Problem 25 - The Calculus with Analytic Geometry

Question

In Exercises 21 through 30, show that the value of the line integral is independent of the path and compute the value in any convenient manner. In each exercise, \(C\) is any sectionally smooth curve from the point \(A\) to the point \(B\). \(\int_{C} \frac{2 y}{(x y+1)^{2}} d x+\frac{2 x}{(x y+1)^{2}} d y ; A\) is \((0,2)\) and \(B\) is \((1,0)\)

Step-by-Step Solution

Answer

The line integral is 0, and it's path-independent.

Step 1: Identify and Verify the Vector Field

The given line integral is y = \frac{2 y}{(x y+1)^{2}} d x+\frac{2 x}{(x y+1)^{2}} d yi.e., the vector field is \(\mathbf{F} = \left( \frac{2y}{(xy+1)^2}, \frac{2x}{(xy+1)^2} \right)\). Determine whether this vector field is conservative by checking if \(\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}\).

Step 2: Calculate Partial Derivatives

Find \(\frac{\partial F_1}{\partial y}\) and \(\frac{\partial F_2}{\partial x}\): \(F_1 = \frac{2y}{(xy+1)^2}\) and \(F_2 = \frac{2x}{(xy+1)^2}\) Calculate: \(\frac{\partial F_1}{\partial y} = \frac{2(xy+1)^{2} - 2y(2y)(xy+1)}{(xy+1)^4}\) \(\frac{\partial F_2}{\partial x} = \frac{2(xy+1)^{2} - 2x(2x)(xy+1)}{(xy+1)^4}\)

Step 3: Simplify the Partial Derivatives

Simplify both partial derivatives to verify if they are equal:\(\frac{\partial F_1}{\partial y} = \frac{2x-2xy}{(xy+1)^3}\) \(\frac{\partial F_2}{\partial x} = \frac{2x-2xy}{(xy+1)^3} \) Since \(\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}\), the vector field is conservative.

Step 4: Find the Potential Function

To find the potential function \(\phi(x,y)\) such that \(abla \phi = \mathbf{F}\): Integrate \(F_1 = \frac{2y}{(xy+1)^2}\) with respect to \(x\): \[\phi(x,y) = -\frac{2}{xy+1} + h(y)\]Differentiate \(\phi(x,y)\) with respect to \(y\) and compare with \(F_2 = \frac{2x}{(xy+1)^2}\) to determine \(h(y)\):\[\frac{\partial \phi}{\partial y} = -\frac{2x}{(xy+1)^2}\]So, \(h(y)\) must be a constant.

Step 5: Evaluate the Potential Function at Points A and B

Evaluate \(\phi(x,y)\) at the points A(0,2) and B(1,0): \[\phi(0,2) = -\frac{2}{(0 \cdot 2 + 1)} = -2\]\[\phi(1,0) = -\frac{2}{(1 \cdot 0 + 1)} = -2\]The total line integral is: \(\phi(1,0) - \phi(0,2) = -2 - (-2) = 0\)

Key Concepts

Conservative Vector Fields

A vector field \(\boldsymbol{F} \) is termed conservative if it can be expressed as the gradient of a potential function \(\boldsymbol{\text{φ}} \). A key feature of conservative vector fields is that the line integral of the field between two points is independent of the path taken. This simplification arises because the line integral equates to the difference in the potential function values between the two end points.

These are some important points to remember about conservative vector fields:

• Closed Path Zero Integral: If a vector field is conservative, \(\boldsymbol{\text{∮_C F \cdot dR }} \) over any closed path \(\boldsymbol{C} \) (i.e., the start and end points are the same) will always be zero.
• Path Independence: This property implies that the work done by the field along any path from point \(\boldsymbol{A} \) to point \(\boldsymbol{B} \) only depends on the end points and not on the specific path taken.
• Identifying Conservatives: To identify if a vector field is conservative, check if its curl is zero. Alternatively, ensure that the mixed partial derivatives of its components are equal, that is, \(\boldsymbol{\frac{\text{∂F1}}{\text{∂y}} = \frac{\text{∂F2}}{\text{∂x}} } \) for a vector field \(\boldsymbol{(F1, F2)} \).

In the original exercise, verifying that \(\boldsymbol{\frac{\text{∂F1}}{\text{∂y}} = \frac{\text{∂F2}}{\text{∂x}} } \) told us the vector field was conservative.

Potential Functions

The potential function, \(\boldsymbol{\text{φ}} \), is a scalar function whose gradient returns the given conservative vector field. If \(\boldsymbol{F} \) is conservative, there exists a potential function \(\boldsymbol{\text{φ}} \) such that \(\boldsymbol{∇φ = F} \). This potential function can be used to compute the value of the line integral without directly integrating along a path.

The process to find a potential function involves:

• Integrating one of the components of \(\boldsymbol{F} \) with respect to its corresponding variable and assuming an arbitrary function of the other variables.
• Differentiating the result to find the arbitrary function by comparing it with the other component of \(\boldsymbol{F} \).

In the provided exercise, we started with \(\boldsymbol{F_1(x, y) = \frac{2y}{(xy + 1)^2}} \) and integrated it with respect to \(\boldsymbol{x} \) to get \(\boldsymbol{\text{φ}(x, y) = -\frac{2}{xy + 1} + h(y) } \). Differentiating this potential function with respect to \(\boldsymbol{y} \) helped determine the form of \(\boldsymbol{h(y)} \).

Partial Derivatives

In this context, partial derivatives are used to check the conservative nature of the vector field as well as to find the potential function. Partial derivatives represent the rate at which a function changes as its variables change, indicative of the slope at a certain point in a particular direction.

For a vector field \(\boldsymbol{F = (F1, F2)} \):

• Calculate \(\boldsymbol{\frac{\text{∂F1}}{\text{∂y}} } \): This is achieved by differentiating \(\boldsymbol{F1} \) with respect to \(\boldsymbol{y} \), while treating \(\boldsymbol{x} \) as a constant.
• Calculate \(\boldsymbol{\frac{\text{∂F2}}{\text{∂x}} } \): Here, \(\boldsymbol{F2} \) is differentiated with respect to \(\boldsymbol{x} \), treating \(\boldsymbol{y} \) as a constant.

In the exercise, we computed that:

\(\boldsymbol{\frac{\text{∂F1}}{\text{∂y}} = \frac{2(xy + 1)^2 - 2y(2y)(xy+1)}{(xy+1)^4} } \) and
\(\boldsymbol{\frac{\text{∂F2}}{\text{∂x}} = \frac{2(xy + 1)^2 - 2x(2x)(xy+1)}{(xy+1)^4} } \).

After simplifying, both were \(\boldsymbol{\frac{\text{2x - 2xy}}{(xy+1)^{3}}} \), thus confirming the vector field's conservative nature.