Problem 14 - Advanced Engineering Mathematics

Question

Determine whether the given vector field is a conservative field. If so, find a potential function \(\phi\) for \(\mathbf{F}\). $$ \mathbf{F}(x, y)=\left(x^{2}+y^{2}+1\right)^{-2}(x \mathbf{i}+y \mathbf{j}) $$

Step-by-Step Solution

Answer

Yes, \( \mathbf{F} \) is conservative; \( \phi(x, y) = -\frac{1}{x^2 + y^2 + 1} + k \).

Step 1: Determine If the Field is Conservative

To determine if the vector field \( \mathbf{F}(x, y) \) is conservative, check if it is the gradient of some potential function \( \phi(x,y) \). A vector field is conservative if its curl is zero. Here, the field is two-dimensional, so it is conservative if the partial derivative of \( M \) with respect to \( y \) equals the partial derivative of \( N \) with respect to \( x \), where \( \mathbf{F}(x, y) = M \mathbf{i} + N \mathbf{j} \). In our case, \( M = \frac{x}{(x^2 + y^2 + 1)^2} \) and \( N = \frac{y}{(x^2 + y^2 + 1)^2} \).

Step 2: Check the Mixed Partial Derivatives

Calculate \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}\left( \frac{x}{(x^2 + y^2 + 1)^2} \right) \) and \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}\left( \frac{y}{(x^2 + y^2 + 1)^2} \right) \). Use the chain and product rules to compute these derivatives.

Step 2.1: Compute \( \frac{\partial M}{\partial y} \)

First, rewrite \( M(x, y) = x (x^2 + y^2 + 1)^{-2} \). Applying the chain rule gives \( \frac{\partial M}{\partial y} = x \cdot -2 (x^2 + y^2 + 1)^{-3} \cdot 2y = -2xy(x^2 + y^2 + 1)^{-3} \).

Step 2.2: Compute \( \frac{\partial N}{\partial x} \)

Similarly, rewrite \( N(x, y) = y (x^2 + y^2 + 1)^{-2} \). Applying the chain rule gives \( \frac{\partial N}{\partial x} = y \cdot -2 (x^2 + y^2 + 1)^{-3} \cdot 2x = -2xy(x^2 + y^2 + 1)^{-3} \).

Step 3: Compare Partial Derivatives

Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the vector field \( \mathbf{F} \) is conservative.

Step 4: Find the Potential Function \( \phi(x, y) \)

To find the potential function \( \phi(x, y) \), solve \( \frac{\partial \phi}{\partial x} = M \) and \( \frac{\partial \phi}{\partial y} = N \). Integrate \( M \) with respect to \( x \) to find \( \phi(x, y) = \int \frac{x}{(x^2 + y^2 + 1)^2} \, dx \).

Step 4.1: Integrate \( M \, dx \)

Integrate \( \frac{x}{(x^2 + y^2 + 1)^2} \) with respect to \( x \). Use substitution \( u = x^2 + y^2 + 1 \) leading to the integral \( \int \frac{1}{u^2} \, du = -\frac{1}{u} + C(y) \). Substitute back \( u = x^2 + y^2 + 1 \) to obtain \(-\frac{1}{x^2 + y^2 + 1} + C(y)\).

Step 4.2: Determine \( C(y) \)

To find \( C(y) \), differentiate the expression \(-\frac{1}{x^2 + y^2 + 1} + C(y) \) with respect to \( y \) and set it equal to \( N \). Solve \( \frac{2y}{(x^2 + y^2 + 1)^2} + C'(y) = \frac{y}{(x^2 + y^2 + 1)^2} \), resulting in \( C'(y) = 0 \). Thus, \( C(y) \) is a constant, say \( k \), and the potential function is \( \phi(x, y) = -\frac{1}{x^2 + y^2 + 1} + k \).

Key Concepts

Potential Function

A potential function, often denoted as \( \phi(x, y) \), is a scalar function whose gradient yields a given vector field \( \mathbf{F} \). In simpler terms, if a vector field is conservative, it means there exists a potential function \( \phi \) such that \( abla \phi = \mathbf{F} \). For the vector field to be conservative, it must meet specific criteria. One crucial condition is that the vector field should be path-independent, meaning the line integral of the field depends only on the endpoints and not on the path taken.

In the exercise, the challenge was to determine if \( \mathbf{F}(x, y) = \frac{x \mathbf{i} + y \mathbf{j}}{(x^2 + y^2 + 1)^2} \) has such a potential function. By confirming that the mixed partial derivatives of the field components match, it was concluded that a potential function indeed exists for \( \mathbf{F} \). Integrating the components of \( \mathbf{F} \) led to the potential function \( \phi(x, y) = -\frac{1}{x^2 + y^2 + 1} + k \), where \( k \) is a constant.

Gradient Vector Field

A gradient vector field is a vector field that can be expressed as the gradient of a scalar function. The gradient, denoted \( abla \phi \), represents the rate of change or slope of the function \( \phi \) at every point, pointing in the direction of the steepest ascent. In mathematical terms, if \( \mathbf{F} = abla \phi \), then \( F(x, y) = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y} \right) \).

For the vector field \( \mathbf{F}(x, y) = \frac{x \mathbf{i} + y \mathbf{j}}{(x^2 + y^2 + 1)^2} \), finding \( \phi \) through integration confirmed that \( \mathbf{F} \) is indeed a gradient field. This property is essential because it confirms the vector field is conservative and can be represented by a potential function.

Partial Derivatives

Partial derivatives are fundamental when dealing with functions of multiple variables, as they allow us to analyze how a function changes with respect to one variable while keeping all others constant. For a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted \( \frac{\partial f}{\partial x} \), and with respect to \( y \), it is denoted \( \frac{\partial f}{\partial y} \).

In the problem's context, to verify that the vector field \( \mathbf{F} \) is conservative, the partial derivatives were used to check the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). The functions \( M \) and \( N \) are the components of \( \mathbf{F} \), and through computing these derivatives, it was confirmed that they are equal, thus demonstrating the conservativeness of the field.

Chain Rule

The chain rule is a crucial tool in calculus, especially when differentiating composite functions. It helps in finding derivatives of functions that are built up of other functions. The rule is applicable when we have a function of the form \( h(x) = f(g(x)) \), and it can be expressed as \( h'(x) = f'(g(x)) \cdot g'(x) \).

In this exercise, the chain rule was extensively used to find the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) for the vector field \( \mathbf{F}(x, y) \). These derivatives involved functions like \( (x^2 + y^2 + 1)^{-2} \), requiring careful application of the chain rule to differentiate correctly. By successfully applying the chain rule, one ensures the precision needed for confirming the conservative nature of vector fields.