Problem 28 - Calculus: Early Transcendental Functions

Question

Determine whether or not the vector field is conservative. If it is, find a potential function. $$\left(x^{2}-y\right) \mathbf{i}+(x-y) \mathbf{j}$$

Step-by-Step Solution

Answer

The vector field \((x^2 - y) \mathbf{i}+(x - y) \mathbf{j}\) is not conservative, so there is no potential function.

Step 1: Identifying the Components

First, identify the components of the vector field. \(F = (F1, F2)\) where \(F1 = x^2 - y\) and \(F2 = x - y\).

Step 2: Checking for Conservativeness

Evaluate if \(\frac{∂F2}{∂x} = \frac{∂F1}{∂y}\). This condition, that the derivative of \(F2\) with respect to \(x\) is equal to the derivative of \(F1\) with respect to \(y\), is necessary for the vector field to be conservative. Differentiating both components with respect to the respective variables gives: \(\frac{∂F2}{∂x} = 1\) and \(\frac{∂F1}{∂y} = -1\). The condition does not hold and hence the vector field is not conservative.

Step 3: Explaining the Lack of Potential Function

Since the vector field is not conservative, there does not exist a potential function that the vector field can be derived from.

Key Concepts

Vector Calculus

Vector calculus is an integral part of advanced calculus and mathematical physics, focusing on vectors and differentiating or integrating them through various operations. It allows us to analyze and visualize fields of all dimensions, from the terrestrial electric field to the velocity field of fluid flow. Common operations include:

• Gradient, which represents the rate and direction of change in a scalar field.
• Divergence, indicative of the extent to which there is a source or sink at a point in a vector field.
• Curl, which measures the rotation or angular momentum of the field.

In the context of our exercise, vector calculus helps us examine whether a given vector field—expressed as \( \mathbf{F} = (x^2 - y) \mathbf{i} + (x-y) \mathbf{j} \)—is conservative or not. A conservative vector field typically means that it can be expressed as the gradient of some scalar potential function, which simplifies many aspects of analysis and computation.

Potential Function

The concept of a potential function is central to understanding conservative vector fields. In simple terms, a potential function is a scalar function whose gradient (a vector of its partial derivatives) results in the original vector field. Given a vector field \((F_1, F_2)\), a potential function \( \phi(x,y)\) meets the conditions: \"

• \( \frac{∂\phi}{∂x} = F_1(x, y) \)
• \( \frac{∂\phi}{∂y} = F_2(x, y) \)

These conditions imply that the field is conservative. However, if these conditions don't agree across variable changes, the potential function cannot exist.In our example, the lack of equality between \(\frac{∂F_2}{∂x}\) and \(\frac{∂F_1}{∂y}\) (\(1 eq -1\)), signifies the absence of a potential function, reinforcing the non-conservativeness of the vector field.

Partial Derivatives

Partial derivatives play a crucial role in multivariable calculus, allowing us to explore how a function changes in relation to only one of its variables, holding others constant. This provides a way to understand multidimensional spaces by looking at their projections in one direction at a time. ** Why are Partial Derivatives Important?**

• They are foundational in deriving gradients, divergences, and curls in vector calculus.
• Critical in determining whether a vector field is conservative, through cross-partial derivatives.

In our problem, determining the partial derivatives between components of the \(\mathbf{F}\) field was key to checking its conservativeness. Partial derivative of \(F_2\) with respect to \(x\), noted as \(\frac{∂F2}{∂x}\), shows how \(F_2\) varies as \(x\) changes. Similarly, \(\frac{∂F1}{∂y}\) exhibits the change in \(F_1\) concerning \(y\). Failure of these derivatives to equate (\(1 eq -1\)) indicates \(\mathbf{F}\) isn't conservative, thus no potential function exists for this vector field.