Problem 21 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The work done is \(-\pi\).

Step 1: Understand the problem

We need to find the work done by the force \( \mathbf{F} \) along the given curve parameterized by \( \mathbf{r}(t) \). Work done by a vector field along a path is calculated using a line integral.

Step 2: Formula for work

The work done by \( \mathbf{F} \) is given by the line integral \( W = \int_C \mathbf{F} \cdot d\mathbf{r} \), where \( C \) is the curve described by \( \mathbf{r}(t) \).

Step 3: Calculate \( \mathbf{F}(\mathbf{r}(t)) \)

Substitute \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} + t \mathbf{k} \) into \( \mathbf{F}(x, y, z) = z \mathbf{i} + x \mathbf{j} + y \mathbf{k} \). Here, \( x = \sin t, y = \cos t, \) and \( z = t \). So, \( \mathbf{F}(\mathbf{r}(t)) = t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k} \).

Step 4: Find \( d\mathbf{r} \)

Calculate the derivative of \( \mathbf{r}(t) \): \( \mathbf{r}'(t) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} + \mathbf{k} \). Therefore, \( d\mathbf{r} = \mathbf{r}'(t) \, dt = (\cos t \mathbf{i} - \sin t \mathbf{j} + \mathbf{k}) \, dt \).

Step 5: Set up the integral

The integral becomes \(W = \int_0^{2\pi} (t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k}) \cdot (\cos t \mathbf{i} - \sin t \mathbf{j} + \mathbf{k}) \, dt.\)

Step 6: Compute the dot product

First, compute the dot product: \((t \cdot \cos t) + (\sin t \cdot -\sin t) + (\cos t \cdot 1) = t \cos t - \sin^2 t + \cos t.\)

Step 7: Evaluate the integral

Now evaluate \( W = \int_0^{2\pi} (t \cos t - \sin^2 t + \cos t) \, dt.\)Split the integral:- \( \int_0^{2\pi} t \cos t \, dt \) can be done by parts.- \( \int_0^{2\pi} - \sin^2 t \, dt \) can be transformed using \( \sin^2 t = \frac{1-\cos(2t)}{2} \).- \( \int_0^{2\pi} \cos t \, dt \) vanishes because the integral of a periodic function over a full period from 0 to \( 2\pi \) is zero.Let's simplify and solve each.

Step 8: Solve each integral component

1. Using integration by parts for \( \int_0^{2\pi} t \cos t \, dt \):- Let \( u = t \), \( dv = \cos t \, dt \), then \( du = dt \), \( v = \sin t \).- \( \int u \, dv = uv - \int v \, du = t \sin t \Big|_0^{2\pi} - \int_0^{2\pi} \sin t \, dt \).- Compute boundaries to find this part equals zero after evaluating the integral over \( [0, 2\pi] \).2. For \( \int_0^{2\pi} -\sin^2 t \, dt \):- \( \int_0^{2\pi} \frac{-1 + \cos(2t)}{2} \, dt = -\frac{1}{2} \left[ t + \frac{1}{2}\sin(2t) \right]_0^{2\pi} \), which evaluates to \(-\pi.\)Combine these results: the only remaining nonzero integral is \(-\pi.\)

Step 9: Conclude the solution

The work done by the vector field \( \mathbf{F} \) over the curve \( C \) in the direction of increasing \( t \) is \(-\pi\).

Key Concepts

Vector Field

In mathematics and physics, a vector field represents a spatial configuration in which a vector quantity is assigned to every point in space. This can be thought of as a representation of a force, flow or velocity that varies over different points in a particular region. For instance, consider the vector field \( \mathbf{F} = z \mathbf{i} + x \mathbf{j} + y \mathbf{k} \) given in the problem. Here, the vector field defines a force with components dependent on the coordinates \( x, y, \) and \( z. \)

Vector fields have applications in many areas such as fluid dynamics, electromagnetism, and gravitational fields. Each vector in a vector field can be visualized as a small arrow with a given direction and magnitude depending on its location.

• The vector field \( \mathbf{F} = z \mathbf{i} + x \mathbf{j} + y \mathbf{k} \) implies:
• \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors along the x, y, and z axes respectively.
• 'z', 'x', and 'y' represent variable coefficients in a 3D space.
• The field influences the motion or distribution described by these coordinates at each point.

Understanding vector fields is crucial when calculating line integrals to determine work done by a force over a path.

Parametrized Curve

A parametrized curve is a way to represent a curve in the plane or space using a parameter — typically denoted as \( t. \) This parameter traces the curve as it increases through a particular range.

In the exercise, we have the curve \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} + t \mathbf{k}, \quad 0 \leq t \leq 2 \pi \). Here, this curve represents a helix that traces around a circular path while moving upwards as the parameter \( t \) increases.

Through parametrization, complex curves and surfaces can be more easily handled especially when integrated through vector fields.

• This helix starts at \( t = 0 \) and finishes one complete revolution around a circle by the time \( t \) becomes \( 2\pi. \)
• In our vector field calculation, \( x = \sin t \), \( y = \cos t \), and \( z = t \) are derived from the parametrization.
• Parametrized curves are fundamental in calculus for transforming difficult problems into simpler one-variable problems.

These curves allow us to track the pod's motion within the vector field uniformly over time, ensuring accurate integration along the path.

Work Done

In the context of physics and engineering, the work done by a force along a path is a key quantity. It represents the energy transferred by the force to move an object along a certain trajectory.

To calculate the work done by a force field, a line integral is used. The formula \( W = \int_C \mathbf{F} \cdot d\mathbf{r} \) determines the work done as the dot product of the vector field \( \mathbf{F} \) and the differential path element \( d\mathbf{r} \), integrated over the curve \( C. \)

This particular problem evaluates the energy required to move through a field governed by \( \mathbf{F} = z \mathbf{i} + x \mathbf{j} + y \mathbf{k}.\)

• The differential of the parametrized curve \( \mathbf{r}(t) \), \( d\mathbf{r} = (\cos t \mathbf{i} - \sin t \mathbf{j} + \mathbf{k}) \, dt \), represents a tiny segment along the curve.
• The work integral \( W = \int_0^{2\pi} (t \cos t - \sin^2 t + \cos t) \, dt \) splits into manageable expressions for calculation.
• Key operations include evaluating \( \int_0^{2\pi} t \cos t \, dt \) using integration by parts and recognizing trigonometric identities for simplification.

The resulting work, computed as \(-\pi, \) signifies the cumulative effect of all forces field interactions along the curved path.