Problem 7 - Thomas Calculus

Question

Give the position vectors of particles moving along various curves in the \(x y\) -plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectors on the curve. Motion on the cycloid \(x=t-\sin t, \quad y=1-\cos t\) \begin{equation} \mathbf{r}(t)=(t-\sin t) \mathbf{i}+(1-\cos t) \mathbf{j} ; \quad t=\pi \text { and } 3 \pi / 2 \end{equation}

Step-by-Step Solution

Answer

At \( t=\pi \), velocity is \( 2\mathbf{i} \) and acceleration is \(-\mathbf{j} \). At \( t=\frac{3\pi}{2} \), velocity is \( \mathbf{i} - \mathbf{j} \) and acceleration is \(-\mathbf{i} \).

Step 1: Position Vector

The position vector of the particle, \( \mathbf{r}(t) \), is given by \( x(t) = t - \sin t \) and \( y(t) = 1 - \cos t \). So, the position vector can be represented as: \[ \mathbf{r}(t) = (t - \sin t)\mathbf{i} + (1 - \cos t)\mathbf{j}. \]

Step 2: Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the first derivative of the position vector with respect to time, \( t \). This means we need to differentiate each component of \( \mathbf{r}(t) \):\( v_x(t) = \frac{d}{dt}(t - \sin t) = 1 - \cos t \), and\( v_y(t) = \frac{d}{dt}(1 - \cos t) = \sin t \).Thus, the velocity vector is:\[ \mathbf{v}(t) = (1 - \cos t) \mathbf{i} + \sin t \mathbf{j}. \]

Step 3: Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector, \( \mathbf{v}(t) \):\( a_x(t) = \frac{d}{dt}(1 - \cos t) = \sin t \), and\( a_y(t) = \frac{d}{dt}(\sin t) = \cos t \).Thus, the acceleration vector is:\[ \mathbf{a}(t) = \sin t \mathbf{i} + \cos t \mathbf{j}. \]

Step 4: Compute at \(t = \pi\)

Plug \( t = \pi \) into our velocity and acceleration vectors:For velocity:\[ \mathbf{v}(\pi) = (1 - \cos \pi) \mathbf{i} + \sin \pi \mathbf{j} = (1 + 1) \mathbf{i} + 0 \mathbf{j} = 2 \mathbf{i}. \]For acceleration:\[ \mathbf{a}(\pi) = \sin \pi \mathbf{i} + \cos \pi \mathbf{j} = 0 \mathbf{i} - 1 \mathbf{j} = -\mathbf{j}. \]

Step 5: Compute at \(t = \frac{3\pi}{2}\)

Plug \( t = \frac{3\pi}{2} \) into our velocity and acceleration vectors:For velocity:\[ \mathbf{v}\left(\frac{3\pi}{2}\right) = (1 - \cos \frac{3\pi}{2}) \mathbf{i} + \sin \frac{3\pi}{2} \mathbf{j} = (1 - 0) \mathbf{i} - 1 \mathbf{j} = \mathbf{i} - \mathbf{j}. \]For acceleration:\[ \mathbf{a}\left(\frac{3\pi}{2}\right) = \sin \frac{3\pi}{2} \mathbf{i} + \cos \frac{3\pi}{2} \mathbf{j} = -1 \mathbf{i} + 0 \mathbf{j} = -\mathbf{i}. \]

Step 6: Sketch the Vectors

On the coordinate plane, plot the position of the particle at times \( t = \pi \) and \( t = \frac{3\pi}{2} \), and then draw vectors for velocity and acceleration:At \( t = \pi \), position is \((\pi, 2)\) and vectors are \(2\mathbf{i}\) and \(-\mathbf{j}\).At \( t = \frac{3\pi}{2} \), position is \(\left(\frac{3\pi}{2}, 2\right)\) and vectors are \(\mathbf{i} - \mathbf{j}\) and \(-\mathbf{i}\).

Key Concepts

Position Vector

The position vector is a fundamental concept in vector calculus used to describe the position of a point in space relative to an origin. In the context of this exercise, the position vector is denoted by \( \mathbf{r}(t) \) and is expressed as a function of time \( t \). It is given by two components:

• The \( x \)-component: \( x(t) = t - \sin t \).
• The \( y \)-component: \( y(t) = 1 - \cos t \).

This represents a parametric equation of motion for a particle moving along a cycloid. The position vector \( \mathbf{r}(t) = (t - \sin t)\mathbf{i} + (1 - \cos t)\mathbf{j} \) is crucial for deriving other vectors related to motion, such as velocity and acceleration.

Velocity Vector

Velocity is a key concept when analyzing motion. It describes how the position of a particle changes over time. Mathematically, it is defined as the first derivative of the position vector with respect to time.
For the problem at hand, this means differentiating \( \mathbf{r}(t) \):

• The \( x \)-component of velocity: \( v_x(t) = \frac{d}{dt}(t - \sin t) = 1 - \cos t \).
• The \( y \)-component of velocity: \( v_y(t) = \frac{d}{dt}(1 - \cos t) = \sin t \).

Thus, the velocity vector \( \mathbf{v}(t) \) becomes \( \mathbf{v}(t) = (1 - \cos t) \mathbf{i} + \sin t \mathbf{j} \). This vector provides both direction and magnitude of motion at any time \( t \). It is an essential tool to understand how fast and in which direction a particle is moving.

Acceleration Vector

Acceleration is a measure of how velocity changes over time. It is an important part of understanding the dynamics of motion. Acceleration is the derivative of the velocity vector with respect to time.
In this exercise, to find the acceleration vector \( \mathbf{a}(t) \), we differentiate the velocity vector:

• For the \( x \)-component: \( a_x(t) = \frac{d}{dt}(1 - \cos t) = \sin t \).
• For the \( y \)-component: \( a_y(t) = \frac{d}{dt}(\sin t) = \cos t \).

So, the acceleration vector is \( \mathbf{a}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} \). This vector tells us the rate and direction at which velocity is changing. This is fundamental in predicting how a particle will accelerate under varying conditions.

Derivative in Calculus

The derivative is a cornerstone concept in calculus, representing the rate of change of a function with respect to a variable. It is essential in vector calculus for determining velocity and acceleration of moving particles. In our context:

• The derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \) yields the velocity vector \( \mathbf{v}(t) \).
• Consequently, differentiating the velocity vector gives us the acceleration vector \( \mathbf{a}(t) \).

Derivatives give us a mathematical way to understand how a system evolves over time. They show us not just where a particle is, but how it's changing its position and speed. Understanding derivatives helps us grasp the deeper mechanics behind motion, such as predicting future states of motion in dynamic systems.