Problem 32 - Calculus: Early Transcendental Functions

Question

Determine whether or not the vector field is conservative. If it is, find a potential function. $$\left(z^{2}+2 x y, x^{2}-z, 2 x z-1\right\rangle$$

Step-by-Step Solution

Answer

Yes, the vector field is conservative and a potential function is \( f(x, y, z) = z^{2}x + x^{2}y + xz^{2} - yz - z \).

Step 1: Compute the Curl of the Vector Field

The curl of a vector field \( F = (z^{2}+2 x y, x^{2}-z, 2 x z-1) \) is given by the determinant of the following 3x3 matrix: \[\nabla \times F = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\z^{2}+2xy & x^{2}-z & 2xz-1 \\end{vmatrix}\]

Step 2: Compute the Determinant

Expanding the determinant along the top row, we get \[\nabla \times F = \mathbf{i} \begin{vmatrix}\frac{\partial}{\partial y} & \frac{\partial}{\partial z}\x^{2}-z & 2xz-1 \\end{vmatrix} - \mathbf{j} \begin{vmatrix}\frac{\partial}{\partial x} & \frac{\partial}{\partial z}\z^{2}+2xy & 2xz-1 \\end{vmatrix} + \mathbf{k} \begin{vmatrix}\frac{\partial}{\partial x} & \frac{\partial}{\partial y}\z^{2}+2xy & x^{2}-z \\end{vmatrix}\]

Step 3: Compute the Partial Derivatives

Computing the partial derivatives and subtracting as necessary, we find that \[\nabla \times F = (2z - 2z) \mathbf{i} + (2x - 2x) \mathbf{j} + (2x - 2x) \mathbf{k} = 0\], which means that the vector field is conservative.

Step 4: Find the Potential Function

To find a potential function \( f \) for the conservative vector field \( F \), we solve the following system of partial differential equations: \[\begin{align*}\frac{\partial f}{\partial x} &= z^{2}+2xy \\frac{\partial f}{\partial y} &= x^{2}-z \\frac{\partial f}{\partial z} &= 2xz-1\end{align*}\]Integrating the first equation with respect to x, the second equation with respect to y, and the third equation with respect to z, we find that a potential function is \( f(x, y, z) = z^{2}x + x^{2}y + xz^{2} - yz - z \).

Key Concepts

Conservative Vector Fields

A conservative vector field is one where the line integral between two points is independent of the path taken. In simpler terms, if you can move around in the field and always return to the same point without gaining or losing energy, then the field is conservative.
To determine if a vector field is conservative, you need to check its curl. If the curl is zero everywhere, then the vector field is conservative. This was exactly what was done in the original problem:

• A vector field is considered conservative when \( abla \times \mathbf{F} = 0 \).
• Checking this involves taking partial derivatives and applying them in a cross-product-like determinant calculation.

This is crucial because it tells us that such vector fields have potential functions, meaning we can represent the field as the gradient of some scalar function.

Potential Function

A potential function, often noted as \`f(x, y, z)\`, is a scalar function that helps in describing the vector field when it is conservative. The gradient of this scalar function gives you the original vector field back.
Finding a potential function involves integrating the components of the vector field with respect to their respective variables. It sounds complex, but let's break it down:

• For the given vector field, find \( f(x, y, z) \) such that \( abla f = \mathbf{F} \).
• This typically involves solving a set of partial differential equations derived from the components of the vector field as demonstrated in the exercise.

In general, knowing whether a potential function exists and finding it can be immensely beneficial, especially in physics, as it helps in simplifying problems related to force fields and energy.

Curl of a Vector Field

The curl of a vector field essentially tells you how much the vector field 'rotates' around each point. It's an important concept in vector calculus, especially in understanding the nature of the vector field.
The mathematical operation involves calculating the determinant of a matrix constructed from both the vector field components and partial derivatives, as shown in the solution.

• If the curl is zero (i.e., \( abla \times \mathbf{F} = 0 \)), the vector field has no circular rotation and is deemed conservative.
• A non-zero curl indicates the presence of rotation, and thus, the field is non-conservative.

This characteristic helps in determining whether certain operations or theorems, like Stokes' theorem or Green's theorem, can be applied to the field. In summary, understanding the curl is invaluable for analyzing the behavior of topological and vector fields.