Problem 20 - 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}=2 \mathrm{yi}+3 x \mathbf{j}+(x+y) \mathbf{k}} \\\ {\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin 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 is \(-2\pi\).

Step 1: Understand the Problem

We need to find the work done by the vector field \( \mathbf{F} = 2y\mathbf{i} + 3x\mathbf{j} + (x+y)\mathbf{k} \) over the curve described by \( \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + (t/6) \mathbf{k} \), for \( 0 \leq t \leq 2\pi \). Work is calculated as the line integral of \( \mathbf{F} \) along the curve \( \mathbf{r}(t) \).

Step 2: Set Up the Line Integral

The work done by \( \mathbf{F} \) over the path \( \mathbf{r}(t) \) is given by the line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \), where \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt \).

Step 3: Compute \( d\mathbf{r} \)

Take the derivative of \( \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + (t/6) \mathbf{k} \) with respect to \( t \). This yields \( \mathbf{r}'(t) = -\sin(t) \mathbf{i} + \cos(t) \mathbf{j} + 1/6 \mathbf{k} \).

Step 4: Substitute into the Line Integral

Substitute \( \mathbf{F} \) and \( \mathbf{r}(t) \) into the line integral. It becomes \( \int_{0}^{2\pi} (2\sin(t) \mathbf{i} + 3\cos(t) \mathbf{j} + (\cos(t) + \sin(t)) \mathbf{k}) \cdot (-\sin(t) \mathbf{i} + \cos(t) \mathbf{j} + 1/6 \mathbf{k}) dt \).

Step 5: Evaluate the Dot Product

Evaluate the dot product: \((2\sin(t) \cdot -\sin(t)) + (3\cos(t) \cdot \cos(t)) + ((\cos(t) + \sin(t)) \cdot 1/6) \). Simplify to get \(-2\sin^2(t) + 3\cos^2(t) + 1/6(\cos(t) + \sin(t))\).

Step 6: Integrate Over the Interval

Integrate the expression from Step 5 with respect to \( t \) from \( 0 \) to \( 2\pi \): \( \int_{0}^{2\pi} (-2\sin^2(t) + 3\cos^2(t) + 1/6(\cos(t) + \sin(t))) dt \).

Step 7: Simplify the Integral

Use the identity \( \sin^2(t) = (1 - \cos(2t))/2 \) and \( \cos^2(t) = (1 + \cos(2t))/2 \) to simplify: \( \int_{0}^{2\pi} (-1 + \cos(2t) + 1/6 \cos(t) + 1/6 \sin(t)) dt \).

Step 8: Calculate the Integral

Integrate each term separately. The integrals of \( \cos(2t) \), \( \cos(t) \), and \( \sin(t) \) over \( \left[0, 2\pi\right] \) yield zero. Only the integral of \(-1\) contributes: \(-1 \times (2\pi) = -2\pi\).

Step 9: Conclude with Result

Add up the contributions from each part of the integrand to find the total work done. The total work done by the force field \( \mathbf{F} \) along the path given by \( \mathbf{r}(t) \) is \( -2\pi \).

Key Concepts

Vector Field

A vector field is a mathematical construct in which each point on a certain space is associated with a vector. When dealing with problems in physics or engineering, these fields often represent phenomena like velocity of flux across a surface, magnetic fields, or in our particular case, a force field. The vector field \( \mathbf{F} = 2y\mathbf{i} + 3x\mathbf{j} + (x+y)\mathbf{k} \) describes such a scenario. Here, the vectors vary with the spatial position \( (x, y, z) \), specifying the magnitude and direction of a force at each point. Understanding how to interpret these vectors can aid in visualizing the field as a whole and determining subsequent work done or potential energy changes across the field.

Work Done

In physics, the work done by a force on an object along a path is the energy transferred by that force to move the object. Mathematically, this is the line integral of the force along that path. In our problem, we want to find the work done by the vector field \( \mathbf{F} \) along a curve described by \( \mathbf{r}(t) \). The integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) is used to calculate this work. It encapsulates how the vector field "pushes" along the path. When you evaluate the line integral, you're essentially adding up

• All the tiny efforts exerted by the force.
• As the parameter \( t \) varies along the defined curve, it accumulates these contributions.

Thus, the result measures how effectively the force field moves something along the directed path in space.

Curve Parameterization

The path taken in a vector field can be tricky, so we use parameterization to make things easier. Curve parameterization involves expressing the path of a curve as functions of a single variable, usually \( t \). For instance, in our scenario, the curve is expressed by \( \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + (t/6) \mathbf{k} \). Here, as \( t \) goes from \( 0 \) to \( 2\pi \), it traces out a helical path. The parameter \( t \) serves as a "time-like" variable that tells us where on the curve we are. To evaluate the line integral, we first need to find the derivative of this parametrized curve, \( \mathbf{r}'(t) \), which guides us to understand how the curve changes direction and length with respect to \( t \). This technique translates complex spatial paths into simpler, more manageable expressions.

Dot Product

The dot product is a key concept when computing work done by a vector field because it essentially measures how much one vector projects along another. This projection is important because only the component of force acting along the path (i.e., the component in the direction of the displacement) contributes to the work done. In our exercise:

• Two vectors are involved: \( \mathbf{F} \) (the force vector) and \( d\mathbf{r} \) (the infinitesimal path vector).
• The dot product \( \mathbf{F} \cdot \mathbf{r}'(t) \) transforms the vector situation into a scalar one.

This transformation allows us to form an integral over a scalar field, simplifying calculation. The resultant integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) is the accumulation of tiny amounts of work along the entire path, each derived from the locally relevant projection of force onto direction.