Problem 23 - The Calculus with Analytic Geometry

Question

In Exercises 21 through 34 , find the total work done in moving an object along the given arc \(C\) if the motion is caused by the given force field. Assume the arc is measured in inches and the force is measured in pounds. \(F(x, y)=(y-x) i+x^{2} y j ; C:\) the line segment from the point \((1,1)\) to \((2,4)\)

Step-by-Step Solution

Answer

The work done is \(\frac{83}{4}\).

Step 1: Parameterize the Line Segment

From \((1, 1)\) to \((2, 4)\):
\(x = 1 + t\), \(y = 1 + 3t\), \(0 \leq t \leq 1\).
\(dx = dt\), \(dy = 3\,dt\).

Step 2: Substitute into the Integral

\(\int_C (y - x)\,dx + x^2 y\,dy\)
\(= \int_0^1 [(1+3t) - (1+t)]\,dt + (1+t)^2(1+3t) \cdot 3\,dt\)
\(= \int_0^1 [2t + 3(1+t)^2(1+3t)]\,dt\)

Step 3: Expand and Integrate

\(3(1+t)^2(1+3t) = 3(1 + 2t + t^2)(1 + 3t)\)
\(= 3(1 + 3t + 2t + 6t^2 + t^2 + 3t^3) = 3(1 + 5t + 7t^2 + 3t^3)\)
\(= 3 + 15t + 21t^2 + 9t^3\)
So the integrand is \(2t + 3 + 15t + 21t^2 + 9t^3 = 3 + 17t + 21t^2 + 9t^3\).
\(\int_0^1 (3 + 17t + 21t^2 + 9t^3)\,dt = [3t + \frac{17t^2}{2} + 7t^3 + \frac{9t^4}{4}]_0^1\)
\(= 3 + 8.5 + 7 + 2.25 = 20.75 = \frac{83}{4}\)

Key Concepts

Parametrization

The first step to solving the problem is to parametrize the path along which the work is done. Parametrization converts the geometric path into a mathematical form that we can work with. For the line segment from \((1,1)\) to \((2,4)\), we apply the linear interpolation formula:

\[ \textbf{r}(t) = (1-t)\textbf{A} + t\textbf{B}, \]

where \( \textbf{A} \) and \( \textbf{B} \) are the endpoints, \((1,1)\) and \((2,4)\) respectively. By performing the interpolation, we get:

\[ \textbf{r}(t) = (1-t)(1,1) + t(2,4) = (1+t, 1+3t). \]

This parametric representation helps us define the path clearly in terms of a parameter \( t \) which ranges from 0 to 1.

Force Field

A force field describes the influence that a vector field exerts on objects within it. In this exercise, the given force field is: \( F(x, y) = (y - x)\textbf{i} + x^2 y\textbf{j} \).

Here's a quick breakdown:

• \( (y - x)\textbf{i} \) represents the component of the force in the x-direction.
• \( x^2 y\textbf{j} \) represents the component of the force in the y-direction.

For each point along the path, the force field specifies the force experienced by the object. Understanding this force field is crucial for calculating the work done. The next step involves evaluating the force along the parametrized path.

Line Integrals

To calculate the work done by the force field along the path, we use a line integral. Work done is the integral of the force along the path:

\[ W = \int_C \textbf{F} \, \cdot \textbf{dr} \]

In parametric form, \( \textbf{r}(t) = (1+t, 1+3t) \), and the bounds for \( t \) are from 0 to 1. We first find the derivative of \( \textbf{r}(t) \):

\[ \textbf{r}'(t) = (1, 3). \]

Then, the differential vector \( \textbf{dr} = \textbf{r}'(t) dt \). Substituting these back into the integral, the line integral becomes an integral over the parameter, making it computationally feasible.

Vector Calculus

Vector calculus provides the tools we need to deal with vector fields and line integrals. Concepts like gradient, divergence, and curl in vector calculus explain how fields behave and interact with paths and surfaces. For instance,

• The dot product in the line integral helps quantify the work done by the force field.
• Integration converts local information over an interval into a total measure, which gives the total work done along the path.

These concepts transform abstract mathematical definitions into practical computations, vital for solving physics and engineering problems. By understanding parametrization, force fields, and line integrals, we use vector calculus to arrive at the solution.