Question

In Exercises \(13-16,\) find the work performed by the force field \(\vec{F}\) moving a particle along the path \(C\). \(\vec{F}=\left\langle 2 x y, x^{2}, 1\right\rangle\) lbs; \(C\) is the path from (0,0,0) to (2,4,8) via \(\vec{r}(t)=\langle t, 2 t, 4 t\rangle\) on \(0 \leq t \leq 2,\) where distance are measured in feet.

Short answer

The work performed is 24 ft-lbs.

Step-by-step solution

Parameterize the Path

Given the path as \( \vec{r}(t) = \langle t, 2t, 4t \rangle \) for \( 0 \leq t \leq 2 \), parameterize the position of the particle in terms of \( t \). Here, the initial position \( \vec{r}(0) = \langle 0, 0, 0 \rangle \) and final position \( \vec{r}(2) = \langle 2, 4, 8 \rangle \) confirms the path boundaries.

Find the Velocity Vector

Differentiate the position vector \( \vec{r}(t) \) to find the velocity vector: \( \vec{r}'(t) = \left\langle \frac{d}{dt}(t), \frac{d}{dt}(2t), \frac{d}{dt}(4t) \right\rangle = \langle 1, 2, 4 \rangle \).

Evaluate the Force Field Along the Path

Substitute \( x = t, \ y = 2t, \ z = 4t \) into the force field \( \vec{F} = \langle 2xy, x^2, 1 \rangle \). Thus, \( \vec{F}(t) = \langle 2(t)(2t), t^2, 1 \rangle = \langle 4t^2, t^2, 1 \rangle \).

Compute the Dot Product

Calculate the dot product of the force field and the velocity vector: \( \vec{F}(t) \cdot \vec{r}'(t) = \langle 4t^2, t^2, 1 \rangle \cdot \langle 1, 2, 4 \rangle = 4t^2 \cdot 1 + t^2 \cdot 2 + 1 \cdot 4 = 4t^2 + 2t^2 + 4 = 6t^2 + 4 \).

Integrate the Dot Product over the Interval

Integrate the dot product from \( t=0 \) to \( t=2 \) to find the work done: \[W = \int_{0}^{2} (6t^2 + 4) \, dt = \left[ 2t^3 + 4t \right]_0^2 = \left( 2(2)^3 + 4(2) \right) - \left( 2(0)^3 + 4(0) \right) = 16 + 8 = 24 \text{ ft-lbs}.\]

Solution Verification

The integration was performed accurately, confirming that the work done by the force field along the path \( C \) is \( 24 \text{ ft-lbs} \).

Key concepts

Line Integrals

Line integrals are a fundamental concept when dealing with vector fields and paths. Essentially, a line integral helps calculate the total effect of a field, such as a force field, along a path. This can be thought of as adding up the contributions of the field at each point along the path. In the context of work, the line integral of a force field along a path measures how much work is done by the force in moving an object along that path.

To compute a line integral, we follow several steps:

• Parameterize the path, which means representing the path through one or more parameters. This allows us to evaluate the vector field along the path.
• Find the dot product of the force field and the velocity vector along the path.
• Integrate this dot product over the interval of the parameter, which gives us the total work done.

Line integrals are widely used in physics and engineering to solve problems related to force, energy, and flux across different path forms.

Parameterization of Paths

Parameterizing a path is a crucial step in calculating line integrals, as it allows us to express the trajectory of an object as a function of a parameter, typically time. In this exercise, the path is given by

\( extbf{r}(t) = \langle t, 2t, 4t \rangle\)
for \(0 \leq t \leq 2\). This means that the object's position changes in a linear fashion, starting from the origin (0,0,0) and ending at (2,4,8) as time progresses from 0 to 2.

By parameterizing the path, we translate the 3D path into a one-dimensional problem that is easier to work with, especially for integration. This transformation helps us take a vector space and compress it into a simpler form, allowing for straightforward application of calculus tools. Understanding how to parameterize paths effectively is essential for any vector calculus operation involving motion or trajectories.

Dot Product

The dot product, also known as the scalar product, is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors), multiplies them pairwise, and then sums the results. It's an essential tool in vector calculus, often used to determine angles between vectors or project one vector onto another. In terms of work done by a force, the dot product tells you how much of the force actually contributes to moving an object along a path.

In the exercise, we compute the dot product by multiplying the components of the force vector and the velocity vector at each moment:\(\langle 4t^2, t^2, 1 \rangle \cdot \langle 1, 2, 4 \rangle = 4t^2 + 2t^2 + 4\).

This results in a scalar expression \(6t^2 + 4\) that represents the strength of the force acting along the direction of motion. Computing the dot product is crucial for finding the work done, as it isolates the effective part of the force that contributes along the path.

Integration in Calculus

Integration is a fundamental concept in calculus that involves finding the whole from knowing the part, like summing up infinitesimal contributions to get a total value. When involving vector fields, integration allows us to calculate quantities like work by accumulating effects across a path.

In the context of this exercise, after computing the dot product, we need to integrate\(6t^2 + 4\) from \(t=0\) to \(t=2\):

\[W = \int_{0}^{2} (6t^2 + 4) \, dt = \left[ 2t^3 + 4t \right]_0^2 = 24 \text{ ft-lbs}\].

The result of the integration gives us the total work done. Thus, integration translates our function of infinitesimal work into an accumulative sum, showing the power of calculus in computing real-world quantities. Understanding integration's role is vital for handling various problems in mechanics and physics.