Problem 22 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The total work done by \( \mathbf{F} \) along the path is 0.

Step 1: Understand the Problem

We need to find the work done by the vector field \( \mathbf{F} = 6z \mathbf{i} + y^2 \mathbf{j} + 12x \mathbf{k} \) along the path defined by \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} + (t/6) \mathbf{k} \) from \( t = 0 \) to \( t = 2\pi \).

Step 2: Compute Derivative of the Curve

First, we find the derivative \( \mathbf{r}'(t) \) of the parametrization \( \mathbf{r}(t) \). Thus,\[ \mathbf{r}'(t) = \frac{d}{dt}(\sin t) \mathbf{i} + \frac{d}{dt}(\cos t) \mathbf{j} + \frac{d}{dt}(t/6) \mathbf{k} = \cos t \mathbf{i} - \sin t \mathbf{j} + (1/6) \mathbf{k}. \]

Step 3: Evaluate the Force Field at \( \mathbf{r}(t) \)

We substitute \( x = \sin t \), \( y = \cos t \), and \( z = t/6 \) into \( \mathbf{F} \), resulting in:\[ \mathbf{F}(t) = 6(t/6) \mathbf{i} + (\cos t)^2 \mathbf{j} + 12(\sin t) \mathbf{k} = t \mathbf{i} + (\cos t)^2 \mathbf{j} + 12 \sin t \mathbf{k}. \]

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

Now find the dot product of \( \mathbf{F}(t) \) and \( \mathbf{r}'(t) \):\[ \mathbf{F}(t) \cdot \mathbf{r}'(t) = (t \mathbf{i} + (\cos t)^2 \mathbf{j} + 12 \sin t \mathbf{k}) \cdot (\cos t \mathbf{i} - \sin t \mathbf{j} + 1/6 \mathbf{k}) = t \cos t - (\cos t)^3 + 2\sin t. \]

Step 5: Integrate Over the Interval \( [0, 2\pi] \)

Calculate the integral to find the work done:\[ \int_0^{2\pi} (t \cos t - (\cos t)^3 + 2\sin t) \, dt. \]This is computed to get the total work done. The integral simplifies as follows:1. \( \int_0^{2\pi} t \cos t \, dt \) can be solved using integration by parts.2. \( \int_0^{2\pi} (\cos t)^3 \, dt \) requires a trigonometric identity and substitution.3. \( 2 \int_0^{2\pi} \sin t \, dt = 0 \), since the integral of sine over a full period is zero.

Step 6: Calculate Values of Integrals

- The integral \( \int_0^{2\pi} t \cos t \, dt \) simplifies to zero after evaluation.- The integral \( \int_0^{2\pi} (\cos t)^3 \, dt \) is handled by using \( \cos^3 t = (3\cos t + \cos 3t)/4 \), which produces zero when evaluated over \([0, 2\pi]\).Thus, the total work done is zero.

Key Concepts

Line integrals

When we talk about the work done by a force field along a curve, we are dealing with line integrals. This type of integral helps us calculate the accumulated impact of a vector field along a certain path or curve.

In simpler terms, a line integral sums up small contributions from a vector field along a given path. Imagine walking along a trail while the wind (vector field) pushes against you — the line integral of this force measures how much work it takes you to complete the walk.

For a curve defined parametrically as \( \mathbf{r}(t) \) and a vector field \( \mathbf{F} \), the line integral is given by:
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt. \]
Here, \( \mathbf{r}'(t) \) is the derivative of the parameterized curve with respect to \( t \), describing the path's direction at each point. The expression \( \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \) refers to the dot product of the vector field \( \mathbf{F} \) at the path with the path's direction vector. This dot product essentially measures how aligned the vector field is with the path, impacting the net work calculated by the line integral.

Parametric equations

To comprehend how line integrals work, it's essential to understand parametric equations. These equations enable us to describe curves and paths in two or three dimensions using a parameter, commonly denoted as \( t \).

Think of parametric equations as instructions or formulas that tell you how to travel along a path over time. For example, a parametric equation \( \mathbf{r}(t) = (\sin t)\mathbf{i} + (\cos t)\mathbf{j} + (t/6)\mathbf{k} \), describes a path that twists in three-dimensional space.

Using parameters makes it easier to incorporate time and motion into mathematical models. Here, as \( t \) varies from 0 to \( 2\pi \), the point \( \mathbf{r}(t) \) traces out the path in space.

Parametric equations are also vital for finding derivatives and tangents to these paths, which help in calculating the line integrals by giving the direction vector \( \mathbf{r}'(t) \). This vector has components indicative of how fast and in what direction the path is changing as \( t \) changes.

Vector calculus

Vector calculus is a branch of mathematics that deals with vector fields and operations on them, such as differentiation and integration.

One key operation in vector calculus relevant to the problem concerns how vector fields are integrated along curves — a vital concept when calculating work done.

In our problem, the vector field \( \mathbf{F} = 6z \mathbf{i} + y^2 \mathbf{j} + 12x \mathbf{k} \) is evaluated along a parameterized path given by \( \mathbf{r}(t) \). The evaluation of \( \mathbf{F} \) at points along the path transforms it into a function of \( t \), facilitating the calculation of a line integral.

Vector calculus further includes calculating dot products of vectors. In our case, it's used to compute \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \), representing the force in the direction of the path.

The ultimate goal in problems like ours is to learn about the net effect of a field on a path, which is efficiently achieved through line integrals facilitated by vector calculus concepts, combining force fields and travel directions to reach a meaningful physical quantity: work done.