Problem 65 - Calculus 3

Question

The position vector of a particle is \(\mathbf{r}(t)=5 \sec (2 t) \mathbf{i}-4 \tan (t) \mathbf{j}+7 t^{2} \mathbf{k}\) a. Graph the position function and display a view of the graph that illustrates the asymptotic behavior of the function. b. Find the velocity as \(t\) approaches but is not equal to \(\pi / 4\) (if it exists).

Step-by-Step Solution

Answer

The velocity does not exist at \( t = \pi/4 \) due to divergences in \( \sec(2t) \tan(2t) \) and \( \sec^2(t) \).

Step 1: Understand the Position Function

The position vector is given as \( \mathbf{r}(t) = 5 \sec(2t) \mathbf{i} - 4 \tan(t) \mathbf{j} + 7t^2 \mathbf{k} \). We need to analyze its components to understand their behavior as \( t \) changes, especially near the asymptotes created by \( \sec \) and \( \tan \) functions.

Step 2: Identify Asymptotic Behavior

The asymptotes in \( 5 \sec(2t) \) occur when \( \cos(2t) = 0 \), which corresponds to \( t = \frac{\pi}{4} + \frac{k \pi}{2} \), \( k \in \mathbb{Z} \). Similarly, the asymptotes for \( -4 \tan(t) \) occur when \( \cos(t) = 0 \), that is \( t = \frac{\pi}{2} + k\pi \). These identify potential vertical lines in the graph where the function could go to infinity.

Step 3: Graph the Position Function

To illustrate, graph \( \mathbf{r}(t) \) using a computational tool, showing plots of \( x(t) = 5 \sec(2t) \), \( y(t) = -4 \tan(t) \), and \( z(t) = 7t^2 \). Observe how \( x(t) \) and \( y(t) \) behave near their asymptotes, especially around \( t = \pi/4 \). The behavior should show diverging curves close to these asymptotes.

Step 4: Determine the Velocity Vector

Differentiate the position vector \( \mathbf{r}(t) \) with respect to \( t \) to find the velocity vector \( \mathbf{v}(t) \). We find: \( \frac{d}{dt}[5 \sec(2t) \mathbf{i}] = 10 \sec(2t) \tan(2t) \mathbf{i} \) from chain rule, \( \frac{d}{dt}[-4 \tan(t) \mathbf{j}] = -4 \sec^2(t) \mathbf{j} \), and \( \frac{d}{dt}[7t^2 \mathbf{k}] = 14t \mathbf{k} \). Thus, \( \mathbf{v}(t) = 10 \sec(2t) \tan(2t) \mathbf{i} - 4 \sec^2(t) \mathbf{j} + 14t \mathbf{k} \).

Step 5: Evaluate Velocity Near Asymptote

To find the behavior of \( \mathbf{v}(t) \) as \( t \) approaches \( \frac{\pi}{4} \), substitute \( t = \frac{\pi}{4} \) into \( \mathbf{v}(t) \) for \( t eq \frac{\pi}{4} \). \( \tan(2t) \) diverges as \( t \rightarrow \frac{\pi}{4}, \) thus \( 10 \sec(2t) \tan(2t) \rightarrow \infty \). Additionally, \( \sec^2(t) \rightarrow \infty \) as \( t \rightarrow \frac{\pi}{2} \). Therefore, the velocity vector becomes undefined at \( t=\pi/4 \) due to these infinities.

Key Concepts

Position Vector

In particle motion, the position vector is a way to express the location of a particle in space relative to a reference point, typically the origin. For our problem, the position vector is given by \( \mathbf{r}(t) = 5 \sec(2t) \mathbf{i} - 4 \tan(t) \mathbf{j} + 7t^2 \mathbf{k} \). Each component of this vector represents a function of \( t \), indicating how the position in each spatial dimension changes with time.
Understanding the position vector involves analyzing these functions:

• \( x(t) = 5 \sec(2t) \) shows how the position changes along the x-axis.
• \( y(t) = -4 \tan(t) \) gives the y-component.
• \( z(t) = 7t^2 \) reflects motion along the z-axis.

The vector allows us to visualize the particle's trajectory in three-dimensional space by plugging in different values of \( t \). Each component is essential to predicting the particle's path.

Asymptotic Behavior

Asymptotic behavior in calculus refers to the tendency of a function as it approaches a certain value, often infinity. In the context of our position vector, both \( \sec \) and \( \tan \) functions result in vertical asymptotes, where the function tends toward infinity.
Specifically, asymptotes occur for:

• \( \sec(2t) \) when \( \cos(2t) = 0 \), leading to \( t = \frac{\pi}{4} + \frac{k\pi}{2} \), \( k \in \mathbb{Z} \).
• \( \tan(t) \) when \( \cos(t) = 0 \), or \( t = \frac{\pi}{2} + k\pi \).

These asymptotes are integral to predicting where the behavior of the function could become unbounded, indicating potential breakpoints in the graph of the particle's path. Recognizing these patterns helps in understanding the extreme behaviors that the functions exhibit.

Velocity Vector

The velocity vector of a particle describes its rate of change of position concerning time. It's effectively the derivative of the position vector. For \( \mathbf{r}(t) \), the velocity vector is derived by differentiating each of its components:

• The x-component yields \( 10 \sec(2t) \tan(2t) \), applying the chain rule.
• The y-component results in \( -4 \sec^2(t) \).
• The z-component becomes \( 14t \).

Thus, the velocity vector is expressed as \( \mathbf{v}(t) = 10 \sec(2t) \tan(2t) \mathbf{i} - 4 \sec^2(t) \mathbf{j} + 14t \mathbf{k} \).
This vector provides insight into how fast the particle is moving in each direction and at which points it might be accelerating or decelerating, helping in the analysis of its overall motion.

Differentiation

Differentiation in calculus is the process of finding a derivative, a critical tool for understanding how functions change. In the context of particle motion, differentiation allows us to derive functions like the velocity and acceleration vectors from the position vector.
In our exercise, we differentiate:

• \( 5 \sec(2t) \) to get \( 10 \sec(2t) \tan(2t) \), using the chain rule.
• \( -4 \tan(t) \) to obtain \( -4 \sec^2(t) \).
• \( 7t^2 \) to produce \( 14t \).

These derivatives result in the velocity vector, providing crucial information on the particle's momentum and speed variations over time. Mastery of differentiation is essential for solving many physics problems involving motion.

Trigonometric Functions

Trigonometric functions like \( \sec \) and \( \tan \) play a crucial role in describing periodic behaviors and oscillations, often seen in wave functions and circular motion. Comprehending these functions is essential to understand the position vector in our exercise.
Here’s how each operates:

• \( \sec(\theta) = \frac{1}{\cos(\theta)} \) plays a role in defining the amplitude or scale of the position.
• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \) is important for depicting angles and slopes.

Differences in their values and their manipulation through identities can influence the particle's position and create asymptotes, where the function becomes unbounded. These functions are foundational mathematical tools for analyzing oscillations and repetitive patterns in motion dynamics.