Problem 5 - Advanced Engineering Mathematics

Question

Show that the given line integral is independent of the path. Evaluate in two ways: (a) Find a potential function \(\phi\) and then use Theorem 9.9.1, and (b) Use any convenient path between the endpoints of the path. $$ \int_{(4,1)}^{(4,4)} \frac{-y d x+x d y}{y^{2}} \text { on any path not crossing the } x \text {-axis } $$

Step-by-Step Solution

Answer

The line integral evaluates to 3 for any path not crossing the x-axis.

Step 1: Understanding the Problem and Conditions

We need to show that the line integral \( \int_{(4,1)}^{(4,4)} \frac{-y dx + x dy}{y^2} \) is path-independent whenever the path doesn't cross the x-axis. This implies that the vector field \( \mathbf{F} = \left( \frac{-y}{y^2}, \frac{x}{y^2} \right) \) must be conservative. A vector field is conservative if a potential function \( \phi \) exists such that \( abla \phi = \mathbf{F} \). Path independence can also be shown by direct integration if it is conservative.

Step 2: Determine if Vector Field is Conservative

For \( \mathbf{F} \) to be conservative, its curl must equal zero: \( abla \times \mathbf{F} = ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} ) \hat{k} = 0 \), where \( P = \frac{-y}{y^2} = \frac{-1}{y} \) and \( Q = \frac{x}{y^2} \). Calculate the derivatives: \( \frac{\partial Q}{\partial x} = \frac{1}{y^2} \) and \( \frac{\partial P}{\partial y} = \frac{1}{y^2} \). Therefore, the curl is zero, confirming \( \mathbf{F} \) is conservative.

Step 3: Finding the Potential Function \( \phi \)

The components of \( \mathbf{F} \) give us equations for \( \phi \). We have \( \frac{\partial \phi}{\partial x} = \frac{-1}{y} \) and \( \frac{\partial \phi}{\partial y} = \frac{x}{y^2} \). Integrate the first with respect to \( x \) to get \( \phi(x,y) = \frac{-x}{y} + g(y) \). Differentiate with respect to \( y \) to match the second component: \( \frac{d}{dy} \left( \frac{-x}{y} + g(y) \right) = \frac{x}{y^2} - g'(y) = \frac{x}{y^2} \). Solve for \( g(y) \) to find \( g'(y) = 0 \), hence \( g(y) = C \). Thus, \( \phi(x,y) = \frac{-x}{y} \).

Step 4: Evaluating the Integral Using Theorem 9.9.1

Theorem 9.9.1 states that for a conservative vector field, the line integral \( \int_{C} abla \phi \cdot d\mathbf{r} \) is equal to \( \phi(4,4) - \phi(4,1) \). Calculate the values: \( \phi(4,4) = \frac{-4}{4} = -1 \) and \( \phi(4,1) = \frac{-4}{1} = -4 \). Therefore, the integral evaluates to \( -1 - (-4) = 3 \).

Step 5: Choosing a Convenient Path to Evaluate the Integral

Choose the path consisting of two straight segments: from \( (4,1) \) to \( (4,4) \) along the x-axis, and vertically to \( (4,4) \). For \( (4,1) \) to \( (4,4) \), only \( dy \) is non-zero, so the integral becomes \( \int_{1}^{4} \frac{4}{y^2} \, dy \). Evaluate this: \( \left[ -\frac{4}{y} \right]_{1}^{4} = -\frac{4}{4} + \frac{4}{1} = 3 \). Thus, the integral also evaluates to 3.

Key Concepts

Conservative Vector Fields

In mathematics, a vector field is called conservative if it can be represented as the gradient of a potential function. But why does that matter? Well, one of the exciting properties of a conservative vector field is that the line integral, or the work done by moving an object within this field, is independent of the path taken between two points. This means that regardless of whether you take a winding path or the straightest path possible, the same amount of work is done.

For a vector field \( \mathbf{F} = (P, Q) \), it is conservative if there exists a function \( \phi \) such that \( abla \phi = \mathbf{F} \). More practically, this condition translates to the curl of the field being zero: \( abla \times \mathbf{F} = 0 \).

In the given exercise, the vector field is \( \mathbf{F} = \left( \frac{-y}{y^2}, \frac{x}{y^2} \right) \). By checking the curl, we confirmed it equals zero, establishing that this vector field is indeed conservative.

Potential Functions

A potential function is like a hidden treasure that reveals the secrets of a conservative vector field. It's a scalar function from which a vector field can be derived as its gradient.

To find a potential function \( \phi \) for a field \( \mathbf{F} = (P, Q) \):

• Integrate \( P \) with respect to \( x \) to get a preliminary form of \( \phi \).
• Integrate \( Q \) with respect to \( y \) and adjust to match the previously found expressions.

In our exercise, we needed to find \( \phi \) such that \( abla \phi = \mathbf{F} \). We found \( \phi(x, y) = \frac{-x}{y} \), which matches the vector field's components. Once \( \phi \) is found, it can be used to evaluate the line integral directly by substituting the endpoints into \( \phi(x, y) \).

Theorem on Path Independence

The Theorem on Path Independence gives a powerful tool for calculating line integrals in conservative vector fields. This is justified by the nature of potential functions. For a conservative vector field, the line integral from point \( A \) to point \( B \) depends solely on the values of the potential function at these points.

If \( \phi \) is a potential function for \( \mathbf{F} \), then the line integral along any path \( C \) from \( A \) to \( B \) is simply \( \phi(B) - \phi(A) \).

In the original problem, by using the potential function \( \phi(x, y) = \frac{-x}{y} \) and substituting the coordinates of the target points, the work done evaluates to 3, regardless of the path. This highlights the utility of the theorem: saving us from painstaking calculations for every possible path.