Problem 65 - Thomas Calculus

Question

In Exercises \(63-68,\) use a CAS to perform the following steps for finding the work done by force F over the given path: $$\begin{array}{l}{\mathbf{F}=(y+y z \cos x y z) \mathbf{i}+\left(x^{2}+x z \cos x y z\right) \mathbf{j}+} \\ {(z+x y \cos x y z) \mathbf{k} ; \quad \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(3 \sin t) \mathbf{j}+\mathbf{k}} \\ {0 \leq t \leq 2 \pi}\end{array}$$

Step-by-Step Solution

Answer

The work done is given by the integral of the dot product over \([0, 2\pi]\).

Step 1: Parameterize the Force Field

First, substitute the parameterized path \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + \mathbf{k} \) into the force field \( \mathbf{F}(x, y, z) = (y + yz \cos(xyz)) \mathbf{i} + (x^2 + xz \cos(xyz)) \mathbf{j} + (z + xy \cos(xyz)) \mathbf{k} \). This gives us \( \mathbf{F}(t) = \big((3 \sin t + 3 \sin t \cos(2 \cos t \cdot 3 \sin t)) \mathbf{i} + (4 \cos^2 t + 2 \cos t \cdot \cos(2 \cos t \cdot 3 \sin t)) \mathbf{j} + (1 + 6 \cos t \sin t \cdot \cos(2 \cos t \cdot 3 \sin t)) \mathbf{k}\big).\)

Step 2: Compute the Derivative of the Path

Next, find the derivative of the path \( \mathbf{r}(t) \): \( \mathbf{r}'(t) = \frac{d}{dt}((2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + \mathbf{k}) = (-2 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j}.\)

Step 3: Compute the Dot Product

Compute the dot product \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \). Substitute in the expressions for \( \mathbf{F}(t) \) and \( \mathbf{r}'(t) \). This simplifies to a scalar function that we denote as \( W(t) = \big((3 \sin t + 3 \sin t \cos(6 \cos t \sin t)) \cdot (-2 \sin t) + (4 \cos^2 t + 2 \cos t \cdot \cos(6 \cos t \sin t)) \cdot 3 \cos t\big)\).

Step 4: Integrate Over the Interval

Find the integral of \( W(t) \) from \( t=0 \) to \( t=2\pi \) to determine the work done. Symbolically, this is \( \int_{0}^{2\pi} W(t)\, dt \). Use a CAS to evaluate this integral.

Key Concepts

Work Done by a Force

When discussing vector calculus, one of the fundamental concepts is the work done by a force. Work is calculated as the force applied along a path multiplied by the distance over which it acts. In mathematical terms, work can be represented as an integral of the force dot product with a small segment of the path over which it acts. This is often expressed as: \[ W = \int_{{C}} \mathbf{F} \cdot d\mathbf{r} \].

• \( \mathbf{F} \) represents the force field.
• \( d\mathbf{r} \) denotes the differential vector along the path \( C \).Work is significant because it quantifies the energy transferred by the force along the path.

Understanding work in vector fields helps in analyzing physical systems because it connects force with energy, two pivotal elements in physics. By calculating the work done by a force, we gain insight into the energy dynamics along specified paths.

Parameterization of Force Fields

In vector calculus, parameterization is a method used to represent curves and paths. In our exercise, a path is given by the vector function \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + \mathbf{k} \).
Parameterizing the force field \( \mathbf{F}(x, y, z) \) requires substituting these expressions for \( x, y, \, and \, z \) into our force field. This transforms \( \mathbf{F} \) into a function of \( t \), which is more manageable when performing further calculations like path integrals:

\[ \mathbf{F}(t) = (3 \sin t + 3 \sin t \cos(2 \cos t \cdot 3 \sin t)) \mathbf{i} + (4 \cos^2 t + 2 \cos t \cdot \cos(2 \cos t \cdot 3 \sin t)) \mathbf{j} + (1 + 6 \cos t \sin t \cdot \cos(2 \cos t \cdot 3 \sin t)) \mathbf{k} \].

This substitution step simplifies the calculation of the work done along the curve. It helps to handle what would otherwise be a complex multi-variable function.

Path Integrals

Path integrals are a technique used in mathematics and physics that allows the integration of a function along a specified path. They are particularly useful in computing work done by a force field over a path. In our context, we calculate the integral of the dot product of the force field and the derivative of the parameterized path:

\[ \int_{a}^{b} \mathbf{F}(t) \cdot \mathbf{r}'(t) \, dt \]

where \( a \, and \, b \) are the bounds informing the limits of the path.
Path integrals merge information about both the force acting and the specific trajectory taken, offering valuable insights into physical scenarios where the path influences the outcome, such as electromagnetic fields or gravitational forces. It's a bridge between static analysis of fields and dynamic path-dependent outcomes.

Dot Product

The dot product is a fundamental operation in vector calculus, crucial for finding work done by a force. The dot product \( \mathbf{F}(t) \cdot \mathbf{r}'(t) \) measures the component of one vector in the direction of another. It gives a way to compute the contribution of the force in the direction of the path.

• Mathematically, the dot product of vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).
• The result is a scalar quantity, simplifying further integration.
• Only the parallel component contributes to work done since \( \cos\theta \) (the angle between the vectors) scales the effect.

Computing the dot product reduces complex vector interactions into simpler scalar interactions, integral for tasks like evaluating path integrals to find work done.

Line Integrals

Line integrals extend conventional integrals to functions defined along curves, offering a way to evaluate expressions like \( \int_{{C}} \mathbf{F} \cdot d\mathbf{r} \) over a path \( C \). They generalize single-dimensional integrals to account for vector fields along a path.

• In our exercise, the line integral evaluates the accumulation of work along the curve.
• By integrating \( W(t) \) from \( t=0 \) to \( t=2\pi \), we sum contributions of force over each tiny segment of the path.

We employ line integrals to derive insights that cannot be captured via standard integrals, such as electromagnetic flow or fluid dynamics. Here, we leverage them to capture the impact of a dynamic force along a nonlinear path.