Problem 1 - Calculus Early Transcendentals

Question

Determine whether \(\mathbf{F}\) is a conservative vector field. If so, find a potential function for it. $$ \mathbf{F}(x, y)=x \mathbf{i}+y \mathbf{j} $$

Step-by-Step Solution

Answer

Yes, \( \mathbf{F} \) is conservative with potential function \( f(x, y) = \frac{x^2}{2} + \frac{y^2}{2} + C \).

Step 1: Determine If the Field is Conservative

To determine if a vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \) is conservative, we need to check if the curl of \( \mathbf{F} \) is zero. The condition for a vector field in \( \mathbb{R}^2 \) to be conservative is that \( \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} \). In our case, \( P(x, y) = x \) and \( Q(x, y) = y \). Compute \( \frac{\partial Q}{\partial x} \):\[ \frac{\partial Q}{\partial x} = \frac{\partial y}{\partial x} = 0 \]Compute \( \frac{\partial P}{\partial y} \):\[ \frac{\partial P}{\partial y} = \frac{\partial x}{\partial y} = 0 \]Both derivatives are equal, thus \( \mathbf{F} \) is conservative.

Step 2: Find the Potential Function

Since \( \mathbf{F} \) is a conservative vector field, it has a potential function \( f(x, y) \) such that \( \mathbf{F} = abla f \). We know:\( \frac{\partial f}{\partial x} = P(x, y) = x \)Integrate with respect to \( x \):\[ f(x, y) = \int x \, dx = \frac{x^2}{2} + h(y) \]Where \( h(y) \) is an unknown function of \( y \). Next, differentiate with respect to \( y \):\( \frac{\partial f}{\partial y} = \frac{d}{dy} \left( \frac{x^2}{2} + h(y) \right) = h'(y) \)From \( \frac{\partial f}{\partial y} = Q(x, y) = y \), it follows that \( h'(y) = y \).Integrating \( h'(y) \) with respect to \( y \):\[ h(y) = \int y \, dy = \frac{y^2}{2} + C \]Finally, the potential function is:\[ f(x, y) = \frac{x^2}{2} + \frac{y^2}{2} + C \]

Key Concepts

Potential Function

In vector calculus, the potential function is a vital concept, especially in relation to conservative vector fields. A vector field is said to be conservative if it is the gradient of some potential function. This means that for a vector field \(\mathbf{F}\), there exists a potential function \(f(x, y)\) such that \(\mathbf{F} = abla f\).
When a vector field \(\mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j}\) is conservative, the potential function can be found by integrating the components of the vector field. This involves:

• Integrating \(P(x, y)\) with respect to \(x\) to find the function component involving \(x\).
• Integrating \(Q(x, y)\) with respect to \(y\) to find the function component involving \(y\).

The result of the integration yields the full potential function, often up to an arbitrary constant. In our example, the potential function found is \(f(x, y) = \frac{x^2}{2} + \frac{y^2}{2} + C\). This represents the scalar field from which the vector field is derived, signifying the stored energy or potential at a point \((x, y)\).
The beauty of potential functions lies in their ability to simplify the understanding of vector fields, providing insight into the underlying structure of forces or flows.

Partial Derivatives

Partial derivatives play a crucial role in the analysis of functions of several variables, particularly in vector calculus. They represent the rate at which a function changes as one of the variables changes while the other variables remain fixed. In the context of determining whether a vector field is conservative, partial derivatives provide a key test.
For a vector field \(\mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j}\), the field is conservative if the partial derivative of \(Q\) with respect to \(x\), \(\frac{\partial Q}{\partial x}\), is equal to the partial derivative of \(P\) with respect to \(y\), \(\frac{\partial P}{\partial y}\). If both these derivatives are equal, it indicates that the vector field has no "twirling" or "curling", making it conservative.
In our example, \(P(x, y) = x\) and \(Q(x, y) = y\), yielding both partial derivatives as zero:

• \(\frac{\partial Q}{\partial x} = \frac{\partial y}{\partial x} = 0\)
• \(\frac{\partial P}{\partial y} = \frac{\partial x}{\partial y} = 0\)

This confirms the vector field is conservative. Partial derivatives offer a powerful method to uncover the properties of a vector field, showing whether it can be mapped back to a simpler scalar potential function.

Vector Calculus

Vector calculus is a sophisticated field of mathematics that involves differentiation and integration of vector fields, primarily in two- and three-dimensional space. Key concepts within vector calculus include vector fields, gradients, curls, divergences, and fluxes.
In our context, we've been dealing with a vector field \(\mathbf{F}(x, y) = x \mathbf{i} + y \mathbf{j}\). A critical aspect of vector calculus is determining properties such as conservativeness, which we achieve by examining the curl—a vector operation documenting the rotation or "twist" of the field. For two-dimensional fields like ours, a vector field is conservative if the curl is zero.
Understanding vector calculus enables us to analyze fields in physics, engineering, and other sciences, where they represent directional forces or flows. For example, electric fields, gravitational fields, and fluid flows are often analyzed using vector calculus to comprehend how they act over different regions of space. In our example, identifying the vector field as conservative allows us to find a potential function that unveils the field's fundamental behavior.