Problem 66 - Calculus 3

Question

Find the velocity and the speed of a particle with the position function \(\mathbf{r}(t)=\left(\frac{2 t-1}{2 t+1}\right) \mathbf{i}+\ln \left(1-4 t^{2}\right) \mathbf{j} .\) The speed of a particle is the magnitude of the velocity and is represented by \(\|\dot{r}(t)\|\) A particle moves on a circular path of radius \(b\) according to the function \(\mathbf{r}(t)=b \cos (\omega t) \mathbf{i}+b \sin (\omega t) \mathbf{j}, \quad\) where \(\omega\) is the angular velocity, d\theta/dt.

Step-by-Step Solution

Answer

Velocity: \( \frac{4}{(2t+1)^2} \mathbf{i} - \frac{8t}{1-4t^2} \mathbf{j} \), Speed: \( \sqrt{\frac{16}{(2t+1)^4} + \frac{64t^2}{(1-4t^2)^2}} \). Circular path speed: \( b\omega \).

Step 1: Differentiate the Position Function

To find the velocity of the particle, we need to compute the derivative of the position function with respect to time, \( t \). The given position function is \( \mathbf{r}(t)=\left(\frac{2 t-1}{2 t+1}\right) \mathbf{i}+\ln \left(1-4 t^{2}\right) \mathbf{j} \).Let's differentiate the \( \mathbf{i} \) component: \( \frac{d}{dt} \left( \frac{2t-1}{2t+1} \right) \). Using the quotient rule, \( \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^2} \) where \( u = 2t-1 \) and \( v = 2t+1 \).Calculate \( \frac{du}{dt} = 2 \) and \( \frac{dv}{dt} = 2 \), so the derivative is \( \frac{(2t+1)(2) - (2t-1)(2)}{(2t+1)^2} = \frac{4}{(2t+1)^2} \).Now, differentiate the \( \mathbf{j} \) component: \( \frac{d}{dt} \ln\left(1-4t^2\right) \). Using the chain rule, the derivative is \( \frac{-8t}{1-4t^2} \).Thus, the velocity function is: \( \dot{\mathbf{r}}(t) = \frac{4}{(2t+1)^2} \mathbf{i} - \frac{8t}{1-4t^2} \mathbf{j} \).

Step 2: Find the Speed of the Particle

The speed of the particle is the magnitude of the velocity vector \( \dot{\mathbf{r}}(t) \). To find this, compute \( \|\dot{\mathbf{r}}(t)\| = \sqrt{\left(\frac{4}{(2t+1)^2}\right)^2 + \left(-\frac{8t}{1-4t^2}\right)^2} \).First, compute \( \left(\frac{4}{(2t+1)^2}\right)^2 = \frac{16}{(2t+1)^4} \).Second, compute \( \left(-\frac{8t}{1-4t^2}\right)^2 = \frac{64t^2}{(1-4t^2)^2} \).Thus, the speed is \( \|\dot{\mathbf{r}}(t)\| = \sqrt{\frac{16}{(2t+1)^4} + \frac{64t^2}{(1-4t^2)^2}} \).This represents the instantaneous speed of the particle at time \( t \).

Step 3: Consider Circular Motion Path

Given another position function for circular motion: \( \mathbf{r}(t)=b \cos (\omega t) \mathbf{i}+b \sin (\omega t) \mathbf{j} \). This is a separate scenario and unrelated to the function discussed above.In this context, the velocity is found by differentiating the circular motion path: \( \dot{\mathbf{r}}(t) = -b\omega \sin(\omega t) \mathbf{i} + b\omega \cos(\omega t) \mathbf{j} \).The speed is \( \|\dot{\mathbf{r}}(t)\| = \sqrt{(-b\omega \sin(\omega t))^2 + (b\omega \cos(\omega t))^2} = b\omega \) because the sum of squares of \( \sin \) and \( \cos \) terms is 1.

Key Concepts

Velocity

When discussing particle motion, **velocity** is a crucial concept indicating both the speed and direction of a particle's movement. If you can calculate the derivative of the position function, you can find the velocity of a particle. In our exercise, the position function is given as \[ \mathbf{r}(t)=\left(\frac{2 t-1}{2 t+1}\right) \mathbf{i}+\ln \left(1-4 t^{2}\right) \mathbf{j} . \]To find the velocity, differentiate the position function with respect to time, \( t \). For the \( \mathbf{i} \) component, use the quotient rule to get:\[ \frac{d}{dt} \left( \frac{2t-1}{2t+1} \right) = \frac{4}{(2t+1)^2} \]And for the \( \mathbf{j} \) component, apply the chain rule to obtain:\[ \frac{d}{dt} \ln\left(1-4t^2\right) = \frac{-8t}{1-4t^2} \]Therefore, the velocity vector, \( \dot{\mathbf{r}}(t) \), is given by:\[ \dot{\mathbf{r}}(t) = \frac{4}{(2t+1)^2} \mathbf{i} - \frac{8t}{1-4t^2} \mathbf{j} \]Velocity tells you how quickly and in which direction the particle is moving along its path at any point in time.

Position Function

A **position function** gives the precise location of a particle at any point in time. It's the foundation of understanding how particles travel in a space. In our case, the function provided is:\[ \mathbf{r}(t)=\left(\frac{2 t-1}{2 t+1}\right) \mathbf{i}+\ln \left(1-4 t^{2}\right) \mathbf{j} \]This function describes the trajectory of the particle as time \( t \) changes. It tells us the position of the particle in a two-dimensional space using the \( \mathbf{i} \) and \( \mathbf{j} \) components:

• The \( \mathbf{i} \)-component, \( \frac{2t-1}{2t+1} \), indicates how far left or right the particle is at time \( t \).
• The \( \mathbf{j} \)-component, \( \ln(1-4t^2) \), tells us the particle's position in the up or down directions.

Understanding the position function is crucial because it sets the stage for finding other motion properties, like velocity and speed.

Circular Motion

In scenarios where a particle moves in a **circular motion**, we describe its path using trigonometric functions for the coordinates. A circular path can be expressed with:\[ \mathbf{r}(t)=b \cos (\omega t) \mathbf{i}+b \sin (\omega t) \mathbf{j} \]Here, \( b \) is the radius of the circle, while \( \omega \) represents the angular velocity, which is how fast the particle rotates through the circle.For circular motion:

• Differentiating this function gives the velocity in circular terms:\[ \dot{\mathbf{r}}(t) = -b\omega \sin(\omega t) \mathbf{i} + b\omega \cos(\omega t) \mathbf{j} \]
• The speed of motion in a circle is determined by its constant value:\[ \|\dot{\mathbf{r}}(t)\| = b\omega \].

This process helps us describe every movement of a particle that stuck on a circular track, which is fundamentally different from linear paths.

Speed

The **speed** of a particle at any instant is the magnitude of its velocity vector. It tells us how fast the particle is moving, regardless of its direction.In our problem, we found the velocity:\[ \dot{\mathbf{r}}(t) = \frac{4}{(2t+1)^2} \mathbf{i} - \frac{8t}{1-4t^2} \mathbf{j} \]To find the speed, calculate the magnitude of this velocity:\[ \|\dot{\mathbf{r}}(t)\| = \sqrt{\left(\frac{4}{(2t+1)^2}\right)^2 + \left(-\frac{8t}{1-4t^2}\right)^2} \]

• This gives us \[ \sqrt{\frac{16}{(2t+1)^4} + \frac{64t^2}{(1-4t^2)^2}} \]
• The result represents the instantaneous speed, indicating how fast the particle is traveling at any specific time \( t \).

Speed is important because it provides a measure of motion intensity without regard for direction.