Problem 20 - Thomas Calculus

Question

Find the work done by \(\mathbf{F}\) over the curve in the direction of increasing \(t .\) $$ \begin{array}{l}{\mathbf{F}=2 y \mathbf{i}+3 x \mathbf{j}+(x+y) \mathbf{k}} \\\ {\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+(t / 6) \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Work Done

To find the work done by the force \( \mathbf{F} \) when moving along a curve given by \( \mathbf{r}(t) \) from \( t = 0 \) to \( t = 2\pi \), we need to calculate the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \). Here, \( d\mathbf{r} \) is the differential element of the path, and \( \mathbf{F} \cdot d\mathbf{r} \) denotes the dot product between the force and the differential displacement along the path.

Step 2: Parameters and Components

The force \( \mathbf{F} = 2y \mathbf{i} + 3x \mathbf{j} + (x+y) \mathbf{k} \). The path is parameterized by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + \left( \frac{t}{6} \right) \mathbf{k} \). From \( \mathbf{r}(t) \), we have \( x = \cos t, \ y = \sin t, \ z = \frac{t}{6} \).

Step 3: Calculate Differential Displacement

Find the differential \( d\mathbf{r} \) by differentiating \( \mathbf{r}(t) \) with respect to \( t \): \( d\mathbf{r} = (\frac{d}{dt}\cos t) \mathbf{i} + (\frac{d}{dt}\sin t) \mathbf{j} + \left(\frac{d}{dt}\frac{t}{6}\right) \mathbf{k} \, dt = -\sin t \, \mathbf{i} + \cos t \, \mathbf{j} + \frac{1}{6} \, \mathbf{k} \, dt \).

Step 4: Substitute and Simplify Force

Substitute \( x = \cos t \) and \( y = \sin t \) into \( \mathbf{F} \): \( \mathbf{F} = 2\sin t \, \mathbf{i} + 3\cos t \, \mathbf{j} + (\cos t + \sin t) \, \mathbf{k} \).

Step 5: Compute the Dot Product

Now compute the dot product \( \mathbf{F} \cdot d\mathbf{r} = (2\sin t \mathbf{i} + 3\cos t \mathbf{j} + (\cos t + \sin t) \mathbf{k}) \cdot (-\sin t \, \mathbf{i} + \cos t \, \mathbf{j} + \frac{1}{6} \, \mathbf{k}) \, dt \). This results in \( -2\sin^2 t + 3\cos^2 t + \frac{1}{6}(\sin t + \cos t) \, dt \).

Step 6: Integrate Over the Interval

Integrate the obtained expression with respect to \( t \) from \( 0 \) to \( 2\pi \): \[ \int_0^{2\pi} \left(-2\sin^2 t + 3\cos^2 t + \frac{1}{6}(\sin t + \cos t)\right) \, dt \].

Step 7: Solve the Integral

The integral \( \int_0^{2\pi} (-2\sin^2 t) \, dt + \int_0^{2\pi} (3\cos^2 t) \, dt + \int_0^{2\pi} \frac{1}{6}(\sin t + \cos t) \, dt \) is simplified using trigonometric identities and properties: \( \int_0^{2\pi} \sin t \, dt = 0 \) and \( \int_0^{2\pi} \cos t \, dt = 0 \). Additionally, \( \int_0^{2\pi} \sin^2 t \, dt = \int_0^{2\pi} \cos^2 t \, dt = \pi \). Compute these integrals to obtain the work done as \( -2(\pi) + 3(\pi) = \pi \).

Key Concepts

Vector Calculus

Vector Calculus is essentially the field of mathematics dealing with vectors and how to analyze them within the framework of calculus. It provides the tools necessary to understand varying quantities in multivariable functions. In our problem, we look into the concept of work done by a force field \( \mathbf{F} \) along a path parameterized by a vector function \( \mathbf{r}(t) \).
The concept of a line integral comes into play here, which allows us to calculate things like work done by a force field along a path. This involves integrating the dot product of the force vector \( \mathbf{F} \) and the differential path vector \( d\mathbf{r} \).
Key components of vector calculus used in the exercise include:

• Vector Fields: In our problem, this is represented by \( \mathbf{F} = 2y \mathbf{i} + 3x \mathbf{j} + (x+y) \mathbf{k} \).
• Parameterization: Refers to expressing the curve as a vector function of one variable \( t \).
• Line Integrals: These help specify the cumulative effect of a vector field along a curve; in this case, calculating work.

Parameterization

Parameterization is a technique that involves representing a curve by a set of equations dependent on a parameter, often time \( t \). This makes it easier to perform calculus operations on complex curves. For our exercise, the curve is parameterized by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + \left( \frac{t}{6} \right) \mathbf{k} \).
This effectively transforms the problem of analyzing the curve into analyzing simple trigonometric and linear functions. Using trigonometric functions \( \cos t \) and \( \sin t \) specifies the curve's shape in the xy-plane, while \( \frac{t}{6} \) represents its component in the z-direction, resulting in a helical path.
Parameterization is particularly useful for:

• Tracing Complex Paths: It allows us to describe paths like circles or helices using periodic functions.
• Simplifying Integration: By converting the original problem into a simpler form using a parameter, we ease the process of integration needed to compute line integrals.

Trigonometric Integration

Trigonometric integration is a specialized method employed to solve integrals involving trigonometric functions, which commonly arise in parameterized vector paths like in our exercise. For example, using \( \int \sin^2 t \, dt \) and \( \int \cos^2 t \, dt \) can simplify computations by applying known identities.
In our exercise, trigonometric identities such as \( \sin^2 t + \cos^2 t = 1 \) help in reducing expressions and simplifying the integration process. This is crucial when breaking down integrals of products of trigonometric functions that appear in the dot product calculations.
Steps involved in trigonometric integration:

• Apply Identities: Simplifies the target functions often by reducing the degree or decomplexifying the expression.
• Substitute Values: Known integral values like \( \int_0^{2\pi} \sin t \, dt = 0 \) save computation time and effort.

Dot Product

The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. It provides valuable insights into vector relationships, such as determining parallelism and orthogonality.
In this exercise, the dot product is used to find the work done by force field \( \mathbf{F} \) along the curve \( \mathbf{r}(t) \). This is calculated as \( \mathbf{F} \cdot d\mathbf{r} \), where \( d\mathbf{r} \) is the differential displacement vector along the path.
The importance of the dot product here includes:

• Projection Along the Curve: The computed value shows how much of the force is effective along the path.
• Scalar Result: Produces a single value that simplifies calculations involving integrations.
• Physical Relevance: Directly correlates with physical concepts such as work done, providing practical application of theoretical concepts.