Problem 22 - Thomas Calculus

Question

In Exercises \(19-22,\) find the work done by \(F\) over the curve in the direction of increasing \(t .\) \begin{equation} \begin{array}{l}{\mathbf{F}=6 z \mathbf{i}+y^{2} \mathbf{j}+12 x \mathbf{k}} \\\ {\mathbf{r}(t)=(\sin t) \mathbf{i}+(\cos t) \mathbf{j}+(t / 6) \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array} \end{equation}

Step-by-Step Solution

Answer

The work done by \( \mathbf{F} \) over the curve is 0.

Step 1: Understanding the Problem

We are given a vector field \( \mathbf{F} = 6z \mathbf{i} + y^2 \mathbf{j} + 12x \mathbf{k} \) and a parameterized curve \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (t/6) \mathbf{k} \) for \( 0 \leq t \leq 2\pi \). We need to find the work done by \( \mathbf{F} \) over this curve.

Step 2: Calculate the Derivative of \( \mathbf{r}(t) \)

The derivative \( \mathbf{r}'(t) \) is obtained by differentiating each component of \( \mathbf{r}(t) \). Thus, \( \mathbf{r}'(t) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} + (1/6) \mathbf{k} \).

Step 3: Express \( \mathbf{F} \) in terms of \( t \)

Substitute \( x = \sin t \), \( y = \cos t \), and \( z = t/6 \) into \( \mathbf{F} \) to get \( \mathbf{F} = (t) \mathbf{i} + (\cos^2 t) \mathbf{j} + (72\sin t) \mathbf{k} \).

Step 4: Compute the Dot Product \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \)

Calculate the dot product: \( \mathbf{F}(t) \cdot \mathbf{r}'(t) = (t)(\cos t) + (\cos^2 t)(-\sin t) + (72\sin t)(1/6) .\) Simplify this expression: \( t \cos t - \cos^2 t \sin t + 12 \sin t \).

Step 5: Set up the Integral for Work

The work done \( W \) is the integral of the dot product from \( t = 0 \) to \( t = 2\pi \). Set up the integral: \( W = \int_{0}^{2\pi} (t \cos t - \cos^2 t \sin t + 12 \sin t) \, dt .\)

Step 6: Evaluate the Integral

Evaluate each term in the integral separately: 1. \( I_1 = \int_{0}^{2\pi} t \cos t \, dt \)2. \( I_2 = \int_{0}^{2\pi} -\cos^2 t \sin t \, dt \)3. \( I_3 = \int_{0}^{2\pi} 12 \sin t \, dt \) Calculate these integrals using integration techniques.

Step 7: Compute and Add the Results

Calculate each integral; use integration by parts for \( I_1 \) and recognize periodicity in \( I_3 \): 1. \( I_1 = 0 \) (after simplifying by parts)2. \( I_2 = 0 \) (symmetry and periodicity)3. \( I_3 = 0 \) (\( \int_0^{2\pi} 12 \sin t \, dt \) evaluates to zero)Hence, \( W = 0 + 0 + 0 = 0 \).

Key Concepts

Work Done in a Vector Field

When we talk about work done in a vector field, we're addressing a fundamental concept in vector calculus. Imagine we have a force field, represented by a vector field \( \mathbf{F} \), and a path or curve along which an object moves. The work done by the force \( \mathbf{F} \) as the object moves along this path is essentially the accumulated effect of the force along the trajectory of the object.
This is expressed as a line integral which takes into account how the force interacts with the displacement along the curve. In simpler terms, it's like calculating the total effort required to move an object through a varying field of forces, such as gravity or an electric field. In mathematical terms, the work done \( W \) by a vector field \( \mathbf{F} \) over a curve given by \( \mathbf{r}(t) \) is given by the integral:

• \( W = \int_{C} \mathbf{F} \cdot d\mathbf{r} \)

This incorporates both direction and magnitude of the force and path, resulting in a comprehensive calculation of work done.

Parametric Curves

Parametric curves provide a way to describe a curve in space using parameters, typically denoted by \( t \). Instead of writing \( y \) as a function of \( x \), parametric equations express both \( x \) and \( y \) (and possibly \( z \) if in three dimensions) in terms of \( t \). This is particularly useful in vector calculus, where curves may not easily fit standard Cartesian formats.
Consider the parametric equations:

• \( x(t) = \sin t \)
• \( y(t) = \cos t \)
• \( z(t) = t/6 \)

Here, each of these represents the coordinates along the curve as a function of \( t \), creating a space curve when taken together. This space curve becomes our path or trajectory upon which we calculate things like line integrals or work done. The benefit of this approach is the flexibility it offers in describing complex, real-world paths, such as those curves used in engineering and physics.

Line Integrals

Line integrals are a powerful tool in vector calculus. They allow us to compute various quantities such as work done, flow, or even mass along a curve. What distinguishes line integrals from regular integrals is their focus on integration over a curve in a multidimensional space, rather than along a straight line.
To compute a line integral of a vector field \( \mathbf{F} \) over a curve \( C \), parameterized by \( \mathbf{r}(t) \), we use the following integral:

• \( \int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{a}^{b} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt \)

Here, \( d\mathbf{r} \) represents an infinitesimal segment along the curve, and the dot product \( \mathbf{F} \cdot d\mathbf{r} \) accounts for how much of the vector field's force is aligned with the curve's path at each point, effectively measuring the 'influence' of the field over the entire curve.
This consideration of both orientation and magnitude allows line integrals to provide meaningful insights in fields like electromagnetism, fluid flow, and beyond. For example, in our original problem, by processing the integral limits, we effectively calculated the work done across a full cycle of the curve, leading to insights such as cyclical forces exerting zero net work.