Chapter 16: Problem 28
Find a potential function for \(\mathbf{F}\) $$\mathbf{F}=\left(e^{x} \ln y\right) \mathbf{i}+\left(\frac{e^{x}}{y}+\sin z\right) \mathbf{j}+(y \cos z) \mathbf{k}$$
Short Answer
Expert verified
The potential function is \( f(x, y, z) = e^x \ln y + y \sin z + C \).
Step by step solution
01
Understand the Problem
We need to find a potential function \( f(x, y, z) \) such that \( \mathbf{F} = abla f \), where \( \mathbf{F} = \left(e^{x} \ln y\right) \mathbf{i}+ \left(\frac{e^{x}}{y}+\sin z\right) \mathbf{j}+(y \cos z) \mathbf{k} \).
02
Check and Use Gradient Matching
The components of \( abla f \) are derivative of \( f \) with respect to \( x, y, z \). So, \( f_x = e^x \ln y \), \( f_y = \frac{e^x}{y} + \sin z \), \( f_z = y \cos z \).
03
Integrate with respect to \( x \)
Integrate \( f_x = e^x \ln y \) with respect to \( x \) to get \( f(x, y, z) = e^x \ln y + g(y, z) \) where \( g(y, z) \) is the function of integration constants dependent on \( y \) and \( z \).
04
Use \( f_y \) to solve for \( g(y, z) \)
Differentiate \( f = e^x \ln y + g(y, z) \) with respect to \( y \). Set equal to \( f_y = \frac{e^x}{y} + \sin z \). We get: \( \frac{e^x}{y} + g_y(y, z) = \frac{e^x}{y} + \sin z \). So, \( g_y = \sin z \).
05
Integrate \( g_y \) with Respect to \( y \)
Integrate \( g_y = \sin z \) with respect to \( y \) to get \( g(y, z) = y \sin z + h(z) \), where \( h(z) \) is a function of \( z \) alone.
06
Use \( f_z \) to solve for \( h(z) \)
Differentiate \( f = e^x \ln y + y \sin z + h(z) \) with respect to \( z \). Set equal to \( f_z = y \cos z \). We get: \( y \cos z + h'(z) = y \cos z \). Thus, \( h'(z) = 0 \), so \( h(z) = C \) where \( C \) is a constant.
07
Compile the Potential Function
Since \( h(z) = C \), the potential function is \( f(x, y, z) = e^x \ln y + y \sin z + C \). The function satisfies the conditions, hence it is the potential function for \( \mathbf{F} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient Field
In vector calculus, a gradient field is a special type of vector field. It represents the gradient of a function. Imagine a gradient field as a map of slopes.
Each point in the field has a direction and magnitude, stemming from changes in a function's value over space. The gradient points in the direction of greatest increase.
To better comprehend this, think of a hill. In a physical analogy, the gradient field would show you the steepest path up the hill at every point.
Each point in the field has a direction and magnitude, stemming from changes in a function's value over space. The gradient points in the direction of greatest increase.
To better comprehend this, think of a hill. In a physical analogy, the gradient field would show you the steepest path up the hill at every point.
- The notation, often written as \( abla f \), represents the gradient of a function \( f \).
- In a function with variables \( x, y, \) and thus a vector field \( \textbf{F} = abla f \) equates to finding intense changes at each point \( (x, y) \).
Potential Function
A potential function offers a deeper view of a vector field. It's a function whose gradient is equal to the given vector field.
Think of it as the source function that generates the field. Finding a potential function is about retracing the steps to uncover this source.
The exercise given involves finding a potential function for the vector field \( \mathbf{F} = \left(e^{x} \ln y\right) \mathbf{i}+\left(\frac{e^{x}}{y}+\sin z\right) \mathbf{j}+(y \cos z) \mathbf{k} \). When we solve this, we look for a potential function \( f(x, y, z) \) such that the gradient, \( abla f \), perfectly matches \( \mathbf{F} \).
Think of it as the source function that generates the field. Finding a potential function is about retracing the steps to uncover this source.
The exercise given involves finding a potential function for the vector field \( \mathbf{F} = \left(e^{x} \ln y\right) \mathbf{i}+\left(\frac{e^{x}}{y}+\sin z\right) \mathbf{j}+(y \cos z) \mathbf{k} \). When we solve this, we look for a potential function \( f(x, y, z) \) such that the gradient, \( abla f \), perfectly matches \( \mathbf{F} \).
- This requires integrating the components of \( \mathbf{F} \) to find corresponding pieces of \( f \).
- Piece by piece, the potential function builds up to form a solution which is \( f(x, y, z) = e^x \ln y + y \sin z + C \), where \( C \) is a constant.
Vector Field
A vector field is a function that assigns a vector to each point in space. It can be visualized as a collection of arrows, where each arrow has a direction and magnitude depending on the position in the space.
Vector fields are crucial for modeling a variety of physical phenomena such as fluid flow, electromagnetic fields, and gravitational forces. For example, in the exercise, the vector field \( \mathbf{F} \) is represented as a combination of components along the \( x, y, \) and \( z \) axes.
Key aspects include:
Vector fields are crucial for modeling a variety of physical phenomena such as fluid flow, electromagnetic fields, and gravitational forces. For example, in the exercise, the vector field \( \mathbf{F} \) is represented as a combination of components along the \( x, y, \) and \( z \) axes.
Key aspects include:
- The field \( \mathbf{F} \) in our exercise combines distinct functions like \( e^x \ln y, \frac{e^x}{y}, \) and \( y \cos z \).
- Each component captures different variations, with \( i, j, \) and \( k \) denoting the unit vectors along the respective axes.
- Exploring these components individually gives insight into how the field behaves at any given point in space.