Problem 11 - Advanced Engineering Mathematics

Question

Determine whether the given vector field is a conservative field. If so, find a potential function \(\phi\) for \(\mathbf{F}\). $$ \mathbf{F}(x, y)=\left(4 x^{3} y^{3}+3\right) \mathbf{i}+\left(3 x^{4} y^{2}+1\right) \mathbf{j} $$

Step-by-Step Solution

Answer

The vector field is conservative, and the potential function is \( \phi(x,y) = x^4 y^3 + 3x + y + C \).

Step 1: Check Curl of the Vector Field

To determine if a vector field \( \mathbf{F} = P(x,y) \mathbf{i} + Q(x,y) \mathbf{j} \) is conservative, we need to check if its curl is zero. This means we compute \( abla \times \mathbf{F} \) and see if it's equal to zero.The curl in 2D is:\[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\]With \( P(x,y) = 4x^3 y^3 + 3 \) and \( Q(x,y) = 3x^4 y^2 + 1 \), compute the partial derivatives.\( \frac{\partial Q}{\partial x} = 12x^3 y^2 \)\( \frac{\partial P}{\partial y} = 12x^3 y^2 \)Since\[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 12x^3 y^2 - 12x^3 y^2 = 0,\]the curl is zero, indicating that the vector field \( \mathbf{F} \) is conservative.

Step 2: Find the Potential Function

Since the vector field is conservative, there exists a potential function \( \phi(x,y) \) such that \( abla \phi = \mathbf{F} \).Given \( abla \phi = \left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}\right) = \left(4x^3 y^3 + 3, 3x^4 y^2 + 1\right) \), we integrate separately with respect to \(x\) and \(y\).First, integrate \( P(x,y) = 4x^3 y^3 + 3 \) with respect to \( x \):\[\phi(x,y) = \int (4x^3 y^3 + 3) \, dx = x^4 y^3 + 3x + g(y)\]where \( g(y) \) is the function of integration with respect to \( y \).

Step 3: Determine the Function of Integration

Now, differentiate \( \phi(x,y) = x^4 y^3 + 3x + g(y) \) with respect to \( y \) to find \( \frac{\partial \phi}{\partial y} \):\[\frac{\partial \phi}{\partial y} = 3x^4 y^2 + g'(y)\]This must equal \( Q(x,y) = 3x^4 y^2 + 1 \). Thus,\[3x^4 y^2 + g'(y) = 3x^4 y^2 + 1\]Solving for \( g'(y) \), we get \( g'(y) = 1 \).Integrating with respect to \( y \), we find \( g(y) = y + C \), where \( C \) is a constant.

Step 4: Write the Potential Function

Now, substitute \( g(y) = y + C \) back into \( \phi(x,y) \):\[\phi(x,y) = x^4 y^3 + 3x + y + C\]This is the potential function for the vector field \( \mathbf{F} \).

Key Concepts

Potential Function

A potential function, often denoted by \( \phi(x, y) \), is a scalar function whose gradient (del operator applied) corresponds to a given vector field. When a vector field is conservative, it means that this vector field can be expressed as the gradient of the potential function. The process of finding the potential function involves integrating the components of the vector field.

• Why is this important? Identifying a potential function simplifies the analysis of the vector field, as potential functions can help in evaluating line integrals conveniently.
• Finding the potential function: Start by integrating the component of the vector field concerning its respective variable. For instance, integrate \( P(x, y) \) with respect to \( x \) and \( Q(x, y) \) with respect to \( y \).

In the problem given, after confirming that the vector field is conservative, we find the potential function through integration, which leads us to \( \phi(x,y) = x^4 y^3 + 3x + y + C \). Here, \( C \) denotes a constant of integration, necessary as antiderivatives are determined up to a constant.

Curl of Vector Field

The curl of a vector field in three dimensions is a vector operation that describes the infinitesimal rotation of the field at a given point. However, in two dimensions, the curl simplifies to a scalar value. For vector fields in the plane, we can check if the vector field is conservative by determining if the curl is zero.

• How to find the curl: For a two-dimensional vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), the curl is given by the partial derivative formula \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \).
• Zero Curl: If the result is zero, the vector field is potentially conservative, implying there exists a potential function.

In our exercise, we computed the respective partial derivatives \( \frac{\partial Q}{\partial x} = 12x^3 y^2 \) and \( \frac{\partial P}{\partial y} = 12x^3 y^2 \), and found their difference to be zero, confirming that the field is conservative and allows further exploration toward finding a potential function.

Partial Derivatives

Partial derivatives are derivatives where the function is differentiated with respect to one variable while keeping other variables constant. They are crucial in multivariable calculus, allowing us to analyze how a function changes as each variable varies.

• Application in vector fields: When dealing with vector fields, calculating partial derivatives helps us understand the rate of change of the field's components.
• Determining conservativeness: In determining whether a vector field is conservative, partial derivatives are used to check the equality required for a zero curl.

For example, in this exercise, we found the partial derivative \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \) to check the conservativeness of the vector field. Evaluating these derivatives and confirming their equality helped demonstrate that the vector field was indeed conservative, allowing us to proceed with determining the potential function. The entire process illustrates the power of partial derivatives in multidimensional calculus.