Problem 6 - Advanced Engineering Mathematics

Question

In Problems, 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. $$ \int_{(1,0)}^{(3,4)} \frac{x d x+y d y}{\sqrt{x^{2}+y^{2}}} \text { on any path not through the origin } $$

Step-by-Step Solution

Answer

The line integral is independent of the path and evaluates to 4.

Step 1: Check Vector Field for Conservativeness

The given line integral represents the work done by the vector field \( \vec{F} = \left( \frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}} \right) \). For the line integral to be path-independent, \( \vec{F} \) must be conservative. A vector field is conservative if its curl is zero. The curl of \( \vec{F} \) in two dimensions is computed as \( abla \times \vec{F} = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \). Since the field is derived from a potential function, computing the partial derivatives: \( \frac{\partial}{\partial x}\left( \frac{y}{\sqrt{x^2+y^2}} \right) = -\frac{xy}{(x^2+y^2)^{3/2}} \) and \( \frac{\partial}{\partial y}\left( \frac{x}{\sqrt{x^2+y^2}} \right) = -\frac{xy}{(x^2+y^2)^{3/2}} \), the curl is zero hence it's conservative.

Step 2: Finding the Potential Function \( \phi \)

To find the potential function, integrate \( F_1 = \frac{x}{\sqrt{x^2+y^2}} \) with respect to \( x \) to get \( \phi(x, y) = \sqrt{x^2+y^2} + g(y) \), where \( g(y) \) is an arbitrary function of \( y \). Now differentiate this result with respect to \( y \) and set it equal to \( F_2 = \frac{y}{\sqrt{x^2+y^2}} \). We find that \( \frac{\partial}{\partial y} (\sqrt{x^2 + y^2} + g(y)) = \frac{y}{\sqrt{x^2+y^2}} + g'(y) = \frac{y}{\sqrt{x^2+y^2}} \). Therefore, \( g'(y) = 0 \), implying \( g(y) \) is constant. Therefore, the potential function is \( \phi(x,y) = \sqrt{x^2+y^2} \).

Step 3: Evaluate Line Integral Using Theorem 9.9.1

By Theorem 9.9.1, the line integral of a conservative vector field \( \vec{F} = abla \phi \) from point \( A \) to point \( B \) is \( \phi(B) - \phi(A) \). Evaluate \( \phi \) at the endpoints \( (3, 4) \) and \( (1, 0) \). We have \( \phi(3, 4) = \sqrt{3^2 + 4^2} = 5 \) and \( \phi(1, 0) = \sqrt{1^2 + 0^2} = 1 \). Therefore, \( \phi(3, 4) - \phi(1, 0) = 5 - 1 = 4 \). The line integral evaluates to 4.

Step 4: Evaluate Using a Direct Path

Choose the simplest path between the points, like the straight line from \((1,0)\) to \((3,4)\). Parameterize this line as \( \vec{r}(t) = (1 + 2t, 4t) \) where \( t \) ranges from 0 to 1. Compute \( \frac{d\vec{r}}{dt} = (2, 4) \). The integral becomes \[ \int_0^1 \left( \frac{(1+2t) \cdot 2 + 4t \cdot 4}{\sqrt{(1+2t)^2 + (4t)^2}} \right) dt \]. Simplify and compute this integral. Substituting \( u = \sqrt{(1+2t)^2 + (4t)^2 } \) simplifies to finding the same result \( u \) ranges from 1 to 5 as \( t \) goes from 0 to 1, therefore integral evaluates directly to \([u]_1^5 = 5 - 1 = 4 \), showing same answer.

Key Concepts

Potential Function

A potential function is a scalar function associated with a vector field that helps make the process of evaluating line integrals simpler. It is like a "map" that shows how energy or a force field changes over a surface. When you find a potential function for a vector field, it means that the vector field is conservative, and line integrals over it are path-independent.

To construct a potential function, integrate the components of the vector field. In our exercise, integrating the first component \( F_1 = \frac{x}{\sqrt{x^2+y^2}} \) with respect to \( x \) helps in finding the potential function \( \phi(x, y) = \sqrt{x^2+y^2} + g(y) \), where \( g(y) \) is some function of \( y \). By differentiating and aligning these results with the vector field's second component, \( g(y) \) can often be determined to be constant, leading to a full expression for the potential function.

Conservative Vector Field

A conservative vector field is one where the circulation or total work done by the field around any closed loop is zero. That means, if you start at one point and travel around to come back to the same point, the total work done is nil.

Mathematically, this occurs when the curl of the vector field is zero, indicating that there's no spinning or curl within the field in two-dimensional space. In our case, doing the math shows that \( abla \times \vec{F} = 0 \), confirming that the vector field is conservative.

One interesting property of conservative vector fields is that they always have a corresponding potential function. Thus, evaluating line integrals becomes very direct and simple.

Path Independence

If a line integral is path-independent, it means the integral's value only depends on the initial and final points, not on the path taken between them. This makes things more straightforward because it often allows you to choose the simplest path possible for evaluation.

In conservative vector fields, path independence is an essential characteristic because the integral across different paths between two points will still yield the same result, thanks to the potential function. This was demonstrated in the original exercise where two different methods, involving different paths, ended with the same integral value.

Vector Field

A vector field is a mathematical construction used to model and describe the spatial variation of a vector quantity. It assigns a vector to each point in space. In practical terms, it can represent things like fluid dynamics or gravitational fields, capturing the direction and magnitude of a vector variable everywhere in that field.

For line integrals, you often consider vector fields to determine how much 'work' is done along a path using concepts of potential functions and path independence. In our example, the vector field \( \vec{F} = \left( \frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}} \right) \) defines how the vector quantity, perhaps imagined as some force, varies across the plane — except at the origin, due to undefined behavior there. Understanding vector fields is crucial as it lays the groundwork for analyzing more complex dynamic systems.