Chapter 16: Problem 28
Find an \(f\) so that \(\nabla f=\langle y z, x z, x y\rangle\), or explain why there is no such \(f\).
Short Answer
Expert verified
The potential function is \(f(x, y, z) = xyz\).
Step by step solution
01
Understand the Problem
We are given a vector field \(abla f = \langle y z, x z, x y \rangle\) and we need to find a function \(f(x, y, z)\) such that this is the gradient of \(f\). A vector field \(\mathbf{F} = \langle P, Q, R \rangle\) is a gradient field if there exists a scalar potential function \(f\) such that \(abla f = \mathbf{F}\).
02
Check Conditions for Solvability
For \(\mathbf{F} = \langle P, Q, R \rangle\) to be a gradient field, it must be conservative. This means that the following conditions must hold: \(\frac{\partial R}{\partial y} = \frac{\partial Q}{\partial z}\), \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\), and \(\frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}\). Substitute \(P = yz\), \(Q = xz\), and \(R = xy\) into these conditions.
03
Check Specific Conditions
Calculate the partial derivatives: \(\frac{\partial R}{\partial y} = x\), \(\frac{\partial Q}{\partial z} = x\), \(\frac{\partial P}{\partial y} = z\), \(\frac{\partial Q}{\partial x} = z\), \(\frac{\partial P}{\partial z} = y\), and \(\frac{\partial R}{\partial x} = y\). All conditions satisfy: \(x = x\), \(z = z\), and \(y = y\), confirming that \(\mathbf{F}\) is conservative.
04
Find Potential Function
Since these conditions are satisfied, integrate each component starting with \(P = yz\) with respect to \(x\) to find \(f(x, y, z) = x y z + g(y, z)\). Then integrate \(Q = xz\) with respect to \(y\), yielding \(f(x, y, z) = x y z + h(x, z)\). Finally, integrate \(R = xy\) with respect to \(z\), resulting in \(f(x, y, z) = x y z + k(x, y)\). Since all yield the same form \(f(x, y, z) = x y z + C\), where \(C\) is a constant, \(f(x, y, z) = x y z\) is a valid potential function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conservative Vector Fields
A vector field is called conservative if it can be expressed as the gradient of a scalar potential function. In simpler terms, this means there exists a function whose gradient matches the given vector field components. For a vector field \ \( \mathbf{F} = \langle P, Q, R \rangle \ \), it is considered conservative if it satisfies certain conditions related to its partial derivatives.These conditions are:
This characteristic can simplify many problems in physics and mathematics.
- \( \frac{\partial R}{\partial y} = \frac{\partial Q}{\partial z} \)
- \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \)
- \( \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x} \)
This characteristic can simplify many problems in physics and mathematics.
Potential Function
In the context of vector fields, a potential function is the original scalar field which yields the given vector field upon taking its gradient. For a conservative vector field \ \( \mathbf{F} = abla f \) \, the potential function \ \( f(x, y, z) \) is the function we aim to find.To determine the potential function, we typically:
- Integrate one component of the vector field with respect to its corresponding variable.
- Add a function of the remaining variables to account for the partial integration.
- Compare results from integrating different components to find the final form of the potential function.
Partial Derivatives
Partial derivatives are a key tool in determining whether a vector field is conservative and in finding a potential function. They measure how a function changes as one of its variables changes while all others are held constant. For a vector field \ \( \mathbf{F} = \langle P, Q, R \rangle \), the relevant partial derivatives are needed to check the conditions for conservativeness.For instance, with \ \( \mathbf{F} = \langle yz, xz, xy \rangle \):
- \( \frac{\partial R}{\partial y} = x \) matches \( \frac{\partial Q}{\partial z} = x \)
- \( \frac{\partial P}{\partial y} = z \) matches \( \frac{\partial Q}{\partial x} = z \)
- \( \frac{\partial P}{\partial z} = y \) matches \( \frac{\partial R}{\partial x} = y \)