Problem 21 - 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}=z \mathbf{i}+x \mathbf{j}+y \mathbf{k}} \\\ {\mathbf{r}(t)=(\sin t) \mathbf{i}+(\cos t) \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array} \end{equation}

Step-by-Step Solution

Answer

The work done is \(\int_0^{2\pi} (t \cos t + \cos 2t) \, dt\).

Step 1: Determine the Differential Element of the Curve

To find the work done by the vector field \(\mathbf{F}\) along the curve \(\mathbf{r}(t)\), we first need the differential element tangential to the curve, given by the derivative of \(\mathbf{r}(t)\). Differentiating the vector function \[\mathbf{r}(t) = \sin t \, \mathbf{i} + \cos t \, \mathbf{j} + t \, \mathbf{k}\] with respect to \(t\), we have: \[\mathbf{r}'(t) = \cos t \, \mathbf{i} - \sin t \, \mathbf{j} + \mathbf{k}\].

Step 2: Express the Force Field at the Point on the Curve

The vector field \(\mathbf{F}\) given is \(z \mathbf{i} + x \mathbf{j} + y \mathbf{k}\). We substitute \(x = \sin t\), \(y = \cos t\), and \(z = t\) from the parametric equations of the curve to express \(\mathbf{F}\) along the curve: \[\mathbf{F}(t) = t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k}\].

Step 3: Establish the Dot Product of \(\mathbf{F}(t)\) and \(\mathbf{r}'(t)\)

The work done is computed as the integral of the dot product \(\mathbf{F}(t) \cdot \mathbf{r}'(t)\) over the parameter \(t\). Calculate this dot product:\[\begin{align*}\mathbf{F}(t) \cdot \mathbf{r}'(t) &= (t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k}) \cdot (\cos t \mathbf{i} - \sin t \mathbf{j} + \mathbf{k}) \&= t \cos t - \sin^2 t + \cos^2 t.\end{align*}\]

Key Concepts

Vector Field

A vector field is a mathematical construct that assigns a vector to every point in space. Imagine a field of arrows, where each arrow represents both a direction and magnitude at a specific point.
In the context of our exercise, the vector field \( \mathbf{F} = z \mathbf{i} + x \mathbf{j} + y \mathbf{k} \) maps a vector to each point defined by the parameters \(x, y, \) and \(z\). Each variable corresponds to a spatial dimension, indicating the vector field is in three-dimensional space.
Understanding vector fields helps in visualizing forces, velocities, or other vector quantities distributed over a region. They are useful in physics and engineering for describing fluid flow, electromagnetic fields, and more.

• The vector \(\mathbf{i}\) corresponds to the x-axis.
• The vector \(\mathbf{j}\) corresponds to the y-axis.
• The vector \(\mathbf{k}\) corresponds to the z-axis.

In this exercise, the field represents certain physical forces or influences acting along a specified curve.

Parametrization

Parametrization is the process of describing a curve or surface in terms of a single parameter, usually denoted by \(t\). It transforms a geometric problem into an algebraic one by representing the coordinates of points on a curve via functions of this parameter.
In our scenario, \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} + t \mathbf{k} \) defines the curve, where \(t\) acts as the parameter. This curve is traced out as \(t\) changes from \(0\) to \(2\pi\).
Parametrization simplifies calculations as it reduces multi-variable problems into single-variable ones:

• \(\sin t\) gives the x-coordinate.
• \(\cos t\) gives the y-coordinate.
• \(t\) itself gives the z-coordinate.

Parametrization is crucial for evaluating line integrals as it provides a way to express all variable dependencies along a curve.

Dot Product

The dot product, also known as the scalar product, is a fundamental operation that takes two equal-length sequences of numbers (usually coordinates of two vectors) and returns a single number.
Formally, if two vectors \( \mathbf{A} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{B} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \), the dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as:
\[a_1b_1 + a_2b_2 + a_3b_3.\]
In our exercise, when we calculate \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \), the result \( t \cos t - \sin^2 t + \cos^2 t \) represents the magnitude of the vector from \(\mathbf{F}\) projected in the direction of \(\mathbf{r}(t)\).
The dot product is utilitarian in many fields, such as in finding the work done by forces, as it multiplies along the direction of the displacement in physics.

• A positive dot product indicates that the vectors are oriented similarly.
• A zero dot product shows they are perpendicular.
• A negative dot product suggests opposite directions.

For line integrals, it helps determine the contribution to work or energy along a path.

Derivative

In calculus, the derivative measures how a function changes as its input changes. It's akin to finding the slope of a curve at any point.
Applying this to vector calculus, if you have a vector-valued function like \( \mathbf{r}(t) \), the derivative \( \mathbf{r}'(t) \) gives a vector tangent to the curve defined by the function at any point \(t\).
For our parametric curve \( \mathbf{r}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} + t \mathbf{k} \), the derivative is \( \mathbf{r}'(t) = \cos t \mathbf{i} - \sin t \mathbf{j} + \mathbf{k} \), providing the rate of change in each component as \(t\) varies.
Understanding derivatives in vector fields is essential for field mappings, motion trajectory analysis, and evaluation of line integrals used in this exercise. Computation of the derivative here essentializes finding tangent vectors which help us understand the path action/state at every infinitesimal segment. Thus, it greatly aids in evaluating the work done by the field by indicating the precise direction along the curve.