Chapter 7: Problem 4
Überprüfen Sie das Vektorfeld $$ \mathbf{w}(x, y, z)=\left(\begin{array}{c} y z \cos (x y z)+2 x z \\ x z \cos (x y z)+2 y z^{2} \\ x y \cos (x y z)+x^{2}+2 y^{2} z \end{array}\right) $$ auf die Potentialfeldeigenschaft und berechnen Sie gegebenenfalls eine Stammfunktion.
Short Answer
Expert verified
The vector field is not conservative; thus, no potential function exists.
Step by step solution
01
Check for Conservativeness
The first step is to determine if the vector field \( \mathbf{w}(x, y, z) \) is conservative by checking if the curl of the vector field is zero. The vector field \( \mathbf{w}(x, y, z) \) has components \( w_1 = yz\cos(xyz) + 2xz \), \( w_2 = xz\cos(xyz) + 2yz^2 \), and \( w_3 = xy\cos(xyz) + x^2 + 2y^2z \). To check for conservative, find \( abla \times \mathbf{w} \), i.e., the curl of \( \mathbf{w} \). If the curl is zero, the field is conservative.
02
Calculate Partial Derivatives for Curl
Calculate the partial derivatives needed for the curl:- \( \frac{\partial w_3}{\partial y} = \frac{\partial}{\partial y}(xy\cos(xyz) + x^2 + 2y^2z) \)- \( \frac{\partial w_2}{\partial z} = \frac{\partial}{\partial z}(xz\cos(xyz) + 2yz^2) \)- \( \frac{\partial w_1}{\partial z} = \frac{\partial}{\partial z}(yz\cos(xyz) + 2xz) \)- \( \frac{\partial w_3}{\partial x} = \frac{\partial}{\partial x}(xy\cos(xyz) + x^2 + 2y^2z) \)- \( \frac{\partial w_2}{\partial x} = \frac{\partial}{\partial x}(xz\cos(xyz) + 2yz^2) \)- \( \frac{\partial w_1}{\partial y} = \frac{\partial}{\partial y}(yz\cos(xyz) + 2xz) \)
03
Compute the Curl
Compute \( abla \times \mathbf{w} \) using the partial derivatives:- \( \left(\frac{\partial w_3}{\partial y} - \frac{\partial w_2}{\partial z}\right), \left(\frac{\partial w_1}{\partial z} - \frac{\partial w_3}{\partial x}\right), \left(\frac{\partial w_2}{\partial x} - \frac{\partial w_1}{\partial y}\right) \).- If all components are zero, \( abla \times \mathbf{w} = \mathbf{0} \), meaning the field is conservative. Here, for the given vector field, \( abla \times \mathbf{w} eq \mathbf{0} \), hence the vector field is not conservative.
04
Conclude Potential Field Property
Since we have determined that the curl is not zero, the field \( \mathbf{w}(x, y, z) \) is not a conservative vector field. This means there is no potential function or "Stammfunktion" for \( \mathbf{w}(x, y, z) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Potential Function
The concept of a potential function is closely tied to conservative vector fields. A vector field is said to be conservative if there exists a scalar potential function, often denoted as \( \ ext{F}(x, y, z) \), such that the vector field is the gradient of this potential. In mathematical terms, that means \( abla \ ext{F} = \mathbf{w} \). This implies that the vector field can be expressed as the derivative of a single scalar function.
Being able to find a potential function is useful because it simplifies the analysis of the vector field. For example, computing work done along a path in a conservative field only involves evaluating the potential function at the endpoints of the path, not throughout the path.
If a vector field did have a potential function, it would mean that the circulation of the vector field around any closed path is zero. In practical terms, if you could find \( \ ext{F} \), it would have greatly simplified many physics and engineering calculations. However, for our vector field \( \mathbf{w}(x, y, z) \), no potential function exists as it is not conservative.
Being able to find a potential function is useful because it simplifies the analysis of the vector field. For example, computing work done along a path in a conservative field only involves evaluating the potential function at the endpoints of the path, not throughout the path.
If a vector field did have a potential function, it would mean that the circulation of the vector field around any closed path is zero. In practical terms, if you could find \( \ ext{F} \), it would have greatly simplified many physics and engineering calculations. However, for our vector field \( \mathbf{w}(x, y, z) \), no potential function exists as it is not conservative.
Curl of a Vector Field
The curl of a vector field is an important operator in vector calculus that helps determine the field's rotation. Specifically, the curl plays a crucial role in checking whether a vector field is conservative. To find the curl, you use the operation \( abla \times \mathbf{w} \), where \( abla \) is the del operator.
The curl of a vector field \( \mathbf{w}(x, y, z) = \langle w_1, w_2, w_3 \rangle \) is defined as:
For our vector field \( \mathbf{w}(x, y, z) \), the curl is not zero, indicating the field is not conservative. This means the field cannot be derived from a potential function and suggests the presence of rotation or non-zero circulation. Understanding the curl is pivotal when working with electromagnetism and fluid dynamics, where rotation is a fundamental characteristic.
The curl of a vector field \( \mathbf{w}(x, y, z) = \langle w_1, w_2, w_3 \rangle \) is defined as:
- \( \left( \frac{\partial w_3}{\partial y} - \frac{\partial w_2}{\partial z} \right), \)
- \( \left( \frac{\partial w_1}{\partial z} - \frac{\partial w_3}{\partial x} \right), \)
- \( \left( \frac{\partial w_2}{\partial x} - \frac{\partial w_1}{\partial y} \right) \).
For our vector field \( \mathbf{w}(x, y, z) \), the curl is not zero, indicating the field is not conservative. This means the field cannot be derived from a potential function and suggests the presence of rotation or non-zero circulation. Understanding the curl is pivotal when working with electromagnetism and fluid dynamics, where rotation is a fundamental characteristic.
Partial Derivatives
Partial derivatives are a backbone of multi-variable calculus and are integral to understanding vector fields. A partial derivative represents the rate of change of a function with respect to one variable while keeping others constant. They're essential for computing the curl of a vector field.
In the context of a vector field \( \mathbf{w}(x, y, z) \), each component \( w_1, w_2, \) and \( w_3 \) depending on multiple variables may need to be differentiated partially to find the curl.
Here's how they are applied:
In the context of a vector field \( \mathbf{w}(x, y, z) \), each component \( w_1, w_2, \) and \( w_3 \) depending on multiple variables may need to be differentiated partially to find the curl.
Here's how they are applied:
- For \( w_3 \) with respect to \( y \) given by \( \frac{\partial}{\partial y}(xy\cos(xyz) + x^2 + 2y^2z) \), you treat \( x \) and \( z \) as constants.
- Similarly, \( \frac{\partial w_2}{\partial z} \), \( \frac{\partial w_1}{\partial z} \), among others, are calculated by taking the derivative of the specific component concerning the variable of interest.