Problem 21 - 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. \(\mathbf{F}(x, y)=2 x y \mathbf{i}+\left(x^{2}+y^{2}\right) \mathfrak{j} ; C\) : the line segment from the origin to the point \((1,1)\).

Step-by-Step Solution

Answer

The total work done is \ \frac{4}{3} \ pounds-inches.

Step 1 - Parameterize the Line Segment

First, parameterize the line segment from the origin \( (0, 0) \) to the point \( (1, 1) \). Let \( \textbf{r}(t) \) be the position vector for this segment. A common parameterization is \(\textbf{r}(t) = (t, t) \) where \( t \) ranges from 0 to 1.

Step 2 - Compute the Derivative of \(\textbf{r}(t)\)

Find the derivative of the parameterization with respect to \( t \). Thus, \(\textbf{r}'(t) = (1, 1) \).

Step 3 - Substitute Parameterization into Force Field

Substitute \( x = t \) and \( y = t \) into the force field \(\textbf{F}(x, y) = 2xy \textbf{i} + (x^2 + y^2) \textbf{j} \). This gives us: \(\textbf{F}(t, t) = 2tt \textbf{i} + (t^2 + t^2) \textbf{j} = 2t^2 \textbf{i} + 2t^2 \textbf{j} \).

Step 4 - Calculate Dot Product

Compute the dot product \( \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t) \). Substituting from previous steps: \( (2t^2 \textbf{i} + 2t^2 \textbf{j}) \cdot (1, 1) = 2t^2 \cdot 1 + 2t^2 \cdot 1 = 4t^2 \).

Step 5 - Integrate the Dot Product

Integrate the dot product over the interval from \( t = 0 \) to \( t = 1 \): \[ \int_{0}^{1} 4t^2 \, dt \].

Step 6 - Perform the Integration

Perform the integration: \[ \int_{0}^{1} 4t^2 \, dt = 4 \int_{0}^{1} t^2 \, dt = 4 \left[ \frac{t^3}{3} \right]_{0}^{1} = 4 \times \frac{1}{3} = \frac{4}{3} \].

Key Concepts

parametrization of a curve

When we talk about the parametrization of a curve, we are essentially describing the path a point takes along that curve using a parameter, usually denoted as \( t \). For instance, in the given problem, the curve (or arc) is a straight line from the origin \( (0,0) \) to point \( (1,1) \). To parameterize this line, we use the vector notation \( \textbf{r}(t) = (t, t) \) where \( t \) varies from 0 to 1.

This means that as \( t \) changes from 0 to 1, \( r(t) \) starts at the origin and moves to \( (1,1) \). This is a simple form of parameterization since the motion is linear and both x and y coordinates change at the same rate, represented by \( t \).
Parameterization makes it easier to perform calculations since it consolidates the curve's details into a single variable.

dot product

The dot product, often written as \( \textbf{A} \cdot \textbf{B} \), is a way of multiplying two vectors to get a scalar (a single number). For vectors \( \textbf{A} = (a_1, a_2) \) and \( \textbf{B} = (b_1, b_2) \), the dot product is calculated as:
\( a_1b_1 + a_2b_2 \).

In our problem, we needed to compute the dot product of the force field vector \( \textbf{F}(\textbf{r}(t)) \) and the derivative of the parameterization \( \textbf{r}'(t) \). This step is essential because the work done by a force field through a path is calculated using this dot product.

For example, we have:
\( \textbf{F}(t, t) = 2t^2 \textbf{i} + 2t^2 \textbf{j} \) and \( \textbf{r}'(t) = (1, 1) \).
The dot product would be:
\((2t^2 \textbf{i} + 2t^2 \textbf{j}) \cdot (1, 1) = 2t^2 \cdot 1 + 2t^2 \cdot 1 = 4t^2 \).
This result is then used in the next steps to calculate work by integrating over the given interval of \( t \).

integration

Integration is a fundamental concept in calculus used to compute areas under curves, among other things. In the context of work done by a force field along a curve, integration helps to sum up the tiny amounts of work done along small segments of the path.

In our example, we have the dot product \( 4t^2 \) and need to integrate it over the interval from \( t = 0 \) to \( t = 1 \):
\[ \int_{0}^{1} 4t^2 \, dt \].
This integral sums the contributions to work from each infinitesimal part of the path.

Performing the integration involves finding the antiderivative of the function 4t^2:
\( 4 \int t^2 \, dt \).
The antiderivative of \( t^2 \) is \( \frac{t^3}{3} \), so we obtain:
\( 4 \left[ \frac{t^3}{3} \right]_{0}^{1} \).
Evaluating this from 0 to 1 gives: \( 4 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = 4 \times \frac{1}{3} = \frac{4}{3} \).
So, the total work done in moving the object along the specified path is \( \frac{4}{3} \) inch-pounds. Integration allows us to combine all the little pieces of work done along the path into a single total value.