Problem 55 - 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 \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}=x y^{6} \mathbf{i}+3 x\left(x y^{5}+2\right) \mathbf{j} ; \quad \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(\sin t) \mathbf{j}} \\\ {0 \leq t \leq 2 \pi}\end{array} $$

Step-by-Step Solution

Answer

The work done by the force is found by computing the line integral using a CAS; the result is obtained after integrating from 0 to \(2\pi\).

Step 1: Find the Differential of the Path

The vector path is given as \(\mathbf{r}(t) = 2 \cos t \mathbf{i} + \sin t \mathbf{j}\). To find \(d\mathbf{r}\), we differentiate this with respect to \(t\).\[\frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt}(2 \cos t \mathbf{i} + \sin t \mathbf{j})\]Computing the derivatives gives:\[\frac{d}{dt}(2 \cos t) = -2 \sin t, \quad \frac{d}{dt}(\sin t) = \cos t\]Thus, \(d\mathbf{r} = (-2 \sin t \mathbf{i} + \cos t \mathbf{j}) dt\).

Step 2: Substitute Path into Force Field

The force \(\mathbf{F}\) is \(xy^6 \mathbf{i} + 3x(xy^5 + 2) \mathbf{j}\). Substitute \(x = 2 \cos t\) and \(y = \sin t\) from the path into \(\mathbf{F}\):\[\mathbf{F} = (2 \cos t)(\sin t)^6 \mathbf{i} + 3(2 \cos t)((2 \cos t)(\sin t)^5 + 2) \mathbf{j}\]Simplifying, we have:\[\mathbf{F} = 2 \cos t \sin^6 t \mathbf{i} + 6 \cos t (2 \cos t \sin^5 t + 2) \mathbf{j}\]

Step 3: Compute the Line Integral

Now compute \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\). The dot product of \(\mathbf{F}\) and \(d\mathbf{r}\) is given by:\[(2 \cos t \sin^6 t)(-2 \sin t) + [6 \cos t (2 \cos t \sin^5 t + 2)](\cos t)\]Simplifying,\[\text{Dot Product} = -4 \cos t \sin^7 t + 12 \cos^2 t (2 \cos t \sin^5 t + 2)\]This needs to be integrated with respect to \(t\) over the interval \([0, 2\pi]\). Applying a computer algebra system (CAS) for efficiency:\[\int_{0}^{2\pi} \left(-4 \cos t \sin^7 t + 12\cos^3 t \sin^5 t + 24 \cos^2 t\right) dt\]The CAS yields the result, which is the work done by the force.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields. A vector field assigns a vector to every point in space. In this context, we're interested in understanding how a force vector field, represented by \( \mathbf{F} = x y^6 \mathbf{i} + 3 x (x y^5 + 2) \mathbf{j} \), acts on a given path. Vector calculus allows us to perform operations like differentiation and integration on these vector fields. This is crucial when analyzing how a field changes along a particular path and in determining quantities like work done. By applying principles of vector calculus, we explore relationships between paths, differential elements (like \( d\mathbf{r} \)), and how vector fields operate over these paths. In practical terms, vector calculus provides the tools needed to compute line integrals, which are used to quantify the work done by a force along a specified path, integral to both physics and engineering applications. We can calculate movement, work done, and various fluxes by manipulating vectors, much like calculating areas with simple calculus.

Line Integral

A line integral is a technique to compute a sum of field values along a path, capturing how a vector field interacts with a curve. For example, to calculate the work done by a force \( \mathbf{F} \) as a particle travels along a path \( \mathbf{r}(t) \), we use a line integral. The path is parametrized, and the force vector field is applied along each differential point of this path by evaluating \( \mathbf{F} \cdot d\mathbf{r} \). This expression involves a dot product between the force vector and the path's differential element \( d\mathbf{r} \). The final calculation, \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \), integrates the dot product along the entire path. It accounts for the force's direction relative to the path, allowing us to compute work precisely. In our exercise, a computer algebra system (CAS) simplifies the undertaking of integrating over complicated expressions that come from the parametrization and the field interaction.

Parametrized Path

A parametrized path provides a way to express a curve as a function of a parameter, typically \( t \). Here, the path is given by \( \mathbf{r}(t) = 2 \cos t \mathbf{i} + \sin t \mathbf{j} \), which describes a particular trajectory in space. This path allows us to evaluate the effects of a vector field along its length. By expressing it in parameter form, we can compute the differential \( d\mathbf{r} \) through differentiation with respect to \( t \). This breakdown helps in calculating how vectors like force actively influence the path incrementally. Parametrization transforms a path from a conceptual trajectory into a form suitable for calculations and integrations, making it easier to substitute into the vector field equations and perform line integrals. In calculus, the power of parametrized paths lies in turning complex motions into manageable mathematical entities, facilitating precise analyses of various physical phenomena such as the work done by forces.