Problem 45 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Use a CAS to evaluate the integral \( \int_0^{2\pi} (\cos 2t \sqrt{\sin 2t} - 75t^2) \sqrt{29} \, dt \).

Step 1: Compute the Velocity Vector

First, we need to determine the velocity vector \( \mathbf{v}(t) \) by differentiating the path \( \mathbf{r}(t) \) with respect to \( t \). The path is given as \( \mathbf{r}(t) = (\cos 2t) \mathbf{i} + (\sin 2t) \mathbf{j} + 5t \mathbf{k} \). Compute: \[ \mathbf{v}(t) = \frac{d}{dt}((\cos 2t) \mathbf{i} + (\sin 2t) \mathbf{j} + 5t \mathbf{k}) = (-2\sin 2t) \mathbf{i} + (2\cos 2t) \mathbf{j} + 5 \mathbf{k} \]

Step 2: Calculate Magnitude of the Velocity Vector

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

Step 3: Express the Integrand as a function of \( t \)

The next step is to express \( f(g(t), h(t), k(t)) |\mathbf{v}(t)| \) in terms of \( t \). The function \( f(x,y,z) = x\sqrt{y} - 3z^2 \) becomes \( f(\cos 2t, \sin 2t, 5t) = \cos 2t \sqrt{\sin 2t} - 3(5t)^2 \). Thus, the integrand is: \[ f(g(t), h(t), k(t)) |\mathbf{v}(t)| = (\cos 2t \sqrt{\sin 2t} - 75t^2)\sqrt{29} \]

Step 4: Evaluate the Line Integral

Finally, evaluate the line integral \( \int_C f \, ds \) over the interval \( 0 \leq t \leq 2\pi \). Substitute our integrand from Step 3: \[ \int_0^{2\pi} (\cos 2t \sqrt{\sin 2t} - 75t^2) \sqrt{29} \, dt \] Calculate this integral using a CAS (Computer Algebra System). You'll likely find that the result is an evaluated expression or a specific number.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector-valued functions and how they can be integrated or differentiated. It is an essential tool in physics and engineering, providing methods to analyze fields like gravitational, electric, or magnetic fields. In this exercise, we look at line integrals, which help us integrate a scalar or vector field over a curve in space.

Line integrals have two types: path integrals (which involve scalar fields) and circulation integrals (which involve vector fields). Using vector calculus, we analyze these integrals by breaking them down into vector components. This is particularly important in physics, where fields are often represented as vector fields.

By incorporating functions that go beyond simple scalar fields, vector calculus allows for the evaluation of integrals in multidimensional spaces. It is vital to understand how to apply vector calculus when dealing with complex paths, especially in the case of curves or surfaces.

Parameterization of Curves

Parameterization is the process of expressing a curve using a set of equations that describe its position in space as a function of one or more parameters. In the context of line integrals, parameterizing a curve allows us to evaluate integrals more efficiently over those curves.

For instance, in this problem, the curve is parameterized by the function \( \mathbf{r}(t) = (\cos 2t) \mathbf{i} + (\sin 2t) \mathbf{j} + 5t \mathbf{k} \). This representation not only informs the shape and orientation of the path but also specifies how a point travels along the curve as \( t \) varies from 0 to \( 2\pi \). By using trigonometric functions and polynomial expressions, we can describe intricate paths that traverse through three-dimensional space.

To successfully parameterize a curve, the chosen functions must continuously cover the interval of interest. For example, using sine and cosine for circular paths ensures smooth traversal across the curve while also facilitating easier computations in later stages. Parameterization thus simplifies integral evaluations by transforming them into an easier-to-analyze form of the parameter \( t \).

Differentiation of Vector Functions

Differentiation is a fundamental concept that measures how a function changes as its inputs change. In this exercise, we differentiate the vector function \( \mathbf{r}(t) \) to find the velocity vector, \( \mathbf{v}(t) \).

Differentiating a vector function involves taking the derivative of each component separately. For the given path \( \mathbf{r}(t) = (\cos 2t) \mathbf{i} + (\sin 2t) \mathbf{j} + 5t \mathbf{k} \), its differentiation yields:

• A derivative of \( \cos 2t \) with respect to \( t \) gives \(-2\sin 2t \),
• A derivative of \( \sin 2t \) gives \( 2\cos 2t \), and
• A derivative of \( 5t \) naturally gives 5.

Thus, the velocity vector becomes \( \mathbf{v}(t) = (-2\sin 2t) \mathbf{i} + (2\cos 2t) \mathbf{j} + 5 \mathbf{k} \).

Understanding how to differentiate vector functions is critical in physics for determining rates of change, such as velocity or acceleration, which in turn can describe motions of particles or objects along a trajectory.

Magnitude of a Vector

Calculating the magnitude of a vector helps in understanding its length or size, which is crucial when dealing with vector representations in space. It is equally important for assessing the speed or energy associated with vector quantities.

In this problem, the magnitude of the velocity vector \( \mathbf{v}(t) = (-2\sin 2t) \mathbf{i} + (2\cos 2t) \mathbf{j} + 5 \mathbf{k} \) is vital for integrating along the curve. We compute it by applying the Pythagorean theorem, which in multiple dimensions involves summing the squares of all components:

• \((-2\sin 2t)^2\) contributes \(4\sin^2 2t\),
• \((2\cos 2t)^2\) gives \(4\cos^2 2t\), and
• The constant \(5^2\) equals 25.

Put together, the magnitude is \( | \mathbf{v}(t) | = \sqrt{4\sin^2 2t + 4\cos^2 2t + 25} \), which simplifies using the identity \( \sin^2 2t + \cos^2 2t = 1 \), resulting in \( \sqrt{29} \).

Understanding vector magnitudes is crucial in scenarios requiring calculations of distances or lengths, such as in physics for traversing paths, or in geometry and graphics for scaling transformations.