Problem 46 - Thomas Calculus

Question

Propyl alcohol with a density of \(\delta=0.2 \mathrm{g} / \mathrm{cm}^{2}\) flows over the closed curve \(\mathbf{r}(t)=(\sin t) \mathbf{i}-(\cos t) \mathbf{j}, 0 \leq t \leq 2 \pi,\) according to the vector field \(\mathbf{F}=\delta \mathbf{v},\) where \(\mathbf{v}=(x-y) \mathbf{i}+x^{2} \mathbf{j}\) is a velocity field measured in centimeters per second. Find the circulation of \(\mathbf{F}\) around the curve \(\mathbf{r}(t) .\)

Step-by-Step Solution

Answer

The circulation of \( \mathbf{F} \) around the curve is zero.

Step 1: Parameterize the Curve

The curve is defined by the vector function \( \mathbf{r}(t) = (\sin t) \mathbf{i} - (\cos t) \mathbf{j} \). This represents a circle of radius 1 centered at the origin, traced counterclockwise as \( t \) increases from 0 to \( 2\pi \).

Step 2: Substitute Curve into Velocity Field

The velocity field is given as \( \mathbf{v} = (x - y) \mathbf{i} + x^2 \mathbf{j} \). Substituting \( x = \sin t \) and \( y = -\cos t \) into \( \mathbf{v} \), we have \( \mathbf{v} = (\sin t + \cos t) \mathbf{i} + (\sin^2 t) \mathbf{j} \).

Step 3: Determine the Force Field

Since \( \mathbf{F} = \delta \mathbf{v} \) and \( \delta = 0.2 \), substitute into the expression for \( \mathbf{F} \) to get \( \mathbf{F} = 0.2(\sin t + \cos t) \mathbf{i} + 0.2(\sin^2 t) \mathbf{j} \).

Step 4: Compute the Line Integral for Circulation

The circulation is given by the line integral \( \oint_{C} \mathbf{F} \cdot d \mathbf{r} \), where \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \ dt \). Therefore, \( \mathbf{F} \cdot d\mathbf{r} = [0.2(\sin t + \cos t)(\cos t) + 0.2(\sin^2 t)(\sin t)] dt \).

Step 5: Integrate over the Parameter Interval

Integrate the expression from Step 4 from \( t = 0 \) to \( t = 2\pi \): \[\int_0^{2\pi} \left[ 0.2 (\sin t + \cos t) \cos t + 0.2 \sin^2 t \sin t \right] dt\]On closer inspection, this integral equates to zero because it represents a conservative field over a closed path.

Key Concepts

Parameterization of Curves

Parameterizing a curve means describing it in terms of a parameter, often related to time. It's like giving the curve a path or a trail to follow.
In this exercise, the curve is represented by the vector function \( \mathbf{r}(t) = (\sin t) \mathbf{i} - (\cos t) \mathbf{j} \). This is a circular path with a radius of 1, centered at the origin of the plane.
The parameter \( t \) ranges from 0 to \( 2\pi \), indicating that the curve is traced out one full time. When you plot this function, it draws out a circle as \( t \) moves through its interval. Parameterizing curves in this way is essential for applying vector calculus techniques like line integrals.

Line Integrals

Line integrals extend the concept of integration to functions along a curve. Instead of summing values over an interval or area, we sum them along a path.
The line integral of a vector field \( \mathbf{F} \) around a curve \( C \) is denoted as \( \oint_{C} \mathbf{F} \cdot d \mathbf{r} \), where \( d\mathbf{r} \) is a small segment along the curve. Think of it like adding up all the tiny steps or impacts along the way.
This integral tells you how much of the vector field \( \mathbf{F} \) follows the curve \( C \) when traveling around it. In practical terms, it gives the 'circulation' or 'flow' of the field along the closed path traced by the parameterized curve.

Conservative Vector Fields

A vector field is conservative if its line integral around any closed path is zero. This is a core property and can be thought of as having zero net circulation.
In this exercise, the field \( \mathbf{F} = 0.2(\sin t + \cos t) \mathbf{i} + 0.2(\sin^2 t) \mathbf{j} \) is shown to yield an integral of zero. This confirms that the field is indeed conservative.
For a vector field to be conservative, there must exist a scalar potential function such that the field is the gradient of this function. Conservative fields imply path independence of the integral, meaning the integral's value is determined solely by the start and end points, not the specific route taken.