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Chapter 16: Problem 40

In Exercises \(37-40, \mathbf{F}\) 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 \(t .\) $$ \begin{array}{ll}{\mathbf{F}=-y \mathbf{i}+x \mathbf{j}+2 \mathbf{k}} \\\ {\mathbf{r}(t)=(-2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{j}+2 t \mathbf{k},} & {0 \leq t \leq 2 \pi}\end{array} $$

Short Answer

Expert verified
The flow along the curve is zero.

Step by step solution

01

Parametrize the Curve

The curve is already given in a parametrized form as \( \mathbf{r}(t) = (-2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + 2t \mathbf{k} \) with \( 0 \leq t \leq 2\pi \). We use this parameterization throughout the solution.
02

Determine the Differential Element

Take the derivative of the curve function to find \( \mathbf{r}'(t) \). \( \mathbf{r}'(t) = \frac{d}{dt}[(-2\cos t) \mathbf{i} + (2\sin t) \mathbf{j} + 2t \mathbf{k}] = (2\sin t) \mathbf{i} + (2\cos t) \mathbf{j} + 2 \mathbf{k}. \)
03

Evaluate the Velocity Field at \( \mathbf{r}(t) \)

To find \( \mathbf{F}(\mathbf{r}(t)) \), substitute \( x = -2\cos t \) and \( y = 2\sin t \) into \( \mathbf{F} = -y \mathbf{i} + x \mathbf{j} + 2 \mathbf{k} \). This results in \( \mathbf{F}(\mathbf{r}(t)) = -2\sin t \mathbf{i} - 2\cos t \mathbf{j} + 2 \mathbf{k}. \)
04

Calculate the Dot Product

Calculate the dot product \( \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \). This gives: \((-2\sin t) \cdot (2\sin t) + (-2\cos t) \cdot (2\cos t) + 2 \cdot 2 = -4\sin^2 t - 4\cos^2 t + 4 = -4(\sin^2 t + \cos^2 t) + 4 = 0. \)
05

Integrate Over the Interval

To find the flow, integrate the result of the dot product over the interval \([0, 2\pi]\). The integral is \( \int_{0}^{2\pi} 0 \, dt = 0. \)
06

Conclusion

Since the integral evaluates to zero, the flow along the curve in the direction of increasing \( t \) is zero throughout the path.

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Key Concepts

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

Velocity Field
A velocity field is a mathematical description of how fluid particles move at every point in a region of space. Imagine a stream of water flowing in a river; every point within the river can have its speed and direction described by vectors. In the exercise, the velocity field is given as \( \mathbf{F} = -y \mathbf{i} + x \mathbf{j} + 2 \mathbf{k} \). This means that the fluid's velocity depends on its position \((x, y)\) in space.
  • \( -y \mathbf{i} \) and \( x \mathbf{j} \) represent the horizontal components of the velocity, changing with location.
  • \( 2 \mathbf{k} \) is constant along the vertical, indicating a uniform upward flow at every point.
Understanding a velocity field helps in visualizing how particles move, crucial for calculating flow rates and directions along specified paths.
Line Integral
A line integral allows us to integrate a vector field along a path or curve. It helps determine the cumulative effect of a field, like force or velocity, along a curve. In this exercise, we need the line integral of the velocity field along a specific parametrized path. The flow along a curve is found by calculating the line integral.
  • Set up the curve's parameterization.
  • Evaluate the vector field at this parameterization.
  • Calculate the dot product with the curve's derivative.
By integrating the resulting function over the curve's interval, we find the total flow. If this integral results in zero, as it does here, the flow along the path has no net accumulation, meaning any fluid flow through the curve cancels out over time.
Parameterization
Parameterization is a way to describe a curve or surface in terms of a single variable, usually \( t \). It boils down complex curves into a form easily manageable for calculus operations like differentiation or integration. A parameter gives us a handle to simplify the vector equation of a path.
Given in the problem is \( \mathbf{r}(t) = (-2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + 2t \mathbf{k} \), describing the curve entirely in terms of \( t \).
  • Each coordinate is expressed as a function of \( t \).
  • Derivatives of these functions provide tangent vectors, crucial for the line integral.
This method transforms understanding curves in 3D space to more manageable 1D intervals, making problems accessible and computationally feasible.
Dot Product
The dot product is a fundamental algebraic operation involving vectors, crucial in this exercise to determine the projection of one vector onto another. It's computed by multiplying corresponding components and summing the results. The dot product helps understand how much of one vector aligns with another.
In this exercise, the calculated dot product is \( \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \), which essentially measures the component of the velocity field that aligns with the curve's direction.
  • If the dot product is zero over the entire interval, as it is here, this implies no net flow along the path.
  • Positive values indicate flow along the path, while negative values indicate flow against it.
This operation is central for calculating line integrals and determining the geometry of curves relative to vector fields.

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Most popular questions from this chapter

In Exercises \(47-52,\) use a CAS to perform the following steps for finding the work done by force \(\mathbf{F}\) over the given path: a. Find \(d \mathbf{r}\) for the path \(\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k}\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{ll}{\mathbf{F}=2 x y \mathbf{i}-y^{2} \mathbf{j}+z e^{x} \mathbf{k} ;} & {\mathbf{r}(t)=-t \mathbf{i}+\sqrt{t} \mathbf{j}+3 t \mathbf{k}} \\ {1 \leq t \leq 4}\end{array} $$

In Exercises \(13-18\) , use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field \(\mathbf{F}\) across the surface \(S\) in the direction of the outward unit normal \(\mathbf{n} .\) $$ \begin{array}{l}{\mathbf{F}=(x-y) \mathbf{i}+(y-z) \mathbf{j}+(z-x) \mathbf{k}} \\ {S : \quad \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+(5-r) \mathbf{k}} \\ {0 \leq r \leq 5, \quad 0 \leq \theta \leq 2 \pi}\end{array} $$

Let \(S\) be the cylinder \(x^{2}+y^{2}=a^{2}, 0 \leq z \leq h,\) together with its top, \(x^{2}+y^{2} \leq a^{2}, z=h .\) Let \(\mathbf{F}=-y \mathbf{i}+x \mathbf{j}+x^{2} \mathbf{k} .\) Use Stokes' Theorem to find the flux of \(\nabla \times \mathbf{F}\) outward through \(S .\)

Find the flux of the field \(\mathbf{F}(x, y, z)=z^{2} \mathbf{i}+x \mathbf{j}-3 z \mathbf{k}\) outward through the surface cut from the parabolic cylinder \(z=4-y^{2}\) by the planes \(x=0, x=1,\) and \(z=0\)

Unit vectors pointing toward the origin Find a field \(\mathbf{F}=\) \(M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) in the \(x y\) -plane with the property that at each point \((x, y) \neq(0,0), \mathbf{F}\) is a unit vector pointing toward the origin. (The field is undefined at \((0,0) . )\)
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