Problem 27 - Thomas Calculus

Question

Work Find the work done by the force \(\mathbf{F}=x y \mathbf{i}+(y-x) \mathbf{j}\) over the straight line from \((1,1)\) to \((2,3) .\)

Step-by-Step Solution

Answer

The work done by the force is \(\frac{25}{6}\).

Step 1: Determine the Parametric Form of the Line

The line segment from \((1,1)\) to \((2,3)\) can be represented parametrically as \(\mathbf{r}(t) = (1 + t)\mathbf{i} + (1 + 2t)\mathbf{j}\), where \(0 \leq t \leq 1\).

Step 2: Find the Derivative of the Parametric Equation

The derivative of \(\mathbf{r}(t)\) is \(\mathbf{r}'(t) = \frac{d}{dt} [(1 + t)\mathbf{i} + (1 + 2t)\mathbf{j}] = \mathbf{i} + 2\mathbf{j}\). This represents the tangent vector along the path.

Step 3: Substitute Parametric Form into Force

Substitute \(x = 1 + t\) and \(y = 1 + 2t\) into the force \(\mathbf{F}(x,y) = x y \mathbf{i} + (y-x) \mathbf{j}\). We get: \(\mathbf{F}(t) = (1+t)(1+2t)\mathbf{i} + ((1+2t)-(1+t))\mathbf{j}\).

Step 4: Simplify the Force Expression

Simplifying, \((1+t)(1+2t) = 1 + 3t + 2t^2\) for the \(\mathbf{i}\)-component and \((1+2t)-(1+t) = t\) for the \(\mathbf{j}\)-component. Thus, \(\mathbf{F}(t) = (1 + 3t + 2t^2)\mathbf{i} + t\mathbf{j}\).

Step 5: Compute the Work Integral

The work done by the force is the line integral over the path, calculated as:\[W = \int_{0}^{1} \mathbf{F}(t) \cdot \mathbf{r}'(t) \, dt.\]Substituting the expressions, we have:\[W = \int_{0}^{1} \left[(1 + 3t + 2t^2)\mathbf{i} + t\mathbf{j}\right] \cdot (\mathbf{i} + 2\mathbf{j}) \, dt.\]

Step 6: Perform Dot Product and Evaluate the Integral

Calculate the dot product:\[(1 + 3t + 2t^2) \cdot 1 + t \cdot 2 = 1 + 3t + 2t^2 + 2t = 1 + 5t + 2t^2.\]Now integrate:\[W = \int_{0}^{1} (1 + 5t + 2t^2) \, dt = \left[t + \frac{5}{2}t^2 + \frac{2}{3}t^3\right]_{0}^{1} = 1 + \frac{5}{2} + \frac{2}{3}.\]

Step 7: Calculate the Result

Add the constants:\[1 + \frac{5}{2} + \frac{2}{3} = \frac{6}{6} + \frac{15}{6} + \frac{4}{6} = \frac{25}{6}.\]Thus, the work done is \(\frac{25}{6}\).

Key Concepts

Line Integral

A line integral is a type of integral that calculates the cumulative effect of a field along a given path. Imagine walking along a trail while a gentle wind pushes against you. The line integral helps us figure out the total work done by this force as you travel from point A to point B. In our example, the line integral determines the work done by the force \(\mathbf{F} = x y \mathbf{i} + (y-x) \mathbf{j}\) along the straight line path from \((1,1)\) to \((2,3)\). To compute a line integral involving a vector field, we need to

• Define a path, usually through a parametric equation.
• Calculate the dot product of the vector field and the path's tangent vector.
• Integrate the result over the path's range.

These steps ensure that we are summing up the tiny bits of work the force performs along the defined path.

Parametric Equation

A parametric equation allows us to represent a curve or line in terms of a parameter, usually \(t\). Instead of expressing \(y\) as a function of \(x\), we write both \(x\) and \(y\) as functions of \(t\). This is helpful in many applications, including the computation of line integrals. In our exercise, the path from \((1,1)\) to \((2,3)\) is described by the parametric equations \(x = 1 + t\) and \(y = 1 + 2t\), with \(t\) ranging from 0 to 1. Benefits of using parametric equations include:

• Handling curves and line segments uniformly.
• Encapsulating the motion along the path in one parameter.
• Facilitating the computation of derivatives and integrals.

The parametric form simplifies the task of describing complex paths and transforms integration over those paths into more manageable tasks.

Dot Product

The dot product is a mathematical operation that takes two vectors and returns a single number. In the context of physics, it measures how much one vector projects onto another. This makes the dot product a crucial part of calculating work done by a force, where it helps quantify the component of the force in the direction of the motion. In our exercise, we compute the dot product of the force \(\mathbf{F}(t) = (1 + 3t + 2t^2)\mathbf{i} + t\mathbf{j}\) and the tangent vector \(\mathbf{r}'(t) = \mathbf{i} + 2\mathbf{j}\). By performing this operation:

• Multiply the components of each vector.
• Add the results to get a single scalar value.

Here, the dot product yields \((1 + 3t + 2t^2) \cdot 1 + t \cdot 2 = 1 + 5t + 2t^2\), which simplifies the integration process that follows.

Integration

Integration is the process of finding the whole from its parts, often visualized as calculating the area under a curve. In our case, it's used to add up the infinitesimal work done by the force along each tiny segment of the path. After obtaining the expression from the dot product, \(1 + 5t + 2t^2\), integration over the interval from \(t = 0\) to \(t = 1\) tells us the total work done. The integration in this case involves:

• Finding the antiderivative of the expression \(1 + 5t + 2t^2\).
• Evaluating it at the bounds \(t = 0\) and \(t = 1\).

This leads us to the calculation \[ \left[t + \frac{5}{2}t^2 + \frac{2}{3}t^3\right]_{0}^{1} = 1 + \frac{5}{2} + \frac{2}{3} \]. Simplifying this gives the final result: \(\frac{25}{6}\), representing the total work done by the force over the specified path.