Chapter 16: Problem 10
Find a potential function \(f\) for the field \(\mathbf{F}.\) \(\mathbf{F}=(y \sin z) \mathbf{i}+(x \sin z) \mathbf{j}+(x y \cos z) \mathbf{k}\)
Short Answer
Expert verified
The potential function is \( f(x, y, z) = xy \sin z \).
Step by step solution
01
Understanding the Potential Function
To find a potential function \( f(x, y, z) \) for a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), we need to find a scalar function \( f \) such that \( abla f = \mathbf{F} \), where \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (P, Q, R) \).
02
Integrate with respect to x
Given \( P = y \sin z \), integrate function with respect to \( x \): \[ f(x, y, z) = \int y \sin z \, dx = xy \sin z + g(y, z) \]where \( g(y, z) \) is an arbitrary function of \( y \) and \( z \).
03
Differentiate with respect to y
Given the integrated function \( f(x, y, z) = xy \sin z + g(y, z) \), differentiate with respect to \( y \):\[ \frac{\partial f}{\partial y} = x \sin z + \frac{\partial g}{\partial y} = x \sin z \]Since this must match the component \( Q = x \sin z \), it implies:\[ \frac{\partial g}{\partial y} = 0 \]Thus, \( g(y, z) \) is actually \( g(z) \), a function of \( z \) only.
04
Integrate with respect to z
Now consider the term \( R = xy \cos z \). Integrate with respect to \( z \):\[ \frac{\partial f}{\partial z} = xy \cos z + \frac{dg}{dz} = xy \cos z \]This implies that \( \frac{dg}{dz} = 0 \), thus \( g(z) \) is a constant. Set \( g(z) = C \).
05
Write the Potential Function
Combine the results: the potential function, up to a constant, is:\[ f(x, y, z) = xy \sin z + C \]The field \( \mathbf{F} \) is conservative, and its potential function is \( f(x, y, z) = xy \sin z \) (ignoring the constant \( C \) which is arbitrary in potential functions).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Field
In mathematics, a vector field is a function that assigns a vector to each point in a subset of space. For example, the vector field in our exercise is given as \( \mathbf{F}(x, y, z) = (y \sin z) \mathbf{i} + (x \sin z) \mathbf{j} + (x y \cos z) \mathbf{k} \). This means that for each point \((x, y, z)\), there is a corresponding vector representing the field. Vector fields can describe various phenomena where direction and magnitude are essential, like the flow of a fluid or the force experienced by a particle in a force field.
- Vectors: These are arrows that have both a magnitude (how long they are) and a direction (which way they point).
- Points: In our case, every point in space \((x, y, z)\) has a different vector associated with it.
Scalar Function
A scalar function, sometimes simply called a scalar, assigns a single real number to each point in space. In the context of vector fields, a scalar function is vital because it helps us understand potential energy, temperatures, heights, etc., in physical systems. In our exercise, we aim to find a potential function \( f(x, y, z) \), which is a scalar function such that its gradient is equal to the vector field \( \mathbf{F} \). This means if we take the gradient of the potential function, we should get back our original field, thereby revealing the underlying scalar function.
- Physical interpretation: For potential energy, this means knowing the potential function describes how energy changes in space.
- Mathematical relation: The scalar function itself is unknown until we perform the calculations needed to identify it, as shown in the step-by-step solution.
Gradient
The gradient, denoted by \( abla f \), is a vector operation that takes a scalar function as input and produces a vector field as output. This vector field indicates the direction and rate of the fastest increase of the scalar function. For a function \( f(x, y, z) \), the gradient \( abla f \) is given by:\[abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\]In our task, we want the gradient to match the vector field \( \mathbf{F} \). This concept is central to identifying potential functions because if the gradient of \( f \) equals the vector field, then \( f \) is a potential function.
- Directional increase: The gradient shows not just any change, but the steepest rate of change.
- Conservative fields: Fields with potential functions, like our exercise’s vector field, are conservative, meaning path-independent in terms of movement across points.