Problem 63 - Thomas Calculus

Question

In Exercises \(63-68,\) use a CAS to perform the following steps for finding the work done by force F over the given path: $$\begin{array}{l}{\mathbf{F}=x y^{6} \mathbf{i}+3 x\left(x y^{5}+2\right) \mathbf{j} ; \quad \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(\sin t) \mathbf{j}} \\\ {0 \leq t \leq 2 \pi}\end{array}$$

Step-by-Step Solution

Answer

The work done is the evaluated integral of \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \) over \([0, 2\pi]\).

Step 1: Parameterize the Path

Begin by identifying that the path provided is already parameterized as \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (\sin t) \mathbf{j} \). This parameterization will be used to substitute into \( \mathbf{F} \) to calculate the work done along this path.

Step 2: Express Force in Terms of Parameter

Express \( \mathbf{F} \) in terms of parameter \( t \). The parameterized path gives \( x = 2 \cos t \) and \( y = \sin t \). Substitute these into \( \mathbf{F} = xy^6 \mathbf{i} + 3x(xy^5 + 2) \mathbf{j} \), leading to: \[ \mathbf{F}(t) = (2 \cos t)(\sin t)^6 \mathbf{i} + 3(2 \cos t)((2 \cos t)(\sin t)^5 + 2) \mathbf{j}. \]

Step 3: Find Derivative of \\( \mathbf{r}(t) \\\)

Compute the derivative of the path \( \mathbf{r}(t) \) with respect to \( t \). This gives: \[ \frac{d}{dt} \mathbf{r}(t) = \mathbf{r}'(t) = (-2 \sin t) \mathbf{i} + (\cos t) \mathbf{j}. \] This derivative represents the velocity vector along the path.

Step 4: Calculate Work Integral

The work done by the force along a path is given by \( \int_{a}^{b} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt \). Substitute \( \mathbf{F}(t) \) and \( \mathbf{r}'(t) \) into the integrand: \[ \int_{0}^{2\pi} \left[(2 \cos t)(\sin t)^6 (-2 \sin t) + 3(2 \cos t)((2 \cos t)(\sin t)^5 + 2)(\cos t)\right] \, dt. \]

Step 5: Simplify and Integrate

Simplify the integrand and calculate the integral over the interval \( [0, 2\pi] \). Perform the integration either by hand, if simple, or utilize a Computer Algebra System (CAS) to find the exact value. This will yield the total work done by \( \mathbf{F} \) over the path.

Key Concepts

Parameterized Path

In the world of vector calculus, paths can often be intimidating.A parameterized path is a way of representing a path using a parameter.In many situations, this parameter is the variable \( t \), often representing time.
The path is expressed as a vector function \( \mathbf{r}(t) \), which describes the position along the path as \( t \) changes.Instead of dealing with \( x \) and \( y \) values directly, you use trigonometric functions to parameterize them.
For example, the path \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (\sin t) \mathbf{j} \) helps simplify calculations.Here, \( x = 2 \cos t \) and \( y = \sin t \).

• This technique streamlines the process of dealing with curves.
• It allows you to represent complex paths with simple equations.
• Understanding parameterization is crucial for solving integrals related to paths.

The interval \( 0 \leq t \leq 2\pi \) serves as the bounds for \( t \), covering the entire path.

Force Vector Field

A force vector field assigns a vector to each point in the plane.This vector tells you the direction and strength of the force at any location.When working with these fields, a common goal is to determine the work done by the force along a path.
The force vector field in this example is \( \mathbf{F} = xy^6 \mathbf{i} + 3x(xy^5 + 2) \mathbf{j} \).Understanding this field requires substituting the parameter values \( x = 2 \cos t \) and \( y = \sin t \) into the field.

• The vector field equation helps determine how the force varies.
• It is vital to express \( \mathbf{F} \) in terms of the path's parameter.
• Accurate substitution makes it easier to evaluate the force at each point along the path.

This preparation is crucial for calculating the work done over the given path.

Path Integral

A path integral computes the cumulative effect of a vector field along a parameterized path.In simpler terms, it is the way to calculate the total work done when moving along the path under the influence of a force vector field.
The work along the path is expressed as \( \int_{0}^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt \).Here, \( \mathbf{F}(t) \) is the force expressed in terms of \( t \), and \( \mathbf{r}'(t) \) is the derivative of the path, representing velocity.

• The dot product captures how much work the force does in the direction of the path.
• This integral encompasses the entire force interaction along the path.
• Simplifying the integrand is a key step before performing the integral.

Solving this integral directly gives the total work done by the force field over the path.

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and their derivatives and integrals.It is fundamental for understanding how forces work in relation to paths and areas.
The exercise combines various elements of vector calculus:

• Parameterized paths using \( \mathbf{r}(t) \).
• The force vector field \( \mathbf{F} \).
• Path integrals to calculate work.
• Derivative of paths, \( \mathbf{r}'(t) \).

These tools allow for a deeper understanding of physical phenomena, such as calculating the work done by a force across a specific path.Understanding how to manipulate and integrate vectors is crucial for physics and engineering fields.In this context, vector calculus helps efficiently solve problems involving forces and motions.