Problem 37 - Thomas Calculus

Question

The flow of a gas with a density of \(\delta=0.001 \mathrm{kg} / \mathrm{m}^{2}\) over the closed curve \(\mathbf{r}(t)=(-\sin t) \mathbf{i}+(\cos t) \mathbf{j}, 0 \leq t \leq 2 \pi,\) is given by the vector field \(\mathbf{F}=\delta \mathbf{v},\) where \(\mathbf{v}=x \mathbf{i}+y^{2} \mathbf{j}\) is a velocity field measured in meters per second. Find the flux of \(\mathbf{F}\) across the curve \(\mathbf{r}(t) .\)

Step-by-Step Solution

Answer

The flux of \( \mathbf{F} \) across the curve is 0.

Step 1: Understand the Problem

We are given a closed curve in parametric form \( \mathbf{r}(t)=(-\sin t) \mathbf{i}+(\cos t) \mathbf{j} \) for \( 0 \leq t \leq 2 \pi \), and a velocity field \( \mathbf{v}=x \mathbf{i}+y^{2} \mathbf{j} \). We need to find the flux of the vector field \( \mathbf{F}=\delta \mathbf{v} \), where \( \delta=0.001 \mathrm{kg} / \mathrm{m}^{2} \), across the curve \( \mathbf{r}(t) \).

Step 2: Find the Vector Field \( \mathbf{F} \)

Substitute \( \mathbf{v}=x \mathbf{i}+y^{2} \mathbf{j} \) and \( \delta=0.001 \) into \( \mathbf{F}=\delta \mathbf{v} \) to get \( \mathbf{F}=0.001(x \mathbf{i} + y^2 \mathbf{j}) \).

Step 3: Parameterize the Curve

We are given \( \mathbf{r}(t)=(-\sin t) \mathbf{i}+(\cos t) \mathbf{j} \) indicating that \( x = -\sin t \) and \( y = \cos t \). Using these, we can rewrite \( \mathbf{F} \) in terms of \( t \):\[\mathbf{F} = 0.001(-\sin t \mathbf{i} + \cos^2 t \mathbf{j}).\]

Step 4: Find the Tangent Vector

The tangent vector to the curve is \( \mathbf{r}'(t)=\frac{d}{dt}((-\sin t) \mathbf{i}+(\cos t) \mathbf{j}) = (-\cos t) \mathbf{i} - (\sin t) \mathbf{j} \).

Step 5: Compute the Normal Vector

The unit normal vector \( \mathbf{n}(t) \) for an outward-oriented curve in the plane can be found by rotating the tangent vector \( \mathbf{r}'(t) \) by 90 degrees clockwise: \( \mathbf{n}(t) = (\sin t) \mathbf{i} - (\cos t) \mathbf{j} \).

Step 6: Compute the Flux Integral

The flux \( \phi \) across the closed curve is given by the line integral:\[ \phi = \int_{0}^{2\pi} \mathbf{F} \cdot \mathbf{n}\, dt. \]Substitute \( \mathbf{F} = 0.001(-\sin t \mathbf{i} + \cos^2 t \mathbf{j}) \) and \( \mathbf{n}(t) = (\sin t) \mathbf{i} - (\cos t) \mathbf{j} \):\[ \phi = \int_{0}^{2\pi} 0.001(-\sin t \cdot \sin t + \cos^2 t \cdot -\cos t) dt. \]

Step 7: Simplify the Expression

Simplify the expression for the integrand: \[ \phi = \int_{0}^{2\pi} 0.001(-\sin^2 t - \cos^3 t)\, dt = -0.001 \int_{0}^{2\pi} (\sin^2 t + \cos^3 t)\, dt. \]

Step 8: Evaluate the Integral

Since the integrals of \( \sin^2 t \) and \( \cos^3 t \) over \( [0, 2\pi] \) both have symmetric limits implying zero contributions:\[ \int_{0}^{2\pi} \sin^2 t\, dt = \int_{0}^{2\pi} \cos^3 t\, dt = 0. \]Thus, the whole integral \( \phi = -0.001 \times 0 = 0. \) Therefore, the flux is zero.

Key Concepts

Vector Field

A vector field is a mathematical structure where each point in space is associated with a vector. This can be visualized as a collection of arrows, each possibly having different magnitudes and directions throughout a region. In our problem, the vector field \( \mathbf{F} = 0.001(x \mathbf{i} + y^2 \mathbf{j}) \) represents the flow of gas across a plane. Each vector within this field gives information about the direction and speed of the gas flow at each point. Understanding how this vector field behaves is crucial as it forms the basis of analyzing the flow of substances across curves or surfaces.

Line Integral

A line integral allows us to evaluate functions over a curve. It's analogous to adding up small contributions along a path. In the context of flow and vector fields, line integrals measure the cumulative effect of the field along the curve.
For example, if you have a vector field representing fluid flow, the line integral can give you the total amount of fluid passing through a curved boundary. In our case, the line integral \[ \phi = \int_{0}^{2\pi} \mathbf{F} \cdot \mathbf{n} \] measures the flux through our closed curve. It considers not just the magnitude of the vector field but also how it aligns with the curve's orientation.

Parametric Curves in Calculus

In calculus, parametric curves offer a powerful way to represent paths by using parameters like \(t\) instead of the traditional \(x\)-\(y\) equation. In essence, the components of the position vector are defined as functions of a parameter.
This approach lends itself well to describing curves for which traditional methods fall short. For instance, a circle can be described as \( \mathbf{r}(t)=(-\sin t) \mathbf{i}+\cos t \mathbf{j} \). Using parametric representations allows for a straightforward calculation of derivatives, such as tangent vectors, which play a key role in computing integrals.

Flux Integral

Flux integrals specifically calculate the amount of vector field passing through a surface or curve. In our example, the flux integral helps determine how much of the vector field \( \mathbf{F} \) flows across a given path.
Computing the flux requires the use of a normal vector to the curve. The formula for the flux, \[ \phi = \int_{0}^{2\pi} \mathbf{F} \cdot \mathbf{n} \, dt, \] combines the vector field and the curve's normal vector, providing a scalar result that represents the total flow rate. Interpreting this result tells us much about the behavior of the field and flow across the boundary.

Tangent Vector

The tangent vector illustrates the direction in which a curve proceeds at any given point. Determined by differentiating the parametric equations of a curve, it serves to guide us in understanding the local behavior of a curve.
For the curve \( \mathbf{r}(t)=(-\sin t) \mathbf{i}+\cos t \mathbf{j} \), the tangent vector is \( \mathbf{r}'(t)=(-\cos t) \mathbf{i} - (\sin t) \mathbf{j} \). This vector points in the immediate direction of increasing \(t\), offering a snapshot of the curve's trajectory. The tangent vector is not only essential in understanding the curve but also in deriving the normal vector, which is pivotal for calculating the flux.