Problem 5 - Thomas Calculus

Question

Find the velocity and acceleration vectors in terms of \(\mathbf{u}_{r}\) and \(\mathbf{u}_{\theta} .\) \begin{equation}r=2 \cos 4 t \quad \text { and } \quad \theta=2 t\end{equation}

Step-by-Step Solution

Answer

Velocity: \(-8\sin(4t)\mathbf{u}_r + 4\cos(4t)\mathbf{u}_\theta\), acceleration: \(-40\cos(4t)\mathbf{u}_r - 32\sin(4t)\mathbf{u}_\theta\)."

Step 1: Identify the Parametric Equations

Given the functions for polar coordinates:\[ r = 2 \cos(4t) \text{ and } \theta = 2t \] These equations describe the position of a particle in polar coordinates as functions of time \(t\).

Step 2: Convert to Cartesian Coordinates

Using the polar to Cartesian coordinate transformations, \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), we can find:\[ x = 2 \cos(4t) \cos(2t) \]\[ y = 2 \cos(4t) \sin(2t) \].

Step 3: Differentiate the Position Functions

To find the velocity vector, differentiate the position functions \(r\) and \(\theta\) with respect to time \(t\):For \(r(t) = 2\cos(4t)\):\[ \frac{dr}{dt} = -8\sin(4t) \]For \(\theta(t) = 2t\):\[ \frac{d\theta}{dt} = 2 \].

Step 4: Use Polar Coordinate Formulas for Velocity

The velocity in polar coordinates is given by:\[ \mathbf{v} = \frac{dr}{dt}\mathbf{u}_r + r\frac{d\theta}{dt}\mathbf{u}_\theta \]Substitute the derivatives and original functions:\[ \mathbf{v} = (-8\sin(4t))\mathbf{u}_r + (2\cos(4t) \times 2)\mathbf{u}_\theta \]\[ \mathbf{v} = -8\sin(4t) \mathbf{u}_r + 4\cos(4t)\mathbf{u}_\theta \].

Step 5: Find the Acceleration Vector

The acceleration in polar coordinates is given by:\[ \mathbf{a} = \left(\frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 \right)\mathbf{u}_r + \left(r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt}\right)\mathbf{u}_\theta \]Compute the second derivatives:\[ \frac{d^2r}{dt^2} = -32\cos(4t) \]\[ \frac{d^2\theta}{dt^2} = 0 \]Substitute these into the acceleration formula:\[ \mathbf{a} = \left(-32\cos(4t) - 2\cos(4t)(2)^2 \right)\mathbf{u}_r + \left(0 + 2(-8\sin(4t))\times 2\right)\mathbf{u}_\theta \]\[ \mathbf{a} = (-32\cos(4t) - 8\cos(4t))\mathbf{u}_r - 32\sin(4t)\mathbf{u}_\theta \]\[ \mathbf{a} = -40\cos(4t)\mathbf{u}_r - 32\sin(4t)\mathbf{u}_\theta \].

Key Concepts

Velocity Vector

In the world of physics, the velocity vector is a crucial concept to understand. It describes the direction and speed of a moving object. When dealing with polar coordinates, the velocity vector has two components – one in the direction of the radial unit vector \( \mathbf{u}_r \) and one in the direction of the angular unit vector \( \mathbf{u}_{\theta} \). This makes polar coordinates particularly useful for problems with cyclic or rotational movement, where these directions change as the object moves.

To find the velocity vector in polar coordinates, you need to calculate the rate of change of both the radial and angular positions over time. This involves taking the derivatives of the given parametric equations \( r(t) \) and \( \theta(t) \) with respect to time \( t \). In the example provided, the derivatives are:

• \( \frac{dr}{dt} = -8\sin(4t) \) for the radial component
• \( \frac{d\theta}{dt} = 2 \) for the angular component

These derivatives are then inserted into the formula for the velocity vector, \( \mathbf{v} = \frac{dr}{dt}\mathbf{u}_r + r\frac{d\theta}{dt}\mathbf{u}_{\theta} \). For the given parametric equations, this results in the velocity \( \mathbf{v} = -8\sin(4t) \mathbf{u}_r + 4\cos(4t) \mathbf{u}_{\theta} \). Understanding this derivation helps in visualizing how the speed and path of the object vary over time.

Acceleration Vector

The acceleration vector extends the concept of velocity by describing how it changes over time. In polar coordinates, acceleration has two main components much like velocity: radial acceleration, which is along the radial vector \( \mathbf{u}_r \), and angular acceleration, which is along the angular vector \( \mathbf{u}_{\theta} \). Computing the acceleration vector involves taking the second derivatives of \( r(t) \) and \( \theta(t) \), as well as utilizing the first derivatives obtained for velocity.

The formula to find the acceleration vector \( \mathbf{a} \) in polar coordinates is:

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

For the given exercise, the second derivatives are calculated as:

• \( \frac{d^2r}{dt^2} = -32\cos(4t) \)
• \( \frac{d^2\theta}{dt^2} = 0 \)

By substituting these into the equation, the acceleration vector obtained is \( \mathbf{a} = -40\cos(4t)\mathbf{u}_r - 32\sin(4t)\mathbf{u}_{\theta} \). This vector gives us a clear picture of how the velocity of the particle itself is changing.

Parametric Equations

Parametric equations are powerful tools in expressing the path of a particle through different coordinate systems. They define the coordinates of a point in space as functions of a variable, typically time \( t \). In the context of the original problem, we have the parametric equations \( r = 2 \cos(4t) \) and \( \theta = 2t \), which position the particle in polar coordinates as functions of \( t \).

These equations allow you to track the particle's motion as it spirals out, because \( r \) varies with \( \cos(4t) \), producing an oscillating radial distance. Meanwhile, the angular component \( \theta = 2t \) increases linearly with time, showing constant rotational advancement. Parametric equations like these are not just abstract procedures, but useful for applications where understanding how each coordinate behaves separately over time is crucial.

One practical step to better understand these equations is to convert them to Cartesian coordinates, using the relations:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

In this case, these transformations provide a clear visual representation of the particle's path in the familiar Cartesian plane, making it easier to interpret the motion dynamics and analyze the entire movement story.