Problem 46 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The integral is evaluated using a CAS for an approximate solution.

Step 1: Find Velocity Vector

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

Step 2: Calculate Magnitude of Velocity

Now, calculate the magnitude \( |\mathbf{v}(t)| \) as follows: \[ |\mathbf{v}(t)| = \sqrt{(-2\sin 2t)^2 + (2\cos 2t)^2 + \left(\frac{5}{2}t^{3/2}\right)^2} = \sqrt{4 \sin^2 2t + 4 \cos^2 2t + \frac{25}{4} t^3} = \sqrt{4 + \frac{25}{4} t^3}. \] The simplification is due to the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \).

Step 3: Express Integrand as a Function of t

Next, express the integrand \( f(g(t), h(t), k(t)) \cdot |\mathbf{v}(t)| \) in terms of \( t \). Given that \( f(x, y, z) = \left(1 + \frac{9}{4} z^{1/3}\right)^{1/4} \), substitute \( z = t^{5/2} \) to get \( f(\cos 2t, \sin 2t, t^{5/2}) = \left(1 + \frac{9}{4} (t^{5/2})^{1/3}\right)^{1/4} = \left(1 + \frac{9}{4} t^{5/6}\right)^{1/4}. \) Therefore, the integrand becomes: \[ \left(1 + \frac{9}{4} t^{5/6} \right)^{1/4} \cdot \sqrt{4 + \frac{25}{4} t^3}. \]

Step 4: Set up Line Integral

The line integral is expressed as \( \int_{0}^{2\pi} f(g(t), h(t), k(t)) |\mathbf{v}(t)| \ dt \). Substitute our expression for the integrand: \[ \int_{0}^{2\pi} \left(1 + \frac{9}{4} t^{5/6} \right)^{1/4} \sqrt{4 + \frac{25}{4} t^3} \, dt. \]

Step 5: Evaluate the Integral

Using a CAS (Computer Algebra System), evaluate the integral: \[ \int_{0}^{2\pi} \left(1 + \frac{9}{4} t^{5/6} \right)^{1/4} \sqrt{4 + \frac{25}{4} t^3} \, dt. \] The exact computation might involve numerical methods as it's difficult to determine analytically. By entering this into a CAS tool, we can find the numerical approximation.

Key Concepts

Multivariable Calculus

In multivariable calculus, we work with functions of several variables and explore concepts such as gradients, divergence, and curl. These topics extend the principles of single-variable calculus to higher dimensions, which is incredibly useful for analyzing physical phenomena.

A line integral is a crucial tool in this field. It is essentially an integration performed over a path or a curve, rather than over an interval as in single-variable calculus. When dealing with vector fields, a line integral helps us compute quantities like fluid flow or electromagnetic work along a specific path.

To understand how to evaluate a line integral along a path, you start by parameterizing the curve. This involves expressing the three-dimensional path as a function of a single parameter, often denoted by \( t \). Once parameterized, you can separate the line integral problem into parts, involving calculations such as finding the velocity vector and integrating functions over the specified path using the determined parameter.

Velocity Vector Magnitude

The velocity vector describes the rate of change of a point moving along a path, providing insight into direction and speed. For a parametric curve defined by \( \mathbf{r}(t) = g(t) \mathbf{i} + h(t) \mathbf{j} + k(t) \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to the parameter \( t \).

Calculating the magnitude of the velocity vector, \( |\mathbf{v}(t)| \), is vital for understanding the path's arc length and setting up line integrals. It represents the curve's instantaneous speed at any point \( t \) and plays a pivotal role in computing the line integral due to its presence in the differential \( ds = |\mathbf{v}(t)| dt \).

To find \( |\mathbf{v}(t)| \), use the formula:

• Differentiate each component function \( g(t), h(t), \) and \( k(t) \) with respect to \( t \).
• Find the square of each derivative.
• Add them together, and then take the square root.

This process reveals both the speed and the orientation of a particle moving along the path, further contributing to the line integral’s evaluation.

Parametric Equations

Parametric equations are a powerful way to describe curves using parameters. In the context of multivariable calculus, they simplify complex functions by representing coordinates \( (x, y, z) \) as functions of a parameter \( t \).

This approach allows for flexibility, particularly when working with line integrals. For example, in the given problem, the path is expressed as \( \mathbf{r}(t) = (\cos 2t) \mathbf{i} + (\sin 2t) \mathbf{j} + t^{5/2} \mathbf{k} \). Here, the parameter \( t \) sweeps through values, forming a 3D curve from start (\( t = 0 \)) to end (\( t = 2\pi \)).

Parametrization is crucial for converting a line integral along a path \( C \) into an integral over an interval. It enables us to use calculus techniques in complex paths that are otherwise hard to integrate directly. The adjustment of \( t \) ensures that every point on the curve is systematically covered, facilitating the evaluation of integral functions or vector fields over specified paths. By working in terms of \( t \), the integrand in the original variable \((x, y, z)\) transforms into one with the parameter \( t \), reflecting any change across the path accurately.