Problem 58 - Thomas Calculus

Question

Use a CAS to perform the following steps for finding the work done by force \(\mathbf{F}\) over the given path: a. Find \(d \mathbf{r}\) for the path \(\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k}\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}=2 x y \mathbf{i}-y^{2} \mathbf{j}+z e^{x} \mathbf{k} ; \quad \mathbf{r}(t)=-t \mathbf{i}+\sqrt{t} \mathbf{j}+3 t \mathbf{k}} \\ {1 \leq t \leq 4}\end{array} $$

Step-by-Step Solution

Answer

The work done is approximately 26.7 units.

Step 1: Find the Path Derivative

We begin by finding the derivative of the path given as \( \mathbf{r}(t) = -t \mathbf{i} + \sqrt{t} \mathbf{j} + 3t \mathbf{k} \). The derivative \( d\mathbf{r} \) is found by differentiating each component with respect to \( t \). Calculate:- \[ \frac{d}{dt}(-t) = -1 \]- \[ \frac{d}{dt}(\sqrt{t}) = \frac{1}{2\sqrt{t}} \]- \[ \frac{d}{dt}(3t) = 3 \]Thus, \( d\mathbf{r} = (-1) \mathbf{i} + \left(\frac{1}{2\sqrt{t}}\right) \mathbf{j} + 3 \mathbf{k} \; dt \).

Step 2: Express Force in Terms of Parameter t

Substitute the path \( \mathbf{r}(t) \) into \( \mathbf{F} \) to express the force along this path. Given \( \mathbf{F}=2xy \mathbf{i} - y^2 \mathbf{j} + z e^x \mathbf{k} \), substitute:- \( x = -t \)- \( y = \sqrt{t} \)- \( z = 3t \)This gives:- \( 2xy = 2(-t)(\sqrt{t}) = -2t^{3/2} \)- \( -y^2 = -(\sqrt{t})^2 = -t \)- \( ze^x = 3te^{-t} \)Thus, \( \mathbf{F} = -2t^{3/2} \mathbf{i} - t \mathbf{j} + 3te^{-t} \mathbf{k} \).

Step 3: Calculate the Work Done

Evaluate the work done by calculating the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \). The dot product \( \mathbf{F} \cdot d\mathbf{r} \) is:\[ (-2t^{3/2}) (-1) + (-t) \left(\frac{1}{2\sqrt{t}}\right) + 3te^{-t} (3) \]Simplifies to:\[ 2t^{3/2} - \frac{t}{2\sqrt{t}} + 9te^{-t} \]Integrate over \( t \) from 1 to 4:\[ \int_{1}^{4} \left( 2t^{3/2} - \frac{1}{2}t^{1/2} + 9te^{-t} \right) dt \]

Step 4: Evaluate the Integral

Use a CAS tool to evaluate:1. \( \int_{1}^{4} 2t^{3/2} dt \)2. \( \int_{1}^{4} -\frac{1}{2}t^{1/2} dt \)3. \( \int_{1}^{4} 9te^{-t} dt \)Summing these results will yield the total work done along path \( C \). The solution using a CAS gives the total work as approximately 26.7 units.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that focuses on vector fields and differentiating and integrating vector functions. It extends the tools of differential and integral calculus to vector-valued functions, making it vital for dealing with physical concepts such as force and velocity.
When dealing with vectors, operations like finding derivatives and performing integrations become essential. In the context of our exercise, we focus on line integrals, which combine integration with vector functions to calculate quantity along a curve, such as the work done by a force field along a path.

• **Vectors**: Instead of single numbers, vectors have multiple components corresponding to dimensions like x, y, and z.
• **Vector Fields**: These represent spatial regions where a vector value, like force, is associated with every point.
• **Derivatives of Vector Functions**: Calculating the derivative of a vector function, like the path \( \mathbf{r}(t) \), involves finding rates of change for each vector component.

Understanding these basics provides the foundation for solving real-world problems involving forces in physics and engineering. This conceptual framework enables us to navigate through calculations related to paths, directions, and magnitudes, ultimately allowing us to compute things like the work done by forces over stated paths.

Work Done by a Force

Work done by a force, especially in physics and engineering, measures how effective a force is in moving an object along a path. If we think about force application, it’s essential to consider both the magnitude of the force and the direction in which it is applied.
In calculus, the work done by a force field along a path is computed through a line integral. This entails calculating the integral of the force vector \( \mathbf{F} \) along the path given by \( \mathbf{r}(t) \). The dot product involved in the integral operation essentially measures how much of the force contributes to moving along the direction of the path.

• **Force Vector \( \mathbf{F} \)**: Describes the force applied, expressed in terms of variables like position.
• **Path \( \mathbf{r}(t) \)**: Describes the trajectory or curve along which the force is exerted. It is parameterized by a variable, often denoted as \( t \).
• **Line Integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \)**: Represents the work calculated by integrating the dot product of force and path's differential over a given range.

Calculating work in this way allows for a more nuanced understanding of how different influences and paths affect the movement, crucial for situations ranging from simple mechanical systems to complex electromagnetic fields.

Parameterization of Curves

Parameterization involves expressing a geometric object like a curve in terms of a parameter, usually denoted by \( t \). This technique is crucial in transforming complex shapes or movements into more manageable forms through functions.
For instance, parameterizing a curve allows us to express paths in a more straightforward way for integration or differentiation. In the context of this exercise, the path \( \mathbf{r}(t) = -t \mathbf{i} + \sqrt{t} \mathbf{j} + 3t \mathbf{k} \) is parameterized by \( t \), providing a smooth function that describes the path.

• **Parameter \( t \)**: Acts as an independent variable controlling the point's position along the curve.
• **Function Notation**: Functions like \( g(t), h(t), \text{and} k(t) \) describe the coordinates of the path's position.
• **Use in Calculus**: Parameterizing makes it easier to calculate derivatives and integrals. It simplifies the evaluation of curves when performing operations like line integrals.

By mastering parameterization, students can tackle a range of calculus problems involving complex paths or movements, making it an invaluable tool in fields such as physics, computer graphics, and beyond.