work with and understand.

Differential Calculus

Differential calculus focuses on the concept of change and is fundamental for finding linear approximations of functions. When you linearize a function, you're effectively finding its tangent plane at a specific point.

• The linearization offers an approximation that simplifies calculations and predictions near the point of interest by replacing the function with its tangent plane.
• The formula used for linearization: \[ L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \]allows us to construct this linear model using partial derivatives.
• The tangent plane is significant because it best estimates the function's surface in a small area around \((x_0, y_0)\). Beyond this immediate area, however, the approximation may not hold.

In our exercise, the linearization was calculated at the point \((1, -1)\), providing a clear, simplified equation \(L(x, y) = -2x + 2y + 2\), which is easier to interpret and use than the original complex surface.