Problem 60 - Thomas Calculus

Question

Use a CAS to perform the following steps for finding the work done by force \(\mathbf{F}\) over the given path: a. Find \(d \mathbf{r}\) for the path \(\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k}\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}=\left(x^{2} y\right) \mathbf{i}+\frac{1}{3} x^{3} \mathbf{j}+x y \mathbf{k} ; \quad \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+} \\ {\left(2 \sin ^{2} t-1\right) \mathbf{k}, \quad 0 \leq t \leq 2 \pi}\end{array} $$

Step-by-Step Solution

Answer

The work done by the force over the path is 0.

Step 1: Differentiate the Path Vector

First, find the derivative of the given path vector \(\mathbf{r}(t)\). The path vector is \(\mathbf{r}(t) = \cos t \ \mathbf{i} + \sin t \ \mathbf{j} + (2 \sin^2 t - 1) \ \mathbf{k}\). Differentiate each component with respect to \(t\): \[ d\mathbf{r} = \left( -\sin t \right) \mathbf{i} + \cos t \ \mathbf{j} + \left(4 \sin t \cos t \right) \mathbf{k} \, dt\]

Step 2: Substitute Path into Force Field

Substitute \(x = \cos t\), \(y = \sin t\) into the components of the force vector \(\mathbf{F}\). The force field is \(\mathbf{F} = x^2 y \ \mathbf{i} + \frac{1}{3} x^3 \ \mathbf{j} + x y \ \mathbf{k}\). So, substitute to get:\[ \mathbf{F}(t) = (\cos^2 t \sin t) \ \mathbf{i} + \left(\frac{1}{3} \cos^3 t \right) \ \mathbf{j} + (\cos t \sin t) \ \mathbf{k}\]

Step 3: Compute the Dot Product \(\mathbf{F} \cdot d\mathbf{r}\)

Now, compute the dot product of the substituted force vector \(\mathbf{F}(t)\) and \(d\mathbf{r}\):\[\mathbf{F} \cdot d\mathbf{r} = (\cos^2 t \sin t)(-\sin t) + \left(\frac{1}{3} \cos^3 t\right)(\cos t) + (\cos t \sin t)(4 \sin t \cos t) \, dt\]Simplifying:\[= -\cos^2 t \sin^2 t + \frac{1}{3} \cos^4 t + 4 \cos^2 t \sin^2 t \, dt\]

Step 4: Simplify the Expression Before Integration

Combine like terms in the dot product expression:\[\mathbf{F} \cdot d\mathbf{r} = 3 \cos^2 t \sin^2 t + \frac{1}{3} \cos^4 t \, dt\]This is the expression to be integrated over the interval \(0 \leq t \leq 2\pi\).

Step 5: Evaluate the Integral Over the Given Path

Integrate the simplified expression over the interval \([0, 2\pi]\):\[\int_0^{2\pi} \left( 3 \cos^2 t \sin^2 t + \frac{1}{3} \cos^4 t \right) \, dt\]Each term requires separate integration. Calculating these integrals gives the total work done by force \( \int_C \mathbf{F} \cdot d\mathbf{r} \):The integral of \(3 \cos^2 t \sin^2 t\) can be solved using a trigonometric identity:\[= \frac{3}{4} \int_0^{2\pi} \sin^2(2t) \, dt = \frac{3}{8} \int_0^{2\pi} 1 - \cos(4t) \, dt\]The integral of \(\frac{1}{3} \cos^4 t\) is solved by expanding and applying identities: \[\int_0^{2\pi} \frac{1}{3} \left(\frac{1 + \cos(2t)}{2}\right)^2 \, dt \]Both integrals evaluate to 0 over a full period, thus the work done is\(0\).

Key Concepts

Work Done by a Force

The concept of 'Work Done by a Force' in the context of line integrals involves calculating the total effect of a force field as it acts upon an object moving along a specified path. In vector calculus, work is computed as the line integral of the force vector field along a path or curve.

When a force is applied along a path, the work done encompasses both the magnitude of the force and the direction of movement along the path. If the force vector \(\mathbf{F}\) and a differential path element \(d\mathbf{r}\) are given, then the work done by the force as the object traverses the path \(C\) is given by the integral \[W = \int_C \mathbf{F} \cdot d\mathbf{r} \\]This expression involves calculating the dot product of the force vector with the differential path vector and integrating it over the curve. This dot product measures how much of the force vector is aligned with the direction of the path at each point.

Importantly, if the force is perpendicular to the movement path at any point, no work is done at this point. This is a critical concept in physics and engineering as it helps determine how energy is transferred by forces.

Parametrization of a Path

The process of parametrizing a path involves expressing the coordinates of a curve as functions of a single parameter, typically denoted as \(t\). For a three-dimensional curve, the position vector \(\mathbf{r}(t)\) is generally expressed in the form:

• \(x = g(t)\)
• \(y = h(t)\)
• and \(z = k(t)\)

This parameter \(t\) usually varies over an interval, defining the start and end points of the path.

Parametrization is a crucial step in the application of line integrals as it simplifies the integration process by converting it into a problem involving a single independent variable. In the exercise provided, the path has been parametrized with trigonometric functions: \(\mathbf{r}(t) = \cos t \ \, \mathbf{i} + \sin t \, \mathbf{j} + (2 \sin^2 t - 1) \mathbf{k}\).

This set of parametric equations not only traces out a path in space but also allows for easy substitution into other vector fields, such as the force vector field we're studying. Proper parametrization captures the essence of the path while prparing it for practical operations such as integration.

Integration in Vector Calculus

Integration in vector calculus extends the idea of calculating areas and accumulated quantities to vector fields. It is especially useful in physics and engineering to find quantities like work, flux, and circulation, where the path or boundary of integration matters.

In the exercise solution, the integral to evaluate is \[\int_C \mathbf{F} \cdot d\mathbf{r} \\]where \(\mathbf{F}\) is the force vector field and \(d\mathbf{r}\) is the differential path vector. The integration process involves combining the vector field with the path via the dot product, simplifying any resulting expressions, and then integrating with respect to \(t\) over a given interval.

• The purpose is to totalize the contribution of the vector field along the path.
• This can involve trigonometric identities or substitutions that simplify the integral expressions.

Furthermore, in many scenarios like this one, symmetry can be leveraged to simplify calculations — such as realizing that integrals over periodic intervals can yield zero, indicating no net work is done. These methods allow complex physical phenomena to be described with mathematical precision.