Problem 44 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Use a CAS to compute the integral numerically.

Step 1: Find the velocity vector

First, differentiate the parametric equations of \( \mathbf{r}(t) = t \mathbf{i} + \frac{1}{3}t^2 \mathbf{j} + \sqrt{t} \mathbf{k} \) to find the velocity vector \( \mathbf{v}(t) \). \[ \mathbf{v}(t) = \frac{d}{dt}(t \mathbf{i} + \frac{1}{3}t^2 \mathbf{j} + \sqrt{t} \mathbf{k}) = \mathbf{i} + \frac{2}{3}t \mathbf{j} + \frac{1}{2\sqrt{t}} \mathbf{k} \]

Step 2: Calculate the magnitude of the velocity vector

To find the value of \( ds = |\mathbf{v}(t)| dt \), we need the magnitude of the velocity vector.\[ |\mathbf{v}(t)| = \sqrt{(1)^2 + \left(\frac{2}{3}t \right)^2 + \left(\frac{1}{2\sqrt{t}}\right)^2} \]\[ = \sqrt{1 + \frac{4}{9}t^2 + \frac{1}{4t}} \]

Step 3: Express the integrand as a function of t

We need to express the given function in terms of \( t \) using the parameterization for integration. For \( f(x, y, z) = \sqrt{1 + x^3 + 5y^3} \), substitute \( x = t \), \( y = \frac{1}{3}t^2 \), and \( z = \sqrt{t} \) to get:\[ f(g(t), h(t), k(t)) = \sqrt{1 + t^3 + 5\left(\frac{1}{3}t^2\right)^3} \] \[ = \sqrt{1 + t^3 + \frac{5}{27}t^6} \]The full integrand becomes:\[ \sqrt{1 + t^3 + \frac{5}{27}t^6} \times \sqrt{1 + \frac{4}{9}t^2 + \frac{1}{4t}} \]

Step 4: Set up the integral

The integral \( \int_C f \, ds \) over the path can be expressed in terms of \( t \) as:\[ \int_{0}^{2} \sqrt{1 + t^3 + \frac{5}{27}t^6} \times \sqrt{1 + \frac{4}{9}t^2 + \frac{1}{4t}} \, dt \]

Step 5: Evaluate the integral

Use a Computer Algebra System (CAS) to evaluate the definite integral:\[ \int_{0}^{2} \sqrt{1 + t^3 + \frac{5}{27}t^6} \times \sqrt{1 + \frac{4}{9}t^2 + \frac{1}{4t}} \, dt \]This step generally involves inputting the expression into the software and letting it compute the result numerically or analytically.

Key Concepts

Velocity Vector

The velocity vector is a fundamental concept in physics and calculus, especially when dealing with motion along a path given by a parametric curve. For a vector-valued function such as \( \mathbf{r}(t) = t \mathbf{i} + \frac{1}{3}t^2 \mathbf{j} + \sqrt{t} \mathbf{k} \), the velocity vector, \( \mathbf{v}(t) \), is found by differentiating \( \mathbf{r}(t) \) with respect to the parameter \( t \). This differentiation yields the rate of change of the position vector, therefore giving us the velocity at any given point along the path.

• The components of the velocity vector are the derivatives of each component of \( \mathbf{r}(t) \) with respect to \( t \).
• In our case, \( \mathbf{v}(t) = \frac{d}{dt}(t \mathbf{i} + \frac{1}{3}t^2 \mathbf{j} + \sqrt{t} \mathbf{k}) = \mathbf{i} + \frac{2}{3}t \mathbf{j} + \frac{1}{2\sqrt{t}} \mathbf{k} \).
• Each component corresponds to the instantaneous rate of change in that direction: the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) components for the x, y, and z directions respectively.

Understanding the velocity vector is crucial for computing the path's arc length and other motion-related quantities, such as acceleration.

Parameterization

Parameterization refers to expressing a curve or surface using a set of equations that depend on a parameter. Typically, in three dimensions, a path is represented with three equations that depend on a single parameter, often \( t \).

• In our given problem, the path is parameterized by \( \mathbf{r}(t) = t \mathbf{i} + \frac{1}{3}t^2 \mathbf{j} + \sqrt{t} \mathbf{k} \).
• This parameterization helps in converting a three-dimensional problem into a one-dimensional one because computations can be done with respect to \( t \).
• It simplifies the process of integration over curves by transforming space coordinates into a single variable.

A major benefit of parameterization is that it allows for animating curves and paths in simulations, providing a versatile tool for both graphical and analytical purposes.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a software program that facilitates symbolic mathematics. These systems are particularly useful in solving complex integrals, derivatives, and algebraic equations symbolically rather than numerically.

• For instance, in our line integral problem, we've set up an integral: \( \int_{0}^{2} \sqrt{1 + t^3 + \frac{5}{27}t^6} \times \sqrt{1 + \frac{4}{9}t^2 + \frac{1}{4t}} \, dt \), which can be cumbersome to solve by hand.
• With a CAS, such as Wolfram Alpha, Mathematica, or Maple, you can input this integral to get either an analytical solution or a numerical estimate.
• This technology is particularly helpful for students and professionals, offering a valuable tool to verify manual computations and explore mathematical problems deeply without getting caught up in complex numerical calculations.

Thus, a CAS opens up new possibilities for exploring advanced mathematics by providing solutions that are otherwise challenging to compute manually.