Problem 24 - Calculus: Early Transcendental Functions

Question

Determine whether or not the vector field is conservative. If it is, find a potential function. $$\langle 2, y\rangle$$

Step-by-Step Solution

Answer

The vector field is conservative and the potential function is \(f(x, y) = 2x + 0.5y^2\).

Step 1: Establish the coefficients of the vector field

First, identify the coefficients of the vector field given. In this case, if the vector field F is denoted as \( \langle F_1, F_2 \rangle \), then \( F_1 = 2 \) and \( F_2 = y \)

Step 2: Check if the curl of the vector field is zero

Compute the curl of the vector field, which is given by \( \nabla \times F = \left (\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right ) \hat{k} \). Here, \( \hat{k} \) represents the unit vector in the z-direction. On calculating both partial derivatives and subtracting, the result should ideally be zero for the vector field to be conservative.

Step 3: Determine the curl of the vector field

Calculate the derivatives and subtract. So it is \( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = \frac{\partial y}{\partial x} - \frac{\partial 2}{\partial y} = 0 - 0 = 0 \). The curl of the vector field is indeed zero, therefore, the vector field is conservative.

Step 4: Compute the potential function

Since the vector field is conservative, it is possible to calculate the potential function. The potential function is given by \( f(x,y) = \int F_1 dx + \int F_2 dy \). By computing these integrations, the student will get the potential function.

Step 5: Compute the potential function of the vector field

To find the potential function, compute the integral of each coefficient: \(\int F_1 dx = \int 2 dx = 2x\) and \(\int F_2 dy = \int y dy = 0.5y^2\). So, the potential function \(f(x, y) = 2x + 0.5y^2\). There could be an arbitrary constant; in physics this is often set to zero for convenience. Hence, the potential function is \(f(x, y) = 2x + 0.5y^2\).

Key Concepts

Curl of Vector Field

In vector calculus, the "curl" of a vector field gives us a way to measure how much the vector field rotates around a given point. To determine if a vector field is conservative, we calculate its curl. If the curl of the vector field is zero, the field is considered conservative.
When you see terms like \( abla \times F \), it refers to computing the curl. For a vector field \( F \) with components \( F_1 \) and \( F_2 \), its curl is expressed as:

• \( abla \times F = \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \hat{k} \)

In our given exercise with \( F_1 = 2 \) and \( F_2 = y \), computing the partial derivatives from our formula results in zero:

• \( \frac{\partial y}{\partial x} = 0 \)
• \( \frac{\partial 2}{\partial y} = 0 \)

These results mean that the difference is zero, showing the vector field has no rotation at any point, confirming it is conservative.

Potential Function

A conservative vector field implies that there exists a "potential function" such that the vector field is its gradient. The potential function is essentially a scalar field, which provides a value at each point in space.
To find this potential function \( f(x, y) \), you integrate each component of the vector field. For the vector field \( \langle 2, y \rangle \), the potential function is determined by:

• Integrating \( F_1 = 2 \) with respect to \( x \) yields \( \int 2 \, dx = 2x \).
• Integrating \( F_2 = y \) with respect to \( y \) yields \( \int y \, dy = 0.5y^2 \).

Combining these gives us the potential function: \( f(x, y) = 2x + 0.5y^2 \). It's noteworthy that there's often an arbitrary constant added to the potential, but it's typically omitted unless the context specifies its necessity.

Partial Derivatives

Partial derivatives are derivatives where you hold some variables constant while differentiating with respect to others. They are crucial in dealing with multi-variable functions, such as vector fields. In our exercise, these partial derivatives are used to compute the curl of the vector field.
Let's recap their calculation:

• \( \frac{\partial y}{\partial x} \): Derivative of \( y \) with respect to \( x \) is 0 because \( y \) is independent of \( x \).
• \( \frac{\partial 2}{\partial y} \): Derivative of a constant with respect to any variable is zero.

These concepts underpin many calculations in vector calculus. Understanding this process is essential, as it forms the foundation for determining conservativeness in vector fields and taking gradients to find potential functions.