Problem 116 - Calculus Volume 3

Question

For the following exercises, determine whether the vector field is conservative and, if so, find a potential function. $$ \mathbf{F}(x, y)=(12 x y) \mathbf{i}+6\left(x^{2}+y^{2}\right) \mathbf{j} $$

Step-by-Step Solution

Answer

The vector field is conservative. Potential function: \( f(x, y) = 6x^2 y + 2y^3 + C \).

Step 1: Verify the Conservative Condition

For a vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \) to be conservative, the partial derivatives must satisfy \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). In this case, \( P(x, y) = 12xy \) and \( Q(x, y) = 6(x^2 + y^2) \). \( \frac{\partial P}{\partial y} = 12x \) and \( \frac{\partial Q}{\partial x} = 12x \). Since these are equal, the vector field is conservative.

Step 2: Set Up the Potential Function

Since the vector field is conservative, there exists a potential function \( f(x, y) \) such that \( abla f = \mathbf{F} \). Thus, \( \frac{\partial f}{\partial x} = P(x, y) = 12xy \) and \( \frac{\partial f}{\partial y} = Q(x, y) = 6(x^2 + y^2) \).

Step 3: Integrate with Respect to x

To find \( f(x, y) \), first integrate \( P(x, y) = 12xy \) with respect to \( x \):\[ f(x, y) = \int 12xy \, dx = 6x^2 y + g(y) \]where \( g(y) \) is an arbitrary function of \( y \).

Step 4: Differentiate with Respect to y

Differentiate \( f(x, y) = 6x^2 y + g(y) \) with respect to \( y \) to ensure consistency with \( \frac{\partial f}{\partial y} = Q(x, y) \):\[ \frac{\partial f}{\partial y} = 6x^2 + g'(y) \equiv 6(x^2 + y^2) \]Thus, \( g'(y) = 6y^2 \).

Step 5: Integrate with Respect to y

Integrate \( g'(y) = 6y^2 \) with respect to \( y \):\[ g(y) = \int 6y^2 \, dy = 2y^3 + C \] where \( C \) is an arbitrary constant.

Step 6: Combine Results for the Potential Function

Combine the results from the integrals to write the potential function \( f(x, y) \):\[ f(x, y) = 6x^2 y + 2y^3 + C \]

Key Concepts

Potential Function

In vector calculus, a potential function is a scalar function whose gradient yields a given vector field. This concept is crucial in determining whether a vector field is conservative.
For a vector field to be conservative, there must exist a potential function such that the vector field can be expressed as the gradient of this function. In simpler terms, any path you take between two points in a conservative field will have the same "energy" cost, making the computations of line integrals path-independent.
In the exercise, since the vector field \( \mathbf{F}(x, y) \) was shown to be conservative, we sought a potential function \( f(x, y) \). The steps involved integrating components of the vector field to reconstruct this potential function, ensuring that its gradient returns the original vector field components.

Partial Derivatives

Partial derivatives play a key role in determining whether a vector field is conservative and in reconstructing potential functions.
They measure how a function changes as one of its input variables changes, holding other input variables constant.
In the context of vector fields and potential functions, we use partial derivatives to match components of the gradient of the potential function with those of the vector field.

• Given \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), the condition for conservativeness is \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \).
• If this condition holds, then there exists a potential function \( f(x, y) \) such that \( abla f = \mathbf{F} \).

In our exercise, calculating \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \) demonstrated that the vector field was indeed conservative.

Integration

Integration is the process of finding the antiderivative of a function. It allows us to determine the accumulated quantity over an interval.
In this context, integration helps build the potential function from its partial derivatives.
In the exercise, we found the potential function by integrating each component separately:

• Integrating \( P(x, y) = 12xy \) with respect to \( x \) resulted in a function that partially contributed to \( f(x, y) \).
• During this process, we added an arbitrary function \( g(y) \), since it accounts for any dependencies not included in \( P(x, y) \).

The solution of integration in this case involved solving simple polynomial antiderivatives, showcasing how integration functions to reconstruct parts of the potential.

Differentiation

Differentiation, essentially the reverse of integration, involves finding how a function changes at each point, often yielding a rate of change.
In solving for a potential function, differentiation ensures consistency between different parts of the function.
After integrating to find \( f(x, y) = 6x^2 y + g(y) \), we differentiated with respect to \( y \) to validate correctness and find the unknown function \( g(y) \).

• We differentiated \( f(x, y) \) to ensure it aligned with our given \( \frac{\partial f}{\partial y} = Q(x, y) \).
• This step forced \( g(y) \) to be specified as \( 2y^3 \), ensuring that the gradient \( abla f \) matched the vector field \( \mathbf{F}(x, y) \).

Differentiation solidifies the correctness of the potential function by aligning derived expressions with the original vector field conditions.