Problem 125 - Calculus Volume 3

Question

For the following exercises, determine whether the given vector field is conservative and find a potential function. $$ \mathbf{F}(x, y)=\left(e^{x} \cos y\right) \mathbf{i}+6\left(e^{x} \sin y\right) \mathbf{j} $$

Step-by-Step Solution

Answer

The vector field is not conservative.

Step 1: Verify if the vector field is conservative

A vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \) is conservative if its curl is zero. Calculate \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \), where \( P(x, y) = e^{x} \cos y \) and \( Q(x, y) = 6e^{x} \sin y \).

Step 2: Calculate partial derivative of Q with respect to x

Compute \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(6e^{x} \sin y) = 6e^{x} \sin y \).

Step 3: Calculate partial derivative of P with respect to y

Compute \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(e^{x} \cos y) = -e^{x} \sin y \).

Step 4: Compare partial derivatives

Since \( \frac{\partial Q}{\partial x} = 6e^{x} \sin y \) and \( \frac{\partial P}{\partial y} = -e^{x} \sin y \), these are not equal. Therefore, the vector field \( \mathbf{F}(x, y) \) is not conservative.

Key Concepts

Partial Derivatives

Partial derivatives are essential building blocks in calculus, particularly when dealing with functions of several variables. When we take a partial derivative of a function concerning one variable, we treat all other variables as constants. This is a crucial process that allows us to understand how a function changes along one dimension while keeping the others fixed.

In the context of vector fields, partial derivatives are vital to determining properties like conservativeness. For instance, if we have a vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), partial derivatives are used to compute \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \).

These derivatives measure how the field's component functions \( Q \) and \( P \) change in response to varying the \( x \) and \( y \) variables, respectively. In the exercise, these calculations are important for finding if a vector field is conservative. If \( \frac{\partial Q}{\partial x} eq \frac{\partial P}{\partial y} \), we conclude the vector field is not conservative.

Potential Function

A potential function is a scalar function whose gradient yields a given vector field. In simpler terms, it helps us determine if a vector field is conservative by checking if such a potential function exists.

Consider a vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \). If there is a potential function \( f(x, y) \) such that its gradient \( abla f(x, y) \) matches \( \mathbf{F} \), then the field is conservative:

• \( abla f(x, y) = \Big(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\Big) = \mathbf{F}(x, y) \)
• This implies \( \frac{\partial f}{\partial x} = P(x, y) \) and \( \frac{\partial f}{\partial y} = Q(x, y) \).

In the scenario of the original exercise, we seek to find such a function to verify the conservativeness of \( \mathbf{F} \). When the partial derivatives did not equate, it indicated no potential function exists, and thus, \( \mathbf{F} \) is not conservative.

Curl of a Vector Field

The curl of a vector field is a vector operation that describes the rotation or the swirling strength of a field around a point. In two dimensions, a key property of conservative fields relates to the curl, which should be zero or absent.

For a two-dimensional vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), the curl is determined by calculating:

• \( \text{curl} \mathbf{F} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \)

If this resultant is zero, the vector field is conservative, and conversely, if it's non-zero, the field isn't conservative. This makes curl a powerful tool for examining the nature of vector fields.

In the exercise, calculating the curl showed \( \frac{\partial Q}{\partial x} eq \frac{\partial P}{\partial y} \), confirming that the vector field is not conservative because its curl is not zero, thus indicating a lack of rotational invariance around any point.