Problem 46 - Thomas Calculus

Question

In Exercises \(43 - 46 ,\) use a CAS to perform the following steps to evaluate the line integrals. $$ \begin{array} { l } { \text { a. Find } d s = | \mathbf { v } ( t ) | d t \text { for the path } \mathbf { r } ( t ) = g ( t ) \mathbf { i } + h ( t ) \mathbf { j } + k ( t ) \mathbf { k } \text { . } } \\ { \text { b. Express the integrand } f ( g ( t ) , h ( t ) , k ( t ) ) | \mathbf { v } ( t ) | \text { as a function of the parameter } t . } \\ { \text { c. Evaluate } \int _ { C } f d s \text { using Equation } ( 2 ) \text { in the text. } } \end{array} $$ $$ \begin{array} { l } { f ( x , y , z ) = \left( 1 + \frac { 9 } { 4 } z ^ { 1 / 3 } \right) ^ { 1 / 4 } ; \quad \mathbf { r } ( t ) = ( \cos 2 t ) \mathbf { i } + ( \sin 2 t ) \mathbf { j } + } \\ { t ^ { 5 / 2 } \mathbf { k } , \quad 0 \leq t \leq 2 \pi } \end{array} $$

Step-by-Step Solution

Answer

Evaluate the integral using a CAS: \[ \int_{0}^{2\pi} \left( 1 + \frac{3}{4} t^{5/6} \right)^{1/4} \cdot \sqrt{4 + \frac{25}{4}t^3} \ dt \].

Step 1: Determine the Velocity Vector

First, we need to find the velocity vector \( \mathbf{v}(t) \) by differentiating the position vector \( \mathbf{r}(t) = (\cos 2t) \mathbf{i} + (\sin 2t) \mathbf{j} + t^{5/2} \mathbf{k} \). Differentiating each component with respect to \( t \), we obtain:\[ \mathbf{v}(t) = \frac{d}{dt}(\cos 2t) \mathbf{i} + \frac{d}{dt}(\sin 2t) \mathbf{j} + \frac{d}{dt}(t^{5/2}) \mathbf{k} \] \[ \mathbf{v}(t) = (-2\sin 2t) \mathbf{i} + (2\cos 2t) \mathbf{j} + \left(\frac{5}{2}t^{3/2}\right) \mathbf{k} \]

Step 2: Compute the Magnitude of the Velocity Vector

Now, we calculate the magnitude of the velocity vector \(|\mathbf{v}(t)|\). This is given by the formula:\[ |\mathbf{v}(t)| = \sqrt{(-2\sin 2t)^2 + (2\cos 2t)^2 + \left(\frac{5}{2}t^{3/2}\right)^2} \]\[ |\mathbf{v}(t)| = \sqrt{4\sin^2 2t + 4\cos^2 2t + \frac{25}{4}t^3} \]Since \(\sin^2 2t + \cos^2 2t = 1\), we have:\[ |\mathbf{v}(t)| = \sqrt{4 + \frac{25}{4}t^3} \]

Step 3: Express the Integrand as a Function of t

The integrand is \( f(g(t), h(t), k(t)) |\mathbf{v}(t)| \), where \( f(x, y, z) = \left( 1 + \frac{9}{4} z^{1/3} \right)^{1/4} \). With \( z = t^{5/2} \), \( f \) evaluated at \( \mathbf{r}(t) \) becomes:\[ f(t) = \left( 1 + \frac{9}{4} \left(t^{5/2}\right)^{1/3} \right)^{1/4} = \left( 1 + \frac{3}{4} t^{5/6} \right)^{1/4} \]The integrand is:\[ \left( 1 + \frac{3}{4} t^{5/6} \right)^{1/4} \cdot \sqrt{4 + \frac{25}{4}t^3} \]

Step 4: Evaluate the Line Integral

Finally, we use the result from the previous step to evaluate the line integral over the path. We need to compute:\[ \int_{0}^{2\pi} \left( 1 + \frac{3}{4} t^{5/6} \right)^{1/4} \cdot \sqrt{4 + \frac{25}{4}t^3} \ dt \]This integral is evaluated using a Computer Algebra System (CAS), as it involves a complex function that represents the combination of algebraic and trigonometric functions.After using the CAS to perform the integration, you find the final numerical result.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and differentiable functions. In simpler terms, it involves calculations with vectors in 2D or 3D space. A vector is a quantity with both magnitude and direction, such as velocity or force.
Vector calculus provides tools to analyze and understand physical phenomena involving vectors like fluid flows or electromagnetic fields. It includes operations such as differentiation and integration, specifically devised for vectors rather than scalar quantities, enabling us to work more smoothly with vector-valued functions.
An essential aspect of vector calculus is the evaluation of line integrals. These integrals account for a function along a curve in space, which might describe a path or trajectory. When calculating such integrals, we often use the position vector, velocity vector, and other derivatives. This approach allows us to study how these vector fields change and interact along a specified path.

Parametric Equations

Parametric equations are a powerful way to represent curves by using a set of equations, where each coordinate is expressed as a function of one or more independent parameters. This is particularly useful in three-dimensional space, as it provides an efficient method to describe complex shapes and trajectories.
Instead of writing a function in the form of y = f(x), a curve is described using equations like x = g(t), y = h(t), and z = k(t). Here, the parameter t usually represents time, but it can be any independent variable that defines the curve's position at any given instance.
In vector calculus, parametric equations are essential for describing the motion of particles along a path. They provide an alternative method to standard forms and can be particularly useful when analyzing motion or integrating along a curve, as they specifically describe the path through its parameter rather than simply through spatial coordinates.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a software tool designed to handle complex mathematical calculations symbolically rather than numerically. These systems can solve equations, compute integrals, and manipulate algebraic expressions without simplifying them to a mere number.
CAS tools are invaluable in higher math and engineering, as they allow users to perform operations that are computationally exhaustive or even impossible by hand, such as evaluating line integrals with intricate behaviors.
When faced with complex integrals, like the one from the exercise in focus, a CAS performs the necessary symbolic operations to derive a solution. These solutions might be exact symbolic expressions or numeric results, providing insightful answers that aid in comprehending detailed mathematical problems efficiently.

Velocity Vector

The velocity vector is a fundamental concept in physics and mathematics, representing the speed and direction of a moving object at a given instant. In mathematical terms, it is the derivative of the position vector with respect to time.
It tells us how the position of an object changes over time and is denoted as \( \mathbf{v}(t) \) for a given position vector \( \mathbf{r}(t) \). Deriving the velocity vector involves differentiating each component of the position vector function.
In 3D space, it is expressed as \( \mathbf{v}(t) = \frac{d}{dt}(x(t)) \mathbf{i} + \frac{d}{dt}(y(t)) \mathbf{j} + \frac{d}{dt}(z(t)) \mathbf{k} \). The example from the exercise illustrates this process by calculating \( \mathbf{v}(t) = (-2\sin 2t) \mathbf{i} + (2\cos 2t) \mathbf{j} + \left(\frac{5}{2}t^{3/2}\right) \mathbf{k} \), assisting in evaluating line integrals by providing the speed component.