Question

For the force field \(F=-y i+x j+z k\), calculate the work done in moving a particle from \((1,0,0)\) to \((-1,0, \pi)\) (a) along the helix \(x=\cos t, y=\sin t, z=t ;\) (b) along the straight line joining the points. Do you expect your answers to be the same? Why or why not?

Short answer

Along the helix: \( \pi + \frac{\pi^2}{2} \). Along the straight line: \( \frac{\pi^2}{2} \). Answers differ due to path-dependent force field.

Step-by-step solution

Understand the Problem

We need to calculate the work done by the force field \( \mathbf{F} = -y \mathbf{i} + x \mathbf{j} + z \mathbf{k} \) in moving a particle along two different paths from point \( (1,0,0) \) to point \( (-1,0, \pi) \).

Work Done Formula

Work done by a force field along a path \( \mathbf{C} \) is given by \(\text{W} = \int_C \mathbf{F} \cdot d\mathbf{r} \). Here, \( d\mathbf{r} \) is the differential element of the path.

Parametrize the Helix Path

For part (a), the path is given by the parametric equations \( x = \cos t, y = \sin t, z = t \) for \( t \) from 0 to \(\pi \). Calculate the differential element: \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (-\sin t \mathbf{i} + \cos t \mathbf{j} + \mathbf{k}) dt \).

Evaluate the Work Integral for Helix Path

Substitute \( x, y, z, d\mathbf{r} \) in the work integral for part (a): \( \text{W}_a = \int_0^{\pi} \mathbf{F} \cdot d\mathbf{r} = \int_0^{\pi} [ -\sin t \mathbf{i} + \cos t \mathbf{j} + t \mathbf{k}] \cdot [-\sin t \mathbf{i} + \cos t \mathbf{j} + \mathbf{k}] \, dt \. \)

Simplify the Integral for Helix Path

Simplify and evaluate: \[ \text{W}_a = \int_0^{\pi} (\sin^2 t + \cos^2 t + t) \, dt = \int_0^{\pi} (1 + t) \, dt = \left[t + \frac{t^2}{2}\right]_0^{\pi} = \pi + \frac{\pi^2}{2}. \]

Parametrize the Straight Line Path

For part (b), use the parametric equations for a straight line: \( x(t) = 1 - 2t, y(t) = 0, z(t) = \pi t \) for \( t \) from 0 to 1. Calculate the differential element: \( d\mathbf{r} = (-2\mathbf{i} + \pi \mathbf{k}) dt \).

Evaluate the Work Integral for Straight Line Path

Substitute \( x, y, z, d\mathbf{r} \) in the work integral for part (b): \( \text{W}_b = \int_0^1 \mathbf{F} \cdot d\mathbf{r} = \int_0^1 [-0 \mathbf{i} + (1 - 2t) \mathbf{j} + (\pi t) \mathbf{k}] \cdot [-2 \mathbf{i} + \pi \mathbf{k}] dt \. \)

Simplify the Integral for Straight Line Path

Simplify and evaluate: \[ \text{W}_b = \int_0^1 [0 + 0 + (\pi t \cdot \pi)] \, dt = \int_0^1 \pi^2 t \, dt = \left[ \frac{\pi^2 t^2}{2} \right]_0^1 = \frac{\pi^2}{2}. \]

Compare the Results and Explain

The work done is \( \pi + \frac{\pi^2}{2} \) along the helix and \( \frac{\pi^2}{2} \) along the straight line. The answers are not the same because the work done by a non-conservative force field depends on the path taken between two points.

Key concepts

Work Done by Force Field

The work done by a force field on a particle moving along a path is a fundamental concept in physics. In mathematical terms, it is represented by the line integral of the force field along the given path. If \(\textbf{F}\) is the force field, and \(\textbf{C}\) is the path, then the work done \(\text{W}\) is computed as: \[ \text{W} = \int_C \textbf{F} \cdot d\textbf{r} \] Here, \(d\textbf{r}\) is the differential path element, essentially a small displacement vector along the path. This integral sums up the dot product of the force applied and the displacement over the entire path.

Parametric Equations

Parametric equations are a powerful tool for describing paths in space. Instead of expressing \(x, y, z\) as direct functions, we express them in terms of another variable, typically \(t\). The path of the particle in three-dimensional space can then be described as \((x(t), y(t), z(t))\). In step 3 of our solution, for the helix, the parametric equations are: \[ x = \cos t, \; y = \sin t, \; z = t \. \] This describes a helical path corresponding to the coordinates changing as \(t\) ranges from 0 to \(\pi\). Using parametric equations simplifies the calculation of the differential element \(d\textbf{r}\).

Path-Dependence in Non-Conservative Fields

Non-conservative force fields have the quality that the work done depends on the path taken between two points. In conservative fields, the work only depends on the positions of the initial and final points. For instance, gravitational force is conservative, while many actual force fields in electromagnetism and fluid dynamics are non-conservative. Comparing the results from our exercise shows that the work done along the helix (\text{W}_a = \pi + \frac{\pi^2}{2}\text{\}) is different from the work done along the straight line (\text{W}_b = \frac{\pi^2}{2}\text{\}). This variation confirms that the force field \(\textbf{F} = -y \mathbf{i} + x \mathbf{j} + z \mathbf{k}\) is non-conservative.

Differential Elements

Differential elements are tiny pieces of the path taken by the particle and are crucial in line integrals. These elements can be computed using parametric equations. For example, for the helix path: \[ d\textbf{r} = (\frac{d\textbf{r}}{dt}) dt = (-\sin t \mathbf{i} + \cos t \mathbf{j} + \mathbf{k}) dt. \] Here, \(d\textbf{r}\) is the vector of the differentials of \(x, y, z\) along the parameter \(t\). The dot product of \(d\textbf{r}\) with the force field \(\textbf{F}\) is then integrated over the parameter's range to find the work done. Similarly, for the straight line path: \[ d\textbf{r} = (-2 \mathbf{i} + \pi \mathbf{k}) dt. \] Each infinitesimal step along the path is considered, aggregated to compute the total work done by the force field.