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Magazine Chapter 13: Problem 7
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 vectorson the curve. $$\mathbf{r}(t)=(t-\sin t) \mathbf{i}+(1-\cos t) \mathbf{j} ; \quad t=\pi \text { and } 3 \pi / 2$$
Short Answer
Expert verified
The velocity at \( t=\pi \) is \( 2\mathbf{i} \), acceleration is \( -\mathbf{j} \); at \( t=3\pi/2 \), velocity is \( \mathbf{i}-\mathbf{j} \), acceleration is \( -\mathbf{i} \).
Step by step solution
01
Write the position vector
The given position vector for the particle is \( \mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j} \). This represents the path the particle takes in the \( xy \)-plane.
02
Find the velocity vector
The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector with respect to time \( t \). Compute: \[ \mathbf{v}(t) = \frac{d}{dt} [(t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j}] = (1 - \cos t) \mathbf{i} + (\sin t) \mathbf{j} \]
03
Find the acceleration vector
The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector with respect to time \( t \). Compute: \[ \mathbf{a}(t) = \frac{d}{dt} [(1 - \cos t) \mathbf{i} + (\sin t) \mathbf{j}] = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} \]
04
Calculate velocity at \( t = \pi \)
Substitute \( t = \pi \) into the velocity function: \[ \mathbf{v}(\pi) = (1 - \cos \pi) \mathbf{i} + (\sin \pi) \mathbf{j} = (1 - (-1)) \mathbf{i} + (0) \mathbf{j} = 2 \mathbf{i} \]
05
Calculate acceleration at \( t = \pi \)
Substitute \( t = \pi \) into the acceleration function: \[ \mathbf{a}(\pi) = (\sin \pi) \mathbf{i} + (\cos \pi) \mathbf{j} = (0) \mathbf{i} + (-1) \mathbf{j} = - \mathbf{j} \]
06
Calculate velocity at \( t = 3\pi/2 \)
Substitute \( t = 3\pi/2 \) into the velocity function: \[ \mathbf{v}(3\pi/2) = (1 - \cos 3\pi/2) \mathbf{i} + (\sin 3\pi/2) \mathbf{j} = (1 - 0) \mathbf{i} + (-1) \mathbf{j} = \mathbf{i} - \mathbf{j} \]
07
Calculate acceleration at \( t = 3\pi/2 \)
Substitute \( t = 3\pi/2 \) into the acceleration function: \[ \mathbf{a}(3\pi/2) = (\sin 3\pi/2) \mathbf{i} + (\cos 3\pi/2) \mathbf{j} = (-1) \mathbf{i} + (0) \mathbf{j} = - \mathbf{i} \]
08
Sketch the velocity and acceleration vectors
Draw the curve described by \( \mathbf{r}(t) \) and represent the vectors \( \mathbf{v}(\pi) = 2 \mathbf{i} \) and \( \mathbf{a}(\pi) = - \mathbf{j} \), and \( \mathbf{v}(3\pi/2) = \mathbf{i} - \mathbf{j} \) and \( \mathbf{a}(3\pi/2) = - \mathbf{i} \) as arrows originating from their respective points on the curve. These vectors demonstrate the direction and magnitude of velocity and acceleration at \( t = \pi \) and \( t = 3\pi/2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Position Vector
A position vector is a crucial concept in vector calculus, representing the exact position of a particle in space relative to an origin. In two-dimensional space like the xy-plane, it typically has two components.
For example, given the exercise with the position vector \( \mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j} \), it describes how a particle traverses through the plane over time \( t \).
The position vector provides all the necessary information about the particle's location, effectively tracing a path on the plane.
For example, given the exercise with the position vector \( \mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j} \), it describes how a particle traverses through the plane over time \( t \).
The position vector provides all the necessary information about the particle's location, effectively tracing a path on the plane.
- Understanding how a position vector works is fundamental for predicting the motion of particles.
- It鈥檚 represented as \( \mathbf{r}(t) \), with i and j denoting the unit vectors along the x and y axes respectively.
Velocity Vector
The velocity vector is a derivative of the position vector concerning time, which tells us not only where the particle is but also how fast and in which direction it moves.
In the solution provided, the velocity vector is derived as \( \mathbf{v}(t) = (1 - \cos t) \mathbf{i} + (\sin t) \mathbf{j} \).
This vector provides critical insights into the speed and path of the particle as it moves.
In the solution provided, the velocity vector is derived as \( \mathbf{v}(t) = (1 - \cos t) \mathbf{i} + (\sin t) \mathbf{j} \).
This vector provides critical insights into the speed and path of the particle as it moves.
- Velocity vectors show the change in the position vector over small time intervals.
- They help in visualizing the instantaneous rate of change at any given moment \( t \).
Acceleration Vector
Acceleration vectors derive from velocity vectors and indicate how the velocity of a particle changes over time. They鈥檙e all about change, offering a deeper understanding of motion beyond just speed.
In this problem, the acceleration vector is calculated as \( \mathbf{a}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} \).
The acceleration provides clues about the forces acting on the particle.
In this problem, the acceleration vector is calculated as \( \mathbf{a}(t) = (\sin t) \mathbf{i} + (\cos t) \mathbf{j} \).
The acceleration provides clues about the forces acting on the particle.
- Acceleration vectors show how velocity itself changes with time, essential for understanding dynamic motion.
- They can reveal information about forces and energy changes in the system.
Derivative
In calculus, derivatives are essential for understanding changes and are used to calculate velocity and acceleration from positions.
A derivative of a function, like that of the position vector, informs us about its rate of change. For example, deriving the position vector \( \mathbf{r}(t) \) gives the velocity vector \( \mathbf{v}(t) \).
Derivatives play a vital role in physics and engineering, helping to model and solve dynamic systems.
A derivative of a function, like that of the position vector, informs us about its rate of change. For example, deriving the position vector \( \mathbf{r}(t) \) gives the velocity vector \( \mathbf{v}(t) \).
Derivatives play a vital role in physics and engineering, helping to model and solve dynamic systems.
- Calculating derivatives is the process of finding the rate at which one quantity changes concerning another.
- They form the backbone of understanding motion dynamics extensively in vector calculus.
