Question

Zero flux fields. For what values of \(a\) and \(d\) does the vector field \(\mathbf{F}=\langle a x, d y\rangle\) have zero flux across the unit circle centered at the origin and oriented counterclockwise?

Short answer

Answer: For the vector field F to have zero flux across the unit circle, we must have a = d.

Step-by-step solution

Parametric Representation of the Unit Circle

The unit circle centered at the origin can be represented parametrically by the equations \(x(\theta) = \cos \theta\) and \(y(\theta) = \sin \theta\), where \(0 \leq \theta \leq 2\pi\). Then, the velocity vector of this circle is the derivative of its position vector with respect to \(\theta\). Differentiating both equations with respect to \(\theta\), we get: $$ \frac{dx}{d\theta} = -\sin \theta, $$ $$ \frac{dy}{d\theta} = \cos \theta. $$

Line Integral of the Vector Field

As mentioned earlier, the flux is calculated using the line integral of the vector field around the curve. $$ \text{Flux} = \oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C \langle ax, dy \rangle \cdot \langle dx, dy \rangle $$ Substituting the parametric representation of the unit circle and its derivatives into the line integral, we have: $$ \text{Flux} = \int_0^{2\pi} \langle a\cos \theta, d\sin \theta \rangle \cdot \left \langle -\sin \theta, \cos \theta \right \rangle d\theta $$

Evaluate the Line Integral

We now have to compute the dot product and integrate over the range \(0 \leq \theta \leq 2\pi\). Evaluating the dot product, we get $$ \text{Flux} = \int_0^{2\pi} (-a\cos \theta \sin \theta + d\sin \theta \cos \theta) d\theta $$ Using the double-angle formula, \(2\sin \theta \cos \theta = \sin(2\theta)\), this simplifies to $$ \text{Flux} = \frac{1}{2} \int_0^{2\pi} (\sin 2\theta(-a + d)) d\theta $$ Integrating over the range \(0 \leq \theta \leq 2\pi\), we get $$ \text{Flux} = \frac{1}{4}(-a + d)\left[ \cos(2\theta) \right]_0^{2\pi} $$

Find the values of a and d that make the flux zero

Finally, we need to equate the computed flux to zero and find the values of \(a\) and \(d\): $$ 0 = \frac{1}{4}(-a + d)\left[ \cos(2\theta) \right]_0^{2\pi} = 0 $$ From the equation above, we can conclude that \(-a + d = 0\). Therefore, for the vector field \(\mathbf{F}\) to have zero flux across the unit circle, we must have \(a = d\).

Key concepts

Parametric Representation

Understanding the parametric representation of geometric shapes is crucial in vector calculus. Specifically, the unit circle, a fundamental curve, is defined as all the points that are exactly one unit away from the origin in a two-dimensional plane. To represent the unit circle parametrically, we use trigonometric functions to describe its x and y coordinates in terms of a variable \( \theta \), which represents an angle measured from the positive x-axis.

For the unit circle, this is given by \( x(\theta) = \cos \theta \) for the x-coordinate, and \( y(\theta) = \sin \theta \) for the y-coordinate, as the solution outlined. These equations capture the essence of the circle’s shape by mapping every angle \( \theta \) to a point on the circle. The parametric form is particularly useful when calculating various properties of the curve, including lengths, areas, and, as in our exercise, the flux.

Line Integral of Vector Field

The line integral of a vector field along a curve gives us a way to measure the total effect of the vector field as you move along the curve. In the context of flux, which visually represents the quantity of a vector field that 'flows' through a curve, the line integral is the mathematical tool that makes this quantification possible.

The line integral takes into account both the vector field and the direction of the path or curve it follows. This is reflected by the dot product in the integral \( \mathbf{F} \cdot d\mathbf{r} \) which combines the vector field \( \mathbf{F} \) and the differential element of the path \( d\mathbf{r} \)—representing an infinitesimally small segment of the curve. In the solution provided, \( \mathbf{F} \) is the given vector field, and \( d\mathbf{r} \) is derived from the parametric equations of the unit circle. By taking the integral over the entire range of \( \theta \) from 0 to \( 2\pi \)—effectively once around the circle—we calculate the total flux.

Double-Angle Formula

The double-angle formulas in trigonometry are a set of equations that express trigonometric functions of twice an angle in terms of functions of the original angle. They are incredibly useful in simplifying expressions where such scenarios occur.

In our exercise, the double-angle formula \( 2\sin \theta \cos \theta = \sin(2\theta) \) helps in converting the product of a sine and cosine function into a single sine function, which simplifies the integral considerably. This formula is fundamental in trigonometry and calculus as it allows us to integrate more easily and achieve solutions that might otherwise be complex or intractable. The simplification it provides is evident in how it transforms the flux integral into an expression that can be integrated over the interval \( [0, 2\pi] \) without complicated manipulation.

Vector Calculus

Vector calculus is the branch of mathematics that deals with the differentiation and integration of vector fields, often in two or three dimensions. It's pivotal in physics and engineering, where it's applied to problems involving vector quantities like electric and magnetic fields, fluid flow, and force fields.

This field of study involves a range of operations, including grad, div, and curl, which respectively denote gradient, divergence, and curl. In the context of our exercise, we're dealing with the concept of flux—which in vector calculus represents the quantity that passes through a surface. The zero flux condition is a special case which often signifies equilibrium or a field being conservative. The exercise takes you through evaluating such a condition using the tools of vector calculus — from parametric equations to evaluating line integrals — to find specific vector field constants that satisfy the given condition.