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Chapter 5: Problem 13

Find the work done by force field \(\mathbf{F}(x, y, z)=-\frac{1}{2} x \mathbf{i}-\frac{1}{2} y \mathbf{j}+\frac{1}{4} \mathbf{k}\) on a particle as it moves along the helix \(\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+t \mathbf{k}\) from point \((1,0,0)\) to point \((-1,0,3 \pi)\)

Short Answer

Expert verified
The work done by the force field is \(\frac{3\pi}{4}\).

Step by step solution

01

Define the Problem

We need to find the work done by the vector field \( \mathbf{F}(x, y, z) = -\frac{1}{2} x \mathbf{i} - \frac{1}{2} y \mathbf{j} + \frac{1}{4} \mathbf{k} \) on a particle moving along the path given by \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + t \mathbf{k} \). The particle moves from the point \((1,0,0)\) to the point \((-1,0,3\pi)\). This is a line integral problem in vector calculus.
02

Determine Parameter Range for t

The particle starts at \((1,0,0)\) when \(t=0\) because \(\cos(0) = 1\), \(\sin(0) = 0\), and \(t = 0\). The particle ends at \((-1,0,3\pi)\) when \(t=3\pi\) because \(\cos(3\pi) = -1\), \(\sin(3\pi) = 0\), and \(t = 3\pi\). Therefore, \(t\) ranges from \(0\) to \(3\pi\).
03

Find the Derivative of the Path

The derivative of the path \(\mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + t \mathbf{k}\) with respect to \(t\) is the velocity vector: \(\mathbf{r}'(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} + \mathbf{k}\).
04

Calculate the Line Integral

The work done by the force field is given by the line integral \( W = \int_{C} \mathbf{F} \cdot d \mathbf{r} = \int_{0}^{3\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt \). We substitute \(\mathbf{F}(\mathbf{r}(t)) = -\frac{1}{2} \cos t \mathbf{i} - \frac{1}{2} \sin t \mathbf{j} + \frac{1}{4} \mathbf{k}\) and \(\mathbf{r}'(t)\).
05

Evaluate the Dot Product

The dot product \(\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \left(-\frac{1}{2} \cos t \right)(-\sin t) + \left(-\frac{1}{2} \sin t\right)(\cos t) + \frac{1}{4}(1)\) simplifies to \(\frac{1}{2} \cos t \sin t - \frac{1}{2} \sin t \cos t + \frac{1}{4}\), which further simplifies to \(\frac{1}{4}\).
06

Integrate the Dot Product

Since \(\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \frac{1}{4}\), the line integral \( W = \int_{0}^{3\pi} \frac{1}{4} \, dt \) evaluates to \(\frac{1}{4}[t]_{0}^{3\pi} = \frac{1}{4}(3\pi - 0) = \frac{3\pi}{4}\).
07

Conclusion

The work done by the force field \(\mathbf{F}\) on the particle moving along the path \(\mathbf{r}(t)\) from \((1,0,0)\) to \((-1,0,3\pi)\) is \(\frac{3\pi}{4}\).

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Key Concepts

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

Line Integrals
Line integrals are a way to integrate functions along a curve or path in vector calculus. They are especially useful for calculating work done by a force field on a particle moving along a specific trajectory. Unlike regular integrals that span an interval on the x-axis, line integrals account for paths in two or three dimensions and involve vector fields that interact with moving objects.

To compute a line integral, you must:
  • Identify the path, often given by a parameterized curve, \( \mathbf{r}(t) \). For instance, our problem uses \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + t \mathbf{k} \).
  • Determine the vector field \( \mathbf{F}(x, y, z) \) affecting the path. Here, \( \mathbf{F}(x, y, z) = -\frac{1}{2} x \mathbf{i} - \frac{1}{2} y \mathbf{j} + \frac{1}{4} \mathbf{k} \).
  • Compute the derivative of the path, \( \mathbf{r}'(t) \), which gives the tangent vector to the path.
  • Substitute \( \mathbf{r}(t) \) into \( \mathbf{F} \) to get \( \mathbf{F}(\mathbf{r}(t)) \), the force at each point along the path.
  • Compute the dot product \( \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \) at each point, which gives the work done at a specific point.
  • Finally, integrate the dot product over the parameter range to find the total work done.
In our case, the integral simplifies to a constant, making the computation straightforward.
Force Fields
A force field is a vector field that describes the influence that forces exert on objects throughout a space. In physics and vector calculus, force fields are essential for expressing how forces like gravity, electricity, or magnetism act in various regions.

In this context, the force field \( \mathbf{F}(x, y, z) = -\frac{1}{2} x \mathbf{i} - \frac{1}{2} y \mathbf{j} + \frac{1}{4} \mathbf{k} \) is affecting a particle moving along a specific path. Here are some points to consider with force fields:
  • Force fields are usually functions of position, meaning they change as you move through space. They provide both direction and magnitude of force at each point, expressed as vectors.
  • The components of a force field can represent different influences, such as the \( -\frac{1}{2} x \mathbf{i} \) and \( -\frac{1}{2} y \mathbf{j} \) parts, which suggest that the force depends on the position in the x and y coordinates.
  • Understanding how a particle interacts with a force field along its path is crucial for evaluating work, energy, or motion.
This setup enables the computation of how much work a force field performs as a particle moves through space, providing insights into dynamic systems.
Helical Paths
Helical paths are fascinating curves found in nature and physics, looking like the shape of a spring or a slinky. In mathematics, these paths are typically defined by parametric equations that involve circular motion along with a linear component in another direction, giving it a 3D twist or spiral.

For our exercise, the helical path is given by the parametric equations \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + t \mathbf{k} \). Here鈥檚 how the helical path works:
  • The terms \( \cos t \mathbf{i} + \sin t \mathbf{j} \) define a circle in the \( xy \)-plane, moving counterclockwise as \( t \) increases.
  • The \( t \mathbf{k} \) component describes upward linear movement in the z-direction, leading to a spiral as it combines with the circular path.
  • Helical paths are found in structures like DNA and certain mechanical springs, showcasing their importance beyond mere calculus problems.
Understanding helical paths is vital for analyzing motion in fields such as physics and engineering, where objects often move in complex, non-linear ways.

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Most popular questions from this chapter

A flow line (or streamline) of a vector field \(\mathbf{F}\) is a curve \(\mathbf{r}(t)\) such that \(d \mathbf{r} / d t=\mathbf{F}(\mathbf{r}(t))\). If \(\mathbf{F}\) represents the velocity field of a moving particle, then the flow lines are paths taken by the particle. Therefore, flow lines are tangent to the vector field. For the following exercises, show that the given curve \(\mathbf{c}(t)\) is a flow line of the given velocity vector field \(\mathbf{F}(x, y, z)\). $$ \mathbf{c}(t)=\left(e^{2 t}, \ln |t|, \frac{1}{t}\right), t \neq 0 ; \mathbf{F}(x, y, z)=\left\langle 2 x, z,-z^{2}\right\rangle $$

Evaluate the following integrals.\(\int_{C} y d x+x y^{2} d y\), where \(C: x=\sqrt{t}, y=t-1,0 \leq t \leq 1\)

For the following exercises, write formulas for the vector fields with the given properties. Give a formula \(\mathbf{F}(x, y)=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) for the vector field in a plane that has the properties that \(\mathbf{F}=0\) at \((0,0)\) and that at any other point \((a, b), \mathbf{F}\) is tangent to circle \(x^{2}+y^{2}=a^{2}+b^{2}\) and points in the clockwise direction with magnitude \(|\mathbf{F}|=\sqrt{a^{2}+b^{2}}\).

Use the divergence theorem to evaluate \(\iint_{S} \mathbf{F} \cdot d S .$$\mathbf{F}(x, y, z)=(2 x y) \mathbf{i}+\left(-y^{2}\right) \mathbf{j}+\left(2 z^{3}\right) \mathbf{k}\), where \(S\) is bounded by paraboloid \(z=x^{2}+y^{2}\) and plane \(z=2\)

For the following exercises, evaluate \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d s\) for vector field \(F\), where \(\mathbf{N}\) is an outward normal vector to surface \(S\). \(\mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}\), and \(S\) is the portion of plane \(z=y+1\) that lies inside cylinder \(x^{2}+y^{2}=1\)
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