Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let $$ \mathbf{F}=\frac{y}{x^{2}+y^{2}} \mathbf{i}-\frac{x}{x^{2}+y^{2}} \mathbf{j}=M \mathbf{i}+N \mathbf{j} $$ (a) Show that \(\partial N / \partial x=\partial M / \partial y\). (b) Show, by using the parametrization \(x=\cos t, y=\sin t\), that \(\oint_{C} M d x+N d y=-2 \pi\), where \(C\) is the unit circle. (c) Why doesn't this contradict Green's Theorem?

Step-by-Step Solution

Answer

(a) \(\partial N/\partial x = \partial M/\partial y\). (b) Integral equals \(-2\pi\). (c) Green's Theorem doesn't apply due to singularity at the origin.

Step 1: Differentiate N with respect to x

Given that \(N = -\frac{x}{x^2 + y^2}\), differentiate it with respect to \(x\). Use the quotient rule for differentiation, which states that \(\frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}\). Applying it here, we have:\[ \frac{\partial N}{\partial x} = \frac{-(x^2 + y^2)(-1) - (-x)(2x)}{(x^2+y^2)^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2} \]

Step 2: Differentiate M with respect to y

Given that \(M = \frac{y}{x^2 + y^2}\), differentiate it with respect to \(y\). Again, apply the quotient rule:\[ \frac{\partial M}{\partial y} = \frac{(x^2+y^2)(1) - y(2y)}{(x^2+y^2)^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2} \].Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), part (a) is satisfied.

Step 3: Parametrize the integral for the unit circle

To evaluate the line integral \(\oint_C M dx + N dy\), use the parametrization \(x = \cos t\), \(y = \sin t\) for the unit circle, where \(0 \leq t \leq 2\pi\). Then, \(dx = -\sin t \ dt\) and \(dy = \cos t \ dt\).

Step 4: Substitute and evaluate the integral

Substitute \(M = \frac{y}{x^2 + y^2}, \ N = -\frac{x}{x^2 + y^2}\), \(y = \sin t\), \(x = \cos t\), \(dx = -\sin t\ dt\), and \(dy = \cos t\ dt\) into the integral:\[ \oint_C M dx + N dy = \int_0^{2\pi} \left( \frac{\sin t}{1} \cdot (-\sin t) + \left(-\frac{\cos t}{1}\right) \cdot \cos t \right) dt \]This simplifies to:\[ \int_0^{2\pi} (-\sin^2 t - \cos^2 t) \, dt = -\int_0^{2\pi} dt = -2\pi \]

Step 5: Address potential contradiction with Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \(C\) to a double integral over the region \(R\) bounded by \(C\). It requires \(M\) and \(N\) to have continuous partial derivatives over \(R\). Here, \(M\) and \(N\) are singular at the origin, as the point \((0,0)\) is not within their domain, hence Green's Theorem does not apply directly to this vector field without caution.

Key Concepts

Line Integrals

Understanding line integrals is fundamental in vector calculus. A line integral is a way to integrate a function along a curve, often used to sum up quantities like mass or charge that vary along paths in space. In general, for vector fields, the line integral of a vector field \(\mathbf{F} = M \mathbf{i} + N \mathbf{j}\) around a curve \(C\) is expressed as \(\oint_{C} M \, dx + N \, dy\).
When evaluating a line integral over a closed curve, such as a circle, one aims to quantify the function’s cumulative effect along the entire path.
This concept is practically illustrated in problems involving fields like electromagnetism and fluid flow.
For the specific problem given, we employ the parameterization of equations to compute the net effect of the vector field \(\mathbf{F}\) around the unit circle.

Vector Fields

Vector fields are an essential concept in calculus, providing a visual and analytical way to examine physical phenomena characterized by direction and magnitude at every point in space. A vector field assigns a vector to each point in a subset of space.
For example, in this exercise, the field \(\mathbf{F} = \frac{y}{x^{2}+y^{2}} \mathbf{i} - \frac{x}{x^{2} + y^{2}} \mathbf{j}\) is described by its components \(M\) and \(N\), which are functions of \(x\) and \(y\).
The field's behavior helps in visualizing forces like magnetic or gravitational fields. Understanding how these fields interact with curves or regions is crucial in applying Green's Theorem or evaluating line integrals.
In our case, calculating partial derivatives helps us confirm whether specific conditions for certain theorems, like Green’s Theorem, are met.

Parametrization

Parametrization is a technique used to represent a curve within a coordinate system in terms of a single parameter, typically denoted \(t\).
This method allows complex paths in space to be easily analyzed or integrated over, making it an invaluable tool in calculus. For circles and other curves, using trigonometric functions for parameterization helps transform equations into forms suitable for integration.
In this problem, we parameterize the unit circle using \(x = \cos t\) and \(y = \sin t\). This choice simplifies the computation of the line integral by reducing the complex path into a straightforward form based on the parameter \(t\), ranging from 0 to \(2\pi\).
It makes operations such as finding \(dx\) and \(dy\) straightforward so that they can be substituted into the integral to compute the net result around the path.

Calculus of Vector Fields

The calculus of vector fields involves differentiating and integrating vector functions, allowing us to analyze how scalar and vector fields behave under various conditions.
One of the fundamental aspects is finding the partial derivatives of vector components, which helps in evaluating line integrals and applying theorems such as Green's Theorem.
In this exercise, we demonstrated the equivalence of mixed partial derivatives by computing \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) independently, finding them to be equal. This was pivotal in establishing the framework for further analysis.
However, the non-application of Green’s Theorem due to singularities in the vector field at the origin highlights the importance of domain consideration in vector calculus. Understanding these nuances is crucial when handling real-world scenarios where vector fields are not always well-behaved across their supposed domains.