Problem 10 - Thomas Calculus

Question

In Exercises \(7-12,\) find a potential function \(f\) for the field \(\mathbf{F}\). $$\mathbf{F}=(y \sin z) \mathbf{i}+(x \sin z) \mathbf{j}+(x y \cos z) \mathbf{k}$$

Step-by-Step Solution

Answer

The potential function is \(f(x, y, z) = xy \sin z + C\), where \(C\) is a constant.

Step 1: Understand the Problem

We are given a vector field \(\mathbf{F} = (y \sin z) \mathbf{i} + (x \sin z) \mathbf{j} + (xy \cos z) \mathbf{k}\). We need to find a potential function \(f(x, y, z)\) such that \(abla f = \mathbf{F}\), meaning \(\frac{\partial f}{\partial x} = y \sin z\), \(\frac{\partial f}{\partial y} = x \sin z\), and \(\frac{\partial f}{\partial z} = xy \cos z\).

Step 2: Integrate with Respect to x

Integrate \(\frac{\partial f}{\partial x} = y \sin z\) with respect to \(x\). This gives \(f(x, y, z) = xy \sin z + g(y, z)\) where \(g(y,z)\) is an arbitrary function of \(y\) and \(z\) alone.

Step 3: Find g(y,z) by Integrating with Respect to y

Next, use \(\frac{\partial f}{\partial y} = x \sin z\) and substitute \( f(x, y, z) = xy \sin z + g(y, z) \). Partially differentiate this with respect to \(y\), getting \(x \sin z + \frac{\partial g}{\partial y} = x \sin z\). Thus, \(\frac{\partial g}{\partial y} = 0\) implies \(g(y, z) = h(z)\), where \(h\) is a function of \(z\) alone.

Step 4: Determine h(z) by Integrating with Respect to z

Use \(\frac{\partial f}{\partial z} = xy \cos z\) with \( f(x, y, z) = xy \sin z + h(z) \). When differentiating \(xy \sin z + h(z)\) with respect to \(z\), we get \(xy \cos z + \frac{dh}{dz} = xy \cos z\). Thus, \(\frac{dh}{dz} = 0\) implies \(h(z) = C\), where \(C\) is a constant.

Step 5: Write the Potential Function

Combine the results from the earlier steps to write the potential function, \(f(x, y, z) = xy \sin z + C\). This function satisfies all conditions from \(abla f = \mathbf{F}\).

Key Concepts

Vector Fields

Vector fields are an essential concept in vector calculus, and they play a critical role in understanding the behavior of scalar fields and multivariable functions.
A vector field is basically a function that attaches a vector to every point in space.

• Typically expressed as a combination of the unit vectors \ \(\mathbf{i} \ \), \ \(\mathbf{j} \ \), and \ \(\mathbf{k} \ \), representing the directions along the x, y, and z axes.
• This allows us to describe the field as \ \( \mathbf{F}(x, y, z) = P(x, y, z) \mathbf{i} + Q(x, y, z) \mathbf{j} + R(x, y, z) \mathbf{k} \ \).

In our given problem, the vector field is \ \( \mathbf{F} = (y \sin z) \mathbf{i} + (x \sin z) \mathbf{j} + (xy \cos z) \mathbf{k} \ \).
This structure helps to visualize how vectors point and vary across the space, providing a powerful tool for mathematical and physical modeling.

Potential Functions

A potential function, denoted by \ \( f(x, y, z) \ \), is a scalar field from which a vector field can be derived.
If a vector field \ \( \mathbf{F} \ \) is the gradient of some scalar function \ \( f \ \), that function is called the potential function of \ \( \mathbf{F} \ \).

• The relationship is described by \ \( abla f = \mathbf{F} \ \), meaning the vector field is conservative.
• Conservative fields have the path-independent movement property.
• In this problem, finding \ \( f(x, y, z) \ \) requires integrating components of the vector field.

The solution to our exercise yields \ \( f(x, y, z) = xy \sin z + C \ \) as the potential function.
Such a function helps simplify the analysis of vector fields and the associated physical processes.

Gradient

The gradient is a pivotal concept in vector calculus, used to convert a scalar field into a vector field.
For a scalar function \ \( f(x, y, z) \ \), the gradient is denoted by \ \( abla f \ \).

• This operation results in a vector comprising the partial derivatives of \ \( f \ \): \ \( abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) \ \).
• The gradient points in the direction of the greatest rate of increase of the function.
• Its magnitude indicates the steepness of the slope.

In our vector field \ \( \mathbf{F} \ \), each component corresponds to a partial derivative, proving that \ \( \mathbf{F} \ \) is indeed the gradient of some function, precisely what we've found by solving for \ \( f \ \).
Being able to determine this demonstrates the field’s potential nature and significant properties.

Partial Derivatives

Partial derivatives are foundational in the study of multivariable functions, capturing how a function changes with respect to each variable independently.
They form the building blocks for gradients and vector fields.

• For function \ \( f(x, y, z) \ \), partial derivatives are noted as \ \( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \ \).
• These derivatives measure the function's rate of change in the specific direction of each coordinate axis.
• Partial derivatives are used in integrals to reconstruct potential functions from vector fields.

In our specific problem, we calculate these derivatives from each component of \ \( \mathbf{F} \ \) to integrate and find the potential function \ \( f(x, y, z) \ \).
This involves integration with respect to each variable while considering others as constants; an essential task for recognizing dependent and independent behavior among variables in a multivariable function.