Problem 37 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The linear approximation of \( \mathbf{f}(x, y) \) at \((1, 1)\) is \( \begin{bmatrix} 4x + 2y - 4 \\ -x - y + 3 \end{bmatrix} \).

Step 1: Understand the Linear Approximation Formula

The linear approximation of a vector function \( \mathbf{f}(x,y) \) at a point \( (a,b) \) is given by:\[\mathbf{f}(x,y) \approx \mathbf{f}(a,b) + \mathbf{J}_{\mathbf{f}}(a,b) \cdot \begin{bmatrix} x-a \ y-b \end{bmatrix}\]where \( \mathbf{J}_{\mathbf{f}}(a,b) \) is the Jacobian matrix of \( \mathbf{f} \) evaluated at \( (a,b) \). Our task is to find the Jacobian, evaluate the function \( \mathbf{f} \) at \( (1,1) \), and then use these results to estimate \( \mathbf{f}(x,y) \).

Step 2: Evaluate the Function at Point (1,1)

Calculate each component of \( \mathbf{f}(x, y) \) at the point \((1,1)\).1. For the first component: \( f_1(x,y) = 2x^2 y \), substitute \( x = 1 \) and \( y = 1 \): \[ f_1(1,1) = 2(1)^2(1) = 2 \].2. For the second component: \( f_2(x,y) = \frac{1}{xy} \), substitute \( x = 1 \) and \( y = 1 \): \[ f_2(1,1) = \frac{1}{1 \cdot 1} = 1 \].Thus, \( \mathbf{f}(1,1) = \begin{bmatrix} 2 \ 1 \end{bmatrix} \).

Step 3: Calculate the Jacobian Matrix

The Jacobian matrix \( \mathbf{J}_{\mathbf{f}}(x,y) \) is a matrix of the first partial derivatives of the vector function. For \( \mathbf{f}(x,y) = \begin{bmatrix} 2x^2y \ \frac{1}{xy} \end{bmatrix} \):- Partial derivatives for \( f_1(x,y) = 2x^2y \): \( \frac{\partial}{\partial x}(2x^2y) = 4xy \) and \( \frac{\partial}{\partial y}(2x^2y) = 2x^2 \).- Partial derivatives for \( f_2(x,y) = \frac{1}{xy} \): \( \frac{\partial}{\partial x}\left(\frac{1}{xy}\right) = -\frac{1}{x^2y} \) and \( \frac{\partial}{\partial y}\left(\frac{1}{xy}\right) = -\frac{1}{xy^2} \).Thus, the Jacobian matrix is:\[\mathbf{J}_{\mathbf{f}}(x,y) = \begin{bmatrix} 4xy & 2x^2 \ -\frac{1}{x^2y} & -\frac{1}{xy^2} \end{bmatrix}\].

Step 4: Evaluate the Jacobian at Point (1,1)

Substitute \( x=1 \) and \( y=1 \) into the Jacobian matrix:- First row: \([4(1)(1), 2(1)^2] = [4, 2]\).- Second row: \([-\frac{1}{1^2 \cdot 1}, -\frac{1}{1 \cdot 1^2}] = [-1, -1]\).Therefore,\[\mathbf{J}_{\mathbf{f}}(1,1) = \begin{bmatrix} 4 & 2 \ -1 & -1 \end{bmatrix}\].

Step 5: Combine Results for Linear Approximation

Use the linear approximation formula with \( \mathbf{f}(1,1) \) and \( \mathbf{J}_{\mathbf{f}}(1,1) \):\[\mathbf{f}(x,y) \approx \begin{bmatrix} 2 \ 1 \end{bmatrix} + \begin{bmatrix} 4 & 2 \ -1 & -1 \end{bmatrix} \begin{bmatrix} x-1 \ y-1 \end{bmatrix}\]Calculating this, you get:\[\mathbf{f}(x,y) \approx \begin{bmatrix} 2 \ 1 \end{bmatrix} + \begin{bmatrix} 4(x-1) + 2(y-1) \ -1(x-1) -1(y-1) \end{bmatrix} = \begin{bmatrix} 2 + 4x + 2y - 6 \ 1 - x - y + 2 \end{bmatrix}\]Therefore, the linear approximation is:\[\mathbf{f}(x,y) \approx \begin{bmatrix} 4x + 2y - 4 \ -x - y + 3 \end{bmatrix}\].

Key Concepts

Jacobian matrix

The Jacobian matrix is a vital concept in multivariable calculus, particularly when dealing with functions of multiple variables. When you have a vector function, or a function that outputs a vector, the Jacobian matrix helps in understanding how the function behaves near a certain point. It does this by summarizing all the partial derivative information in one neat package.

The Jacobian matrix of a vector function \( \mathbf{f}(x, y) \) consists of all the first-order partial derivatives of each component of \( \mathbf{f} \). It essentially captures how each component of the vector changes as you adjust the variables, \( x \) and \( y \).

For instance, if \( \mathbf{f}(x, y) \) has components \( f_1(x, y) \) and \( f_2(x, y) \), the Jacobian matrix \( \mathbf{J}_{\mathbf{f}}(x, y) \) would be:

• \[ \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix} \]

By evaluating this matrix at a specific point, such as \( (1, 1) \) in our exercise, you get a snapshot of how the function is behaving at that point. This matrix plays a crucial role in linear approximations because it provides the necessary derivative information to describe the tangent plane or hyperplane that best fits the function at that specific location.

Partial derivatives

Partial derivatives extend the concept of derivatives to functions of multiple variables. When dealing with functions of one variable, a derivative tells us how the function changes as you make small changes to the input variable. With multiple variables, a partial derivative focuses on how the function changes as we tweak one variable while holding the others constant.

Let's consider a function \( f(x, y) \). The partial derivative of \( f \) with respect to \( x \), denoted by \( \frac{\partial f}{\partial x} \), indicates the rate of change in \( f \) as only \( x \) changes, with \( y \) held constant. Similarly, \( \frac{\partial f}{\partial y} \) shows the rate of change as \( y \) changes, keeping \( x \) constant.

In our exercise:

• The partial derivative \( \frac{\partial}{\partial x}(2x^2y) = 4xy \) tells us how the first component, involving \( x \), behaves with changes in \( x \).
• Similarly, \( \frac{\partial}{\partial y}(2x^2y) = 2x^2 \) indicates the function's sensitivity to changes in \( y \).

These partial derivatives are key to forming the Jacobian matrix and thus crucial for computing linear approximations at any given point. They highlight the influence and contribution of each variable to the function's overall behavior.

Multivariable calculus

Multivariable calculus is an extension of calculus that deals with functions of more than one variable. It's a powerful tool that helps solve complex problems where single-variable calculus falls short. When you're dealing with a function like \( f(x, y) \), you're in the territory of multivariable calculus.

This field allows us to explore and analyze how functions behave in higher dimensions. For example, instead of working with curves, we can handle surfaces and multidimensional shapes. This has vast applications in fields like physics, engineering, and economics, where the behavior of systems often depends on multiple factors.

Key concepts in multivariable calculus include:

• Partial derivatives: As described earlier, these allow us to focus on the effect of each individual variable.
• Jacobian matrix: A matrix that summarizes all the partial derivatives for a vector function.
• Linear approximation: A technique for estimating complex functions with a linear equation, simplifying computation and analysis around a given point.

Understanding these concepts is essential in solving problems that involve complex relationships between multiple quantities, enhancing the ability to predict and model real-world systems effectively.