Problem 39 - Calculus for Biology and Medicine

Question

Find a linear approximation to each function \(f(x, y)\) at the indicated point. $$ \mathbf{f}(x, y)=\left[\begin{array}{c} e^{2 x-y} \\ \ln (2 x-y) \end{array}\right] \text { at }(1,1) $$

Step-by-Step Solution

Answer

The linear approximations are: \(L_1(x, y) = e(1+x-y)\) and \(L_2(x, y) = 2x-y-1\).

Step 1: Understand Linear Approximation

The linear approximation of a function at a point is a function that approximates the original function near that point. Specifically, for a function of two variables, it can be found using the formula: \( L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \), where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\), respectively.

Step 2: Calculate Partial Derivatives

First, calculate the partial derivatives of each component of \(\mathbf{f}(x, y)\). - For \(f_1(x, y) = e^{2x-y}\): - \(\frac{\partial f_1}{\partial x} = 2e^{2x-y}\) - \(\frac{\partial f_1}{\partial y} = -e^{2x-y}\)- For \(f_2(x, y) = \ln(2x-y)\): - \(\frac{\partial f_2}{\partial x} = \frac{2}{2x-y}\) - \(\frac{\partial f_2}{\partial y} = -\frac{1}{2x-y}\)

Step 3: Evaluate Partial Derivatives at the Point

Evaluate the partial derivatives at the given point \((1,1)\). - For \(f_1(x, y)\): - \(\frac{\partial f_1}{\partial x}(1, 1) = 2e^{1} = 2e\) - \(\frac{\partial f_1}{\partial y}(1, 1) = -e^{1} = -e\)- For \(f_2(x, y)\): - \(\frac{\partial f_2}{\partial x}(1, 1) = \frac{2}{1} = 2\) - \(\frac{\partial f_2}{\partial y}(1, 1) = -\frac{1}{1} = -1\)

Step 4: Function Value at the Point

Evaluate the function components at \((1, 1)\): - \(f_1(1, 1) = e^{2\cdot1-1} = e^{1} = e\)- \(f_2(1, 1) = \ln(2\cdot1-1) = \ln(1) = 0\)

Step 5: Formulate Linear Approximation

Form the linear approximation using the evaluated values and derivatives: - For \(f_1(x, y)\), the linear approximation is: \[ L_1(x, y) = e + 2e(x-1) - e(y-1) \]- For \(f_2(x, y)\), the linear approximation is: \[ L_2(x, y) = 0 + 2(x-1) - (y-1) \]

Key Concepts

Partial Derivatives

Partial derivatives are like the building blocks for understanding how multivariable functions behave in the neighborhood of a particular point. When you're dealing with a function that has more than one variable, each of these variables can influence the function's output. Therefore, we focus on one variable at a time, treating the others as constants, and observe how the function changes as we tweak the chosen variable. This process is what we call finding the partial derivative.
- To calculate a partial derivative, determine how the function changes with respect to one variable while holding the other variables constant.

• For example, if you have a function like \( f(x, y) = e^{2x - y} \), the partial derivative with respect to \( x \) will consider the changes in \( f \) as \( x \) shifts, thus \( \frac{\partial f}{\partial x} \), while treating \( y \) as a constant.
• Similarly, for the partial derivative with respect to \( y \), represented as \( \frac{\partial f}{\partial y} \), the focus is on changes in \( f \) due to shifts in \( y \), with \( x \) held steady.

Both of these derivatives give us a glimpse of the 'slope' of our function in different directions, providing critical insights for linear approximation and other applications.

Multivariable Functions

Multivariable functions are mathematical expressions that have more than one input variable. They are the backbone of many real-life phenomena because most processes depend on more than one factor. These functions can take the form \( f(x, y) \), \( g(x, y, z) \), and so on, with as many variables as needed.
One special aspect of multivariable functions is their ability to capture complex interactions between different factors.

• They help model systems where two or more quantities influence an output, such as in physical sciences, engineering, and economics.
• A common way to visualize these functions is through surfaces or hypersurfaces that represent the relationship between inputs and outputs.

By understanding multivariable functions, you can not only solve for a wide array of problems but also develop more robust models of real-world scenarios.

Linear Approximation and Multivariable Functions

Linear approximation provides a simplified way to understand these functions by approximating the function with a straight plane near a particular point. It's a first step toward comprehending how these complex functions behave in local parts of their domain.

Function Evaluation

Function evaluation involves plugging specific values into a function to see what output is produced. It's an essential skill because it lets you understand how a function reacts to particular inputs. When performing linear approximation, you often evaluate the function and its partial derivatives at a given point to create a simpler, local representation of the function.
- In the context of the exercise with \( \mathbf{f}(x, y) = [e^{2x - y}, \ln(2x - y)] \), evaluating the function at point \((1, 1)\) means you substitute \( x = 1 \) and \( y = 1 \) into both components of the function.

• For \( f_1(x, y) = e^{2x - y} \), evaluation at \( (1,1) \) gives \( e^{2 \cdot 1 - 1} = e^1 = e \).
• For \( f_2(x, y) = \ln(2x - y) \), evaluation at \( (1,1) \) gives \( \ln(2 \cdot 1 - 1) = \ln(1) = 0 \).

This evaluation is crucial for creating the linear approximation equation, which then helps in predicting the function's behavior near the chosen point.