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Chapter 16: Problem 49

\(\mathbf{F}\) is the velocity field of a fluid flowing through a region in space. Find the flow along the given curve in the direction of increasing \(t.\) $$\begin{aligned} &\mathbf{F}=(x-z) \mathbf{i}+x \mathbf{k}\\\ &\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{k}, \quad 0 \leq t \leq \pi \end{aligned}$$

Short Answer

Expert verified
The flow along the curve is \( \pi \).

Step by step solution

01

Identify Parameterization of the Curve

The curve is parameterized by the vector-valued function \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{k} \) for \( 0 \leq t \leq \pi \). This describes a semicircle in the \( xz \)-plane, traversed from \( t=0 \) (point (1,0)) to \( t=\pi \) (point (-1,0)).
02

Calculate the Derivative of \( \mathbf{r}(t) \)

The velocity of a point on the curve as \( t \) changes is given by the derivative \( \mathbf{r}'(t) = \frac{d}{dt}((\cos t) \mathbf{i} + (\sin t) \mathbf{k}) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{k} \).
03

Substitute into the Velocity Field \( \mathbf{F} \)

Substitute \( x = \cos t \) and \( z = \sin t \) into \( \mathbf{F} = (x - z) \mathbf{i} + x \mathbf{k} \). This gives \( \mathbf{F}(t) = (\cos t - \sin t) \mathbf{i} + \cos t \mathbf{k} \).
04

Calculate the Dot Product \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \)

The rate of flow along the curve is given by the dot product \( \mathbf{F}(t) \cdot \mathbf{r}'(t) = ((\cos t - \sin t) \mathbf{i} + \cos t \mathbf{k}) \cdot ((-\sin t) \mathbf{i} + (\cos t) \mathbf{k}) = (\cos t - \sin t)(-\sin t) + \cos t(\cos t) \).
05

Simplify and Integrate the Dot Product Over the Given Interval

Simplify the dot product: \( -\cos t \sin t + \sin^2 t + \cos^2 t = -\cos t \sin t + 1\). Integrate this expression with respect to \( t \) from 0 to \( \pi \):\[\int_{0}^{\pi} (-\cos t \sin t + 1) \, dt = \int_{0}^{\pi} -\frac{1}{2}\sin(2t) \, dt + \int_{0}^{\pi} 1 \, dt.\]
06

Evaluate the Integrals

The first integral, \( \int_{0}^{\pi} -\frac{1}{2}\sin(2t) \, dt \), evaluates to 0 since it is the integral of a full period of the sine function. The second integral, \( \int_{0}^{\pi} 1 \, dt = [t]_{0}^{\pi} = \pi - 0 \).
07

Sum the Results of the Integrals

Add the results of the integrals to get the flow along the curve: \( 0 + \pi = \pi \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line Integrals
Line integrals play a crucial role in vector calculus, especially when analyzing fields such as force or velocity fields along a curve. Consider them as a method to measure how a vector field is aligned with a curve as one traverses the curve. In the context of our problem, the line integral is used to calculate the flow of fluid through a specified path.

The formula for a line integral of a vector field \( \mathbf{F} \) along a curve \( C \) is given by \( \int_C \mathbf{F} \cdot d\mathbf{r} \), where \( d\mathbf{r} \) is the differential element of the path.
  • This dot product \( \mathbf{F} \cdot d\mathbf{r} \) finds the component of the vector field that is tangent to the curve.
  • It helps determine the contribution of the vector field along the curve.
  • To compute this line integral, you need a parameterization of the curve and the vector field defined over this curve.
Understanding this concept allows you to calculate quantities like work done by a force or flow rate of a fluid, as we do in this exercise.
Vector Fields
A vector field assigns a vector to every point in space. Imagine it as a weather map where each point indicates wind speed and direction. In terms of mathematics, vector fields are essential for describing fluid flow, electric and magnetic fields, and other systems.

Our exercise uses the vector field \( \mathbf{F} = (x - z) \mathbf{i} + x \mathbf{k} \), representing a fluid's velocity field. Here are some key points to remember:
  • The first component \( (x - z) \mathbf{i} \) influences the flow in the \( x \)-direction.
  • The second component \( x \mathbf{k} \) influences the flow in the \( z \)-direction.
  • Substituting values for \( x \) and \( z \) from the curve's parameterization updates the vector field over the path.
Understanding the vector field allows you to follow how different components interact and contribute to physical phenomena, such as fluid motion in this scenario.
Parameterization of Curves
Parameterization is the process of defining a curve using a parameter, typically denoted as \( t \). This technique is efficient in assigning coordinates to each point on the curve and is vital in transforming equations of curves into vector form.

In the given exercise, the curve is parameterized by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{k} \) for \( 0 \leq t \leq \pi \). This specific parameterization describes a semicircle in the \( xz \)-plane.
  • As \( t \) varies from 0 to \( \pi \), it traces out a curve from the point (1,0) to (-1,0).
  • Parameterization helps in calculating derivatives, which give information on the direction and speed along the curve, crucial for solving problems involving integration along paths.
  • Understanding such parameterization provides a deeper insight into geometrical interpretations and subsequent calculations of physical quantities.
Mastering the parameterization of curves equips you to tackle complex problems involving various shapes and forms in vector calculus.

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Most popular questions from this chapter

Let \(C\) be the ellipse in which the plane \(2 x+3 y-z=0\) meets the cylinder \(x^{2}+y^{2}=12 .\) Show, without evaluating either line integral directly, that the circulation of the field \(\mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) around \(C\) in either direction is zero.

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{n}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{v}\left(u_{0}, v_{0}\right)\) at \(P_{0} .\) In Exercises \(27-30,\) find an equation for the plane tangent to the surface at \(P_{0} .\) Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. The hemisphere surface \(\mathbf{r}(\phi, \theta)=(4 \sin \phi \cos \theta) \mathbf{i}\) \(+(4 \sin \phi \sin \theta) \mathbf{j}+(4 \cos \phi) \mathbf{k}, 0 \leq \phi \leq \pi / 2,0 \leq \theta \leq 2 \pi\) at the point \(P_{0}(\sqrt{2}, \sqrt{2}, 2 \sqrt{3})\) corresponding to \((\phi, \theta)=\) \((\pi / 6, \pi / 4)\)

Let \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2} .\) Show that the clockwise circulation of the field \(\mathbf{F}=\nabla f\) around the circle \(x^{2}+y^{2}=a^{2}\) in the \(x y\) -plane is zero a. by taking \(\mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi,\) and integrating \(\mathbf{F} \cdot d \mathbf{r}\) over the circle. b. by applying Stokes' Theorem.

Use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D\) \(\mathbf{F}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}\) a. Cube \(D:\) The cube cut from the first octant by the planes $$ x=1, y=1, \text { and } z=1 $$ b. Cube \(D:\) The cube bounded by the planes \(x=\pm 1\) $$ y=\pm 1, \text { and } z=\pm 1 $$ c. Cylindrical can \(D: \quad\) The region cut from the solid cylinder $$ \begin{array}{l} x^{2}+y^{2} \leq 4 \text { by the planes } z=0 \text { and } \\ z=1 \end{array} $$

Find the outward flux of the field $$\mathbf{F}=\left(3 x y-\frac{x}{1+y^{2}}\right) \mathbf{i}+\left(e^{x}+\tan ^{-1} y\right) \mathbf{j}$$ across the cardioid \(r=a(1+\cos \theta), a>0.\)
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