Problem 5 - Thomas Calculus

Question

In Exercises \(1-7,\) find the velocity and acceleration vectors in terms of \(\mathbf{u}_{r}\) and \(\mathbf{u}_{\theta} .\) $$ r=e^{a \theta} \quad \text { and } \quad \frac{d \theta}{d t}=2 $$

Step-by-Step Solution

Answer

Velocity: \(2ae^{a\theta} \mathbf{u}_r + 2e^{a\theta} \mathbf{u}_{\theta}\); Acceleration: \((4a^2e^{a\theta} - 4e^{a\theta}) \mathbf{u}_r + (4ae^{a\theta} + 4e^{a\theta}) \mathbf{u}_{\theta})\).

Step 1: Identify Known Variables

We know that \( r = e^{a \theta} \) and \( \frac{d \theta}{d t} = 2 \). Here, \( \theta \) is a function of \( t \), so \( r \) is also a function of \( t \).

Step 2: Find \( \frac{dr}{dt} \) (Rate of Change of \( r \))

To find the rate of change of \( r \), use the chain rule:\[ \frac{dr}{dt} = \frac{dr}{d\theta} \cdot \frac{d\theta}{dt} \]First, find \( \frac{dr}{d\theta} = \frac{d}{d\theta}e^{a\theta} = ae^{a\theta} \).Then, substitute \( \frac{d\theta}{dt} = 2 \):\[ \frac{dr}{dt} = ae^{a\theta} \cdot 2 = 2ae^{a\theta} \]

Step 3: Express Velocity Vector \( \mathbf{v} \)

The velocity vector \( \mathbf{v} \) in polar coordinates is given by:\[ \mathbf{v} = \frac{dr}{dt} \mathbf{u}_r + r\frac{d\theta}{dt} \mathbf{u}_{\theta} \]Substitute the known values:\[ \mathbf{v} = 2ae^{a\theta} \mathbf{u}_r + e^{a\theta} \cdot 2 \mathbf{u}_{\theta} \]This simplifies to:\[ \mathbf{v} = 2ae^{a\theta} \mathbf{u}_r + 2e^{a\theta} \mathbf{u}_{\theta} \]

Step 4: Find \( \frac{d^2 r}{dt^2} \) (Second Derivative of \( r \))

Differentiate \( \frac{dr}{dt} = 2ae^{a\theta} \) again using the chain rule:\[ \frac{d^2r}{dt^2} = \frac{d}{dt}(2ae^{a\theta}) = 2a \cdot \frac{d}{dt}(e^{a\theta}) = 2a \cdot ae^{a\theta} \cdot 2 = 4a^2e^{a\theta} \]

Step 5: Express Acceleration Vector \( \mathbf{a} \)

The acceleration vector \( \mathbf{a} \) in polar coordinates is given by:\[ \mathbf{a} = \left( \frac{d^2r}{dt^2} - r \left(\frac{d\theta}{dt}\right)^2 \right) \mathbf{u}_r + \left( \frac{d}{dt}(r\frac{d\theta}{dt}) + r\frac{d\theta}{dt} \right) \mathbf{u}_{\theta} \]Substitute the values:\[ \mathbf{a} = \left( 4a^2e^{a\theta} - e^{a\theta} \cdot 4 \right) \mathbf{u}_r + \left( \frac{d}{dt}(2e^{a\theta}) + 2e^{a\theta} \cdot 2 \right) \mathbf{u}_{\theta} \]Compute each term:- \( \frac{d}{dt}(2e^{a\theta}) = 4ae^{a\theta} \) (from Step 2)So,\[ \mathbf{a} = (4a^2e^{a\theta} - 4e^{a\theta}) \mathbf{u}_r + (4ae^{a\theta} + 4e^{a\theta}) \mathbf{u}_{\theta} \]

Key Concepts

Velocity Vector

In polar coordinates, the velocity vector is crucial when describing the movement of a particle or object in terms of its radial and angular components. The radial component represents the rate at which the distance from the origin changes, while the angular component represents how quickly the object is moving around the origin.
For a given function, such as the one in the exercise where \( r = e^{a \theta} \) and \( \frac{d \theta}{dt} = 2 \), we first find the rate of change of \( r \) with respect to time. Using the chain rule, \( \frac{dr}{dt} = 2ae^{a\theta} \).
This forms the first part of the velocity vector: \( 2ae^{a\theta} \mathbf{u}_r \).
The complete velocity vector in polar coordinates is expressed as:

• \( \mathbf{v} = \frac{dr}{dt} \mathbf{u}_r + r\frac{d\theta}{dt} \mathbf{u}_{\theta} \)

Substitute \( \frac{dr}{dt} \) and \( r\frac{d\theta}{dt} = 2e^{a\theta} \), then:

• \( \mathbf{v} = 2ae^{a\theta} \mathbf{u}_r + 2e^{a\theta} \mathbf{u}_{\theta} \)

This vector tells us how fast and in what direction the particle is moving at any point in its path based on \( \mathbf{u}_r \) and \( \mathbf{u}_{\theta} \).

Understanding the velocity vector is key as it reflects the combined effect of both moving closer or farther from the origin and moving along the curve or path.

Acceleration Vector

The acceleration vector in polar coordinates extends the concept of velocity by showing how the velocity of an object changes over time. Like velocity, it breaks down into radial and angular components, helping us understand the dynamics more deeply.
To derive it, we first need the second derivative of \( r \), so we differentiate \( \frac{dr}{dt} = 2ae^{a\theta} \) using the chain rule, resulting in \( \frac{d^2r}{dt^2} = 4a^2e^{a\theta} \).
The acceleration vector formula in polar coordinates is:

• \( \mathbf{a} = \left( \frac{d^2r}{dt^2} - r \left(\frac{d\theta}{dt}\right)^2 \right) \mathbf{u}_r + \left( \frac{d}{dt}(r\frac{d\theta}{dt}) + r\frac{d\theta}{dt} \right) \mathbf{u}_{\theta} \)

Plug in the known values:
- The radial component: \( 4a^2e^{a\theta} - 4e^{a\theta} \)
- The angular component: \( 4ae^{a\theta} + 4e^{a\theta} \)
Which simplifies the vector to:

• \( \mathbf{a} = (4a^2e^{a\theta} - 4e^{a\theta}) \mathbf{u}_r + (4ae^{a\theta} + 4e^{a\theta}) \mathbf{u}_{\theta} \)

This shows how quickly the particle’s speed and direction change, broken down into radial changes (towards or away from the origin) and angular changes (around the origin).

The acceleration vector thus provides a complete picture of the changes in motion dynamics.

Chain Rule

The chain rule is a fundamental technique in calculus that helps us differentiate composite functions. It is especially useful in polar coordinates when variables are functions of other variables, like time.
In our exercise, \( r = e^{a\theta} \) and \( \theta \) varies with time. This means \( r \) itself changes over time according to both \( \theta \) and its own relation to time (\( t \)).
The chain rule allows us to find \( \frac{dr}{dt} \), the derivative of \( r \) with respect to time, as follows:

• First, differentiate \( r \) with respect to \( \theta \): \( \frac{dr}{d\theta} = ae^{a\theta} \)

Then, multiply by \( \frac{d\theta}{dt} \), the rate of change of \( \theta \) with respect to time:

• \( \frac{dr}{dt} = ae^{a\theta} \times 2 = 2ae^{a\theta} \)

This principle is crucial in identifying both the velocity and acceleration vectors in polar coordinates. It helps combine individual rates of change into a single expression, portraying how complex systems evolve over time.

Understanding the chain rule enhances fluency in working with polar functions and enables accurate analysis of objects in motion.