91Ó°ÊÓ

To understand the concept of a velocity vector, it's important to view it as essentially describing how an object's position changes over time. When given a position vector, say \( \mathbf{r}(t) = (\cos 2t) \mathbf{i} + (\sin 2t) \mathbf{j} + t^{5/2} \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) is found by taking the derivative of each of its components with respect to time \( t \).
This means differentiating each function within the components of \( \mathbf{r}(t) \). In mathematical terms, the velocity vector comprises the derivatives of the components \( \cos 2t \), \( \sin 2t \), and \( t^{5/2} \).

• \( v_x(t) = \frac{d}{dt}(\cos 2t) = -2 \sin 2t \)
• \( v_y(t) = \frac{d}{dt}(\sin 2t) = 2 \cos 2t \)
• \( v_z(t) = \frac{d}{dt}(t^{5/2}) = \frac{5}{2} t^{3/2} \)

The assembled result is the velocity vector: \( \mathbf{v}(t) = -2 \sin 2t \mathbf{i} + 2 \cos 2t \mathbf{j} + \frac{5}{2} t^{3/2} \mathbf{k} \). This vector encapsulates the speed and direction of the object's motion.

The magnitude of a velocity vector, \( |\mathbf{v}(t)| \), tells us how fast an object is moving regardless of its direction. You can find it by determining the length of the velocity vector.
This involves calculating the square root of the sum of the squares of its components. Given your velocity vector \( \mathbf{v}(t) \), which consists of the components \( v_x(t) = -2 \sin 2t \), \( v_y(t) = 2 \cos 2t \), and \( v_z(t) = \frac{5}{2} t^{3/2} \), the magnitude is derived as follows:
\[ |\mathbf{v}(t)| = \sqrt{(-2 \sin 2t)^2 + (2 \cos 2t)^2 + \left(\frac{5}{2} t^{3/2}\right)^2} \]
By simplifying, you calculate

• \( (-2 \sin 2t)^2 = 4 \sin^2 2t \) and \( (2 \cos 2t)^2 = 4 \cos^2 2t \)
• The terms \( \sin^2 2t \) and \( \cos^2 2t \) add to 1, since for any angle, \( \sin^2 \theta + \cos^2 \theta = 1 \)
• Thus, \( |\mathbf{v}(t)| = \sqrt{4(1) + \frac{25}{4} t^3} = \sqrt{4 + \frac{25}{4} t^3} \)

Hence, you get a simplified expression for the magnitude, providing clarity on the object's speed.

An important step in evaluating line integrals is establishing the integrand function. This involves expressing the function you're integrating over the path, often in terms of a single parameter, typically \( t \).
For the path given by \( \mathbf{r}(t) \), you want the integrand function \( f(g(t), h(t), k(t)) |\mathbf{v}(t)| \), where

• \( f(x, y, z) = \left(1+\frac{9}{4} z^{1/3}\right)^{1/4} \)

Substitute the path \((g(t), h(t), k(t)) = (\cos 2t, \sin 2t, t^{5/2})\) into the function:
\[ f((\cos 2t), (\sin 2t), t^{5/2}) = \left(1 + \frac{9}{4} (t^{5/2})^{1/3} \right)^{1/4} \]
Combine this with \( |\mathbf{v}(t)| \) from before, leading to the complete integrand:
\[ \left(1 + \frac{9}{4} t^{5/6} \right)^{1/4} \times \sqrt{4 + \frac{25}{4} t^3} \]
This function now depends entirely on \( t \), allowing you to integrate along the path.

A Computer Algebra System (CAS) is a powerful tool used to perform symbolic mathematics on a computer. It's especially useful for evaluating complex expressions that are difficult to solve manually. In the context of line integrals, a CAS can instantly evaluate the integral over specified limits.
In the exercise, you have the line integral \( \int_{0}^{2\pi} \left(1+\frac{9}{4} t^{5/6} \right)^{1/4} \times \sqrt{4 + \frac{25}{4} t^3} \, dt \). This integral can be challenging to solve directly by hand due to its complexity.
Here’s why a CAS is beneficial:

• It automates the handling of intricate algebraic manipulations required during integration.
• It performs exact symbolic integration where possible or provides accurate numerical approximations otherwise.
• It can test different methods or provide step-by-step solutions to improve understanding.

Using a CAS, you apply computational techniques to determine either a precise result or a numerical approximation for your particular integral over that interval from \( t = 0 \) to \( t = 2\pi \). This makes CAS invaluable, especially when dealing with complicated integrand functions like in this exercise.