Chapter 10: Problem 35
Find the Jacobi matrix for each given function. $$ \mathbf{f}(x, y)=\left[\begin{array}{c} 2 x^{2} y-3 y+x \\ e^{x} \sin y \end{array}\right] $$
Short Answer
Expert verified
The Jacobi matrix is \( \begin{bmatrix} 4xy + 1 & 2x^2 - 3 \\ e^x \sin y & e^x \cos y \end{bmatrix} \).
Step by step solution
01
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.
02
Compute Partial Derivatives of \( f_1 \)
Calculate \( \frac{\partial f_1}{\partial x} \) and \( \frac{\partial f_1}{\partial y} \):- \( \frac{\partial f_1}{\partial x} = \frac{\partial}{\partial x} (2x^2y - 3y + x) = 4xy + 1 \)- \( \frac{\partial f_1}{\partial y} = \frac{\partial}{\partial y} (2x^2y - 3y + x) = 2x^2 - 3 \)
03
Compute Partial Derivatives of \( f_2 \)
Calculate \( \frac{\partial f_2}{\partial x} \) and \( \frac{\partial f_2}{\partial y} \):- \( \frac{\partial f_2}{\partial x} = \frac{\partial}{\partial x} (e^x \sin y) = e^x \sin y \)- \( \frac{\partial f_2}{\partial y} = \frac{\partial}{\partial y} (e^x \sin y) = e^x \cos y \)
04
Construct the Jacobi Matrix
The Jacobi matrix \( J \) is constructed using the partial derivatives obtained in Steps 2 and 3. It is organized as:\[ J = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix} = \begin{bmatrix} 4xy + 1 & 2x^2 - 3 \ e^x \sin y & e^x \cos y \end{bmatrix} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
- \( f_1(x, y) = 2x^2y - 3y + x \)
- \( f_2(x, y) = e^x \sin y \)
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:
- \( \frac{\partial f_1}{\partial x} = 4xy + 1 \)
- \( \frac{\partial f_1}{\partial y} = 2x^2 - 3 \)
- \( \frac{\partial f_2}{\partial x} = e^x \sin y \)
- \( \frac{\partial f_2}{\partial y} = e^x \cos y \)
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.