Problem 35 - Calculus for Biology and Medicine

Question

Find the Jacobi matrix for each given function. $$ \mathbf{f}(x, y)=\left[\begin{array}{c} 2 x^{2} y-3 y+x \\ e^{x} \sin y \end{array}\right] $$

Step-by-Step Solution

Answer

The Jacobi matrix is \( \begin{bmatrix} 4xy + 1 & 2x^2 - 3 \\ e^x \sin y & e^x \cos y \end{bmatrix} \).

Step 1: Identify the Function Components

The function \( \mathbf{f}(x, y) \) is a vector-valued function with two components: \( f_1(x, y) = 2x^2y - 3y + x \) and \( f_2(x, y) = e^x \sin y \). We need to find the partial derivatives of these components with respect to \( x \) and \( y \) to construct the Jacobi matrix.

Step 2: Compute Partial Derivatives of \( f_1 \)

Calculate \( \frac{\partial f_1}{\partial x} \) and \( \frac{\partial f_1}{\partial y} \):- \( \frac{\partial f_1}{\partial x} = \frac{\partial}{\partial x} (2x^2y - 3y + x) = 4xy + 1 \)- \( \frac{\partial f_1}{\partial y} = \frac{\partial}{\partial y} (2x^2y - 3y + x) = 2x^2 - 3 \)

Step 3: Compute Partial Derivatives of \( f_2 \)

Calculate \( \frac{\partial f_2}{\partial x} \) and \( \frac{\partial f_2}{\partial y} \):- \( \frac{\partial f_2}{\partial x} = \frac{\partial}{\partial x} (e^x \sin y) = e^x \sin y \)- \( \frac{\partial f_2}{\partial y} = \frac{\partial}{\partial y} (e^x \sin y) = e^x \cos y \)

Step 4: Construct the Jacobi Matrix

The Jacobi matrix \( J \) is constructed using the partial derivatives obtained in Steps 2 and 3. It is organized as:\[ J = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix} = \begin{bmatrix} 4xy + 1 & 2x^2 - 3 \ e^x \sin y & e^x \cos y \end{bmatrix} \]

Key Concepts

Vector-Valued Functions

In mathematics, a vector-valued function is a function that takes in one or more variables and returns a vector. Simply put, it is a function where the output consists of several components, each of which can depend on multiple inputs. In the context of the original exercise, the function \( \mathbf{f}(x, y) \) is a vector-valued function. This means that for every pair of values \((x, y)\), the function outputs a vector with two components. For our function, these components are:

• \( f_1(x, y) = 2x^2y - 3y + x \)
• \( f_2(x, y) = e^x \sin y \)

Understanding vector-valued functions is crucial in multivariable calculus, as they are used to describe situations where multiple outcomes depend on a single set of inputs, like in physics and engineering. They are essential for modeling scenarios where different quantities change according to various conditions.

Partial Derivatives

Partial derivatives are a key concept when dealing with multivariable functions, as they reveal how a function changes as only one of its variables is varied, while others are held constant. For the given function components, the exercise involves calculating several partial derivatives, such as:

• \( \frac{\partial f_1}{\partial x} = 4xy + 1 \)
• \( \frac{\partial f_1}{\partial y} = 2x^2 - 3 \)
• \( \frac{\partial f_2}{\partial x} = e^x \sin y \)
• \( \frac{\partial f_2}{\partial y} = e^x \cos y \)

Calculating the partial derivatives allows us to measure how sensitive the function is with respect to changes in \(x\) or \(y\). This information is particularly useful for various practical applications, such as finding the slope of a surface at a given point, optimizing functions, and more. They provide foundational insights necessary for constructing the Jacobian matrix.

Jacobian Determinant

To understand the Jacobian determinant, we must first explore the Jacobian matrix, which was computed by using the partial derivatives from earlier. The Jacobian matrix of a vector-valued function like \( \mathbf{f}(x, y) \) is organized in such a way that each of its entry consists of a partial derivative:\[J = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix} = \begin{bmatrix} 4xy + 1 & 2x^2 - 3 \ e^x \sin y & e^x \cos y \end{bmatrix}\]The determinant of this matrix, known as the Jacobian determinant, can provide critical information about the function.

• It tells us whether a transformation defined by the vector-valued function is locally invertible at a point.
• A non-zero Jacobian determinant implies that the function is invertible around that point.
• The sign of the determinant can indicate the preservation of orientation in transformations.

Calculating the Jacobian determinant is particularly useful in various applications, including change of variables in multiple integrals, solving systems of nonlinear equations, and assessing the behavior and characteristics of maps and transformations.