Chapter 15: Problem 22
Find a potential \(f\) if it exists. $$\mathbf{F}=y z \mathbf{i}+x z \mathbf{j}+\left(x y+z^{2}\right) \mathbf{k}$$
Short Answer
Expert verified
The potential function is \( f = xyz + \frac{z^3}{3} + C \).
Step by step solution
01
Understand the Vector Field
The vector field is given by \( \mathbf{F} = y z \mathbf{i}+x z \mathbf{j} + (x y + z^2) \mathbf{k} \). We aim to find a scalar potential function \( f \) such that \( abla f = \mathbf{F} \).
02
Check for Conservative Field
To ensure \( \mathbf{F} \) is conservative, verify if \( abla \times \mathbf{F} = \mathbf{0} \). Compute the components of \( abla \times \mathbf{F} \): \( \frac{\partial (xy + z^2)}{\partial y} - \frac{\partial (xz)}{\partial z} = x - x = 0 \)\( \frac{\partial (yz)}{\partial z} - \frac{\partial (xy + z^2)}{\partial x} = y - y = 0 \)\( \frac{\partial (xz)}{\partial x} - \frac{\partial (yz)}{\partial y} = z - z = 0 \)As all components are zero, the field is conservative and \( f \) exists.
03
Find the Scalar Potential
Since \( abla f = \mathbf{F} \), integrate component-wise:1. Integrate \( F_1 = yz \) with respect to \( x \) gives \( f = yzx + g(y, z) \).2. Integrate \( F_2 = xz \) with respect to \( y \) gives \( f = xyz + h(x, z) \).3. Integrate \( F_3 = xy + z^2 \) with respect to \( z \) gives \( f = xyz + \frac{z^3}{3} + k(x, y) \).From these, deduce \( f = xyz + \frac{z^3}{3} + C \) where \( C \) is a constant.
04
Verify the Potential
Verify \( abla f = \mathbf{F} \). Compute:\( \frac{\partial f}{\partial x} = yz \)\( \frac{\partial f}{\partial y} = xz \)\( \frac{\partial f}{\partial z} = xy + z^2 \)These match \( \mathbf{F} \), confirming \( f = xyz + \frac{z^3}{3} + C \) is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Scalar Potential
A scalar potential, often denoted as \( f \), is a function from which a vector field \( \mathbf{F} \) can be derived. It means that the vector field is the gradient of some scalar function, such that \( abla f = \mathbf{F} \).
A scalar potential is crucial because it simplifies the representation of a vector field in terms of a single function rather than components.
To obtain the scalar potential, integrate each component of \( \mathbf{F} \) with respect to its respective variable. This process will provide the function \( f \) where its gradient matches \( \mathbf{F} \).
It's important to note, \( f \) is not unique, and can vary by a constant \( C \), since \( abla C = 0 \).
Thus, a family's functions represent the scalar potential because a constant does not affect derivatives.
A scalar potential is crucial because it simplifies the representation of a vector field in terms of a single function rather than components.
- In the exercise, we aimed to find an \( f \) for our given field.
- The field \( \mathbf{F} = y z \mathbf{i} + x z \mathbf{j} + (xy + z^2) \mathbf{k} \).
- Each component corresponds to a partial derivative with respect to one variable.
To obtain the scalar potential, integrate each component of \( \mathbf{F} \) with respect to its respective variable. This process will provide the function \( f \) where its gradient matches \( \mathbf{F} \).
It's important to note, \( f \) is not unique, and can vary by a constant \( C \), since \( abla C = 0 \).
Thus, a family's functions represent the scalar potential because a constant does not affect derivatives.
Conservative Field
A field is deemed conservative if it is possible to express it as the gradient of some scalar potential function \( f \). This means that the field's curl, \( abla \times \mathbf{F} \), is zero everywhere in its domain.
In our exercise, to determine if \( \mathbf{F} \) is conservative, we computed \( abla \times \mathbf{F} \).
For this field, we calculated each component:
All components being zero confirms \( \mathbf{F} \) is indeed conservative.
Thus, we can confidently search for an appropriate scalar potential \( f \) for \( \mathbf{F} \).
- The term "conservative" suggests that the work done by a force derived from \( \mathbf{F} \) around any closed path is zero.
- Also signifies path independence of the work done between any two points in the field.
In our exercise, to determine if \( \mathbf{F} \) is conservative, we computed \( abla \times \mathbf{F} \).
For this field, we calculated each component:
- \( \frac{\partial (xy + z^2)}{\partial y} - \frac{\partial (xz)}{\partial z} = 0 \)
- \( \frac{\partial (yz)}{\partial z} - \frac{\partial (xy + z^2)}{\partial x} = 0 \)
- \( \frac{\partial (xz)}{\partial x} - \frac{\partial (yz)}{\partial y} = 0 \)
All components being zero confirms \( \mathbf{F} \) is indeed conservative.
Thus, we can confidently search for an appropriate scalar potential \( f \) for \( \mathbf{F} \).
Gradient
A gradient is a vector that represents the rate and direction of change in a scalar function. Given a scalar potential function \( f \), its gradient \( abla f \) is a vector field.
According to the exercise, \( abla f = \mathbf{F} \), revealing that our vector field \( \mathbf{F} \) is the gradient of scalar potential \( f \).
Let's break down how you derive \( \mathbf{F} \) from \( f \):
After obtaining \( f = xyz + \frac{z^3}{3} + C \), you verify the gradient by differentiating \( f \) in terms of \( x, y, \) and \( z \), getting back our original field.
This property shows how the gradient operation in vector calculus serves as a bridge that takes us from scalar fields to their corresponding vector representations.
- The gradient provides the directions and magnitudes of greatest increase of the scalar field.
- Each vector in the gradient points in the direction of steepest ascent from any given point in its domain.
According to the exercise, \( abla f = \mathbf{F} \), revealing that our vector field \( \mathbf{F} \) is the gradient of scalar potential \( f \).
Let's break down how you derive \( \mathbf{F} \) from \( f \):
- Start by integrating each component of \( \mathbf{F} \) along its corresponding axis.
- Ensure that these integrations match each part of the vector field \( \mathbf{F} = y z \mathbf{i} + x z \mathbf{j} + (xy + z^2) \mathbf{k} \).
After obtaining \( f = xyz + \frac{z^3}{3} + C \), you verify the gradient by differentiating \( f \) in terms of \( x, y, \) and \( z \), getting back our original field.
This property shows how the gradient operation in vector calculus serves as a bridge that takes us from scalar fields to their corresponding vector representations.