Problem 45 - Thomas Calculus

Question

In Exercises \(43-46,\) use a CAS to perform the following steps to evaluate the line integrals. $$ \begin{array}{l}{\text { a. Find } d s=|\mathbf{v}(t)| d t \text { for the path } \mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k} \text { . }} \\ {\text { b. Express the integrand } f(g(t), h(t), k(t))|\mathbf{v}(t)| \text { as a function of }} \\ {\text { the parameter } t .} \\ {\text { c. Evaluate } \int_{C} f d s \text { using Equation }(2) \text { in the text. }}\end{array} $$ $$ \begin{array}{l}{f(x, y, z)=x \sqrt{y}-3 z^{2} ; \quad \mathbf{r}(t)=(\cos 2 t) \mathbf{i}+(\sin 2 t) \mathbf{j}+5 t \mathbf{k}} \\ {0 \leq t \leq 2 \pi}\end{array} $$

Step-by-Step Solution

Answer

The integral evaluates to \(-75\sqrt{29} \times \frac{8\pi^3}{3}\).

Step 1: Compute Velocity Vector

For the path \( \mathbf{r}(t) \), compute the velocity vector \( \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} \):\[ \mathbf{v}(t) = \frac{d}{dt} \left( \cos(2t) \mathbf{i} + \sin(2t) \mathbf{j} + 5t \mathbf{k} \right) = -2\sin(2t) \mathbf{i} + 2\cos(2t) \mathbf{j} + 5 \mathbf{k}. \]

Step 2: Find Magnitude of Velocity

Compute the magnitude of the velocity vector \( \mathbf{v}(t) \):\[ |\mathbf{v}(t)| = \sqrt{(-2\sin(2t))^2 + (2\cos(2t))^2 + 5^2} = \sqrt{4\sin^2(2t) + 4\cos^2(2t) + 25}. \]Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), simplify:\[ |\mathbf{v}(t)| = \sqrt{4 + 25} = \sqrt{29}. \]

Step 3: Express the Integrand as a Function of t

Substitute \( g(t), h(t), \) and \( k(t) \) into the function \( f(x, y, z) \):\[ f(g(t), h(t), k(t)) = f(\cos(2t), \sin(2t), 5t) = \cos(2t) \sqrt{\sin(2t)} - 3 \cdot (5t)^2. \]

Step 4: Integrate Over the Path C

Set up the integral \( \int_C f \, ds \) over the interval \( 0 \leq t \leq 2\pi \):\[ \int_{0}^{2\pi} \left( \cos(2t) \sqrt{\sin(2t)} - 75t^2 \right) \sqrt{29} \, dt. \]Compute the integral using a CAS or numerical methods. The integral of the second part simplifies to a straightforward calculation over the interval.

Step 5: Evaluation of the Integral

Use a computer algebra system (CAS) to evaluate\[ \sqrt{29} \int_{0}^{2\pi} \cos(2t) \sqrt{\sin(2t)} \, dt - 75 \sqrt{29} \int_{0}^{2\pi} t^2 \, dt. \]Evaluate each integral separately:\[ \int_{0}^{2\pi} \cos(2t) \sqrt{\sin(2t)} \, dt \text{ requires numerical methods for evaluation.} \]\[ \int_{0}^{2\pi} t^2 \, dt = \frac{(2\pi)^3}{3} = \frac{8\pi^3}{3}. \]

Key Concepts

Velocity Vector

In a line integral problem like the one provided, understanding the velocity vector is crucial. The velocity vector for a path, typically represented as \( \mathbf{v}(t) \), describes how the path progresses through each of the spatial dimensions over time. For our specific path \( \mathbf{r}(t) = (\cos 2t) \mathbf{i} + (\sin 2t) \mathbf{j} + 5t \mathbf{k} \), the velocity vector is determined by differentiating each component of \( \mathbf{r}(t) \) with respect to \( t \). This creates a new vector that represents the rate of change of the path at any given moment.

The differentiation returns:

• \( -2\sin(2t) \mathbf{i} \) showing rate of change along the \( x \)-axis,
• \( 2\cos(2t) \mathbf{j} \) representing movement along the \( y \)-axis,
• \( 5 \mathbf{k} \) indicating a constant rate in the \( z \)-axis direction.

Grasping this concept helps visualize how an object would move through threespace along this specified path, providing insights into both direction and speed at different moments of its trajectory.

Magnitude of Velocity

The magnitude of a velocity vector quantifies the speed at which an object travels along a path, without considering direction. It is the length of the vector \( \mathbf{v}(t) \). To calculate it, we derive the square root of the sum of the squares of each component in the velocity vector:

• For our vector \( \mathbf{v}(t) = -2\sin(2t) \mathbf{i} + 2\cos(2t) \mathbf{j} + 5 \mathbf{k} \), the magnitude is computed as \( \sqrt{(-2\sin(2t))^2 + (2\cos(2t))^2 + 5^2} \).

Using the trigonometric identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), the expression simplifies to \( \sqrt{4 + 25} = \sqrt{29} \).

This simplification means that no matter the value of \( t \), the speed or magnitude of velocity remains \( \sqrt{29} \) units per time interval, hinting a consistent pace along this path.

Computer Algebra System (CAS)

In complex mathematical problems, particularly those involving intricate integrals, a Computer Algebra System (CAS) becomes a valuable tool. These systems, like Mathematica or Maple, are designed to handle symbolic computations, which include not only solving integrals but also differentiating functions, simplifying expressions, and more. In exercises like the one described, a CAS can compute tricky integrals that are hard to solve by hand.
For instance, to evaluate the integral \( \sqrt{29} \int_{0}^{2\pi} \cos(2t) \sqrt{\sin(2t)} \, dt - 75 \sqrt{29} \int_{0}^{2\pi} t^2 \, dt \), a CAS aids in automating the procedure of solving these integrals either symbolically or numerically, depending on the complexity. Particularly, when the integrand is not straightforward and involves functions like \( \cos(2t) \sqrt{\sin(2t)} \), numerical methods might be preferred.

Mastering the use of a CAS not only eases the calculation burdens but also enriches one's ability to explore deeper mathematical relationships and outcomes.