Problem 32
Question
Find the Jacobi matrix for each given function. $$ \mathbf{f}(x, y)=\left[\begin{array}{c} (x-y)^{2} \\ \sin (x-y) \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The Jacobi matrix is \( \begin{bmatrix} 2(x-y) & -2(x-y) \\ \cos(x-y) & -\cos(x-y) \end{bmatrix} \).
1Step 1: Understand the Function
The function \( \mathbf{f}(x, y) \) is a vector-valued function given by two components: \( f_1(x, y) = (x-y)^2 \) and \( f_2(x, y) = \sin(x-y) \). We want to find the Jacobi matrix, which consists of the partial derivatives of these components.
2Step 2: Compute Partial Derivatives of \( f_1 \)
Find the partial derivatives of \( f_1(x, y) = (x-y)^2 \):- \( \frac{\partial f_1}{\partial x} = 2(x-y) \)- \( \frac{\partial f_1}{\partial y} = -2(x-y) \)
3Step 3: Compute Partial Derivatives of \( f_2 \)
Find the partial derivatives of \( f_2(x, y) = \sin(x-y) \):- \( \frac{\partial f_2}{\partial x} = \cos(x-y) \)- \( \frac{\partial f_2}{\partial y} = -\cos(x-y) \)
4Step 4: Form the Jacobi Matrix
The Jacobi matrix \( J \) is formed by arranging the partial derivatives into a matrix. Hence, the matrix is:\[J = \begin{bmatrix}2(x-y) & -2(x-y) \\cos(x-y) & -\cos(x-y)\end{bmatrix}\]
Key Concepts
Vector-Valued FunctionPartial DerivativesMatrix Computation
Vector-Valued Function
Understanding vector-valued functions is crucial as they are a foundational concept in multivariable calculus and linear algebra. A vector-valued function takes in one or more variables (in our case, the variables are \( x \) and \( y \)) and returns a vector. In the given problem, the function \( \mathbf{f}(x, y) \) is defined as:
Each component can represent a different physical quantity or a mathematical function mapped from input variables into a higher dimensional space, making vector-valued functions extremely useful in various applications such as modeling forces in physics or evaluating changes in economics.
- \( f_1(x, y) = (x-y)^2 \)
- \( f_2(x, y) = \sin(x-y) \)
Each component can represent a different physical quantity or a mathematical function mapped from input variables into a higher dimensional space, making vector-valued functions extremely useful in various applications such as modeling forces in physics or evaluating changes in economics.
Partial Derivatives
Partial derivatives are used to understand how a multivariable function changes as one of its input variables changes, while keeping the other variables constant. For a vector-valued function like \( \mathbf{f}(x, y) \), we compute the partial derivatives for each component with respect to each variable.
Let's examine \( f_1(x, y) = (x-y)^2 \):
Let's examine \( f_1(x, y) = (x-y)^2 \):
- Its partial derivative with respect to \( x \) is found by treating \( y \) as a constant: \( \frac{\partial f_1}{\partial x} = 2(x-y) \)
- Its partial derivative with respect to \( y \) is found by treating \( x \) as a constant: \( \frac{\partial f_1}{\partial y} = -2(x-y) \)
- The partial derivative with respect to \( x \) is: \( \frac{\partial f_2}{\partial x} = \cos(x-y) \)
- And with respect to \( y \), it is: \( \frac{\partial f_2}{\partial y} = -\cos(x-y) \)
Matrix Computation
Matrix computation is a method to arrange data or functions in a structured way that facilitates easy calculations and manipulations. When constructing the Jacobi matrix, we use the partial derivatives we obtained from each function component. This matrix is crucial in linear algebra because it provides a linear approximation of the vector-valued function near a given point.
To form the Jacobi matrix \( J \) for the function \( \mathbf{f}(x, y) \), we organize the partial derivatives into rows corresponding to each function component:
This matrix is particularly useful in multivariable calculus and optimization, providing key insights into the behavior of the function near specific points.
To form the Jacobi matrix \( J \) for the function \( \mathbf{f}(x, y) \), we organize the partial derivatives into rows corresponding to each function component:
- First row corresponds to \( f_1 \): \([\frac{\partial f_1}{\partial x}, \frac{\partial f_1}{\partial y}] = [ 2(x-y), -2(x-y) ]\)
- Second row corresponds to \( f_2 \): \([\frac{\partial f_2}{\partial x}, \frac{\partial f_2}{\partial y}] = [ \cos(x-y), -\cos(x-y) ]\)
This matrix is particularly useful in multivariable calculus and optimization, providing key insights into the behavior of the function near specific points.
Other exercises in this chapter
Problem 32
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