Chapter 13: Problem 5
\(3-8\) Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for the specified value of \(t .\) $$\mathbf{r}(t)=3 \cos t \mathbf{i}+2 \sin t \mathbf{j}, \quad t=\pi / 3$$
Short Answer
Expert verified
Velocity: \(-\frac{3\sqrt{3}}{2} \mathbf{i} + 1 \mathbf{j}\), Acceleration: \(-\frac{3}{2} \mathbf{i} - \sqrt{3} \mathbf{j}\), Speed: \(\frac{\sqrt{31}}{2}\).
Step by step solution
01
Find the velocity vector
The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Given \( \mathbf{r}(t) = 3 \cos t \mathbf{i} + 2 \sin t \mathbf{j} \), differentiate each component:\[ \mathbf{v}(t) = \frac{d}{dt} (3 \cos t) \mathbf{i} + \frac{d}{dt} (2 \sin t) \mathbf{j} = -3 \sin t \mathbf{i} + 2 \cos t \mathbf{j}. \]
02
Evaluate the velocity vector at the specified time
Substitute \( t = \frac{\pi}{3} \) into the velocity vector \( \mathbf{v}(t) = -3 \sin t \mathbf{i} + 2 \cos t \mathbf{j} \):\[ \mathbf{v}\left(\frac{\pi}{3}\right) = -3 \sin\left(\frac{\pi}{3}\right) \mathbf{i} + 2 \cos\left(\frac{\pi}{3}\right) \mathbf{j}. \] Since \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \) and \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \), we have:\[ \mathbf{v}\left(\frac{\pi}{3}\right) = -\frac{3\sqrt{3}}{2} \mathbf{i} + 1 \mathbf{j}. \]
03
Find the acceleration vector
The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). So, differentiate the velocity vector:\[ \mathbf{a}(t) = \frac{d}{dt}(-3 \sin t \mathbf{i} + 2 \cos t \mathbf{j}) = -3 \cos t \mathbf{i} - 2 \sin t \mathbf{j}. \]
04
Evaluate the acceleration vector at the specified time
Substitute \( t = \frac{\pi}{3} \) into the acceleration vector \( \mathbf{a}(t) = -3 \cos t \mathbf{i} - 2 \sin t \mathbf{j} \):\[ \mathbf{a}\left(\frac{\pi}{3}\right) = -3 \cos\left(\frac{\pi}{3}\right) \mathbf{i} - 2 \sin\left(\frac{\pi}{3}\right) \mathbf{j}. \] Using \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \), we find:\[ \mathbf{a}\left(\frac{\pi}{3}\right) = -\frac{3}{2} \mathbf{i} - \sqrt{3} \mathbf{j}. \]
05
Find the speed of the particle
The speed of the particle is the magnitude of the velocity vector. Calculate the magnitude of \( \mathbf{v}\left(\frac{\pi}{3}\right) = -\frac{3\sqrt{3}}{2} \mathbf{i} + 1 \mathbf{j} \):\[ \text{Speed} = \sqrt{\left(-\frac{3\sqrt{3}}{2}\right)^2 + 1^2} = \sqrt{\frac{27}{4} + 1} = \sqrt{\frac{31}{4}} = \frac{\sqrt{31}}{2}. \]
06
Sketch the path, velocity, and acceleration vectors
The path given by \( \mathbf{r}(t) = 3 \cos t \mathbf{i} + 2 \sin t \mathbf{j} \) traces an ellipse centered at the origin with a semi-major axis of 3 along the x-direction and a semi-minor axis of 2 along the y-direction.At \( t = \frac{\pi}{3} \), the position of the particle is \( \mathbf{r}\left(\frac{\pi}{3}\right) = 3 \cos\left(\frac{\pi}{3}\right) \mathbf{i} + 2 \sin\left(\frac{\pi}{3}\right) \mathbf{j} = \frac{3}{2} \mathbf{i} + \sqrt{3} \mathbf{j} \).Draw the velocity vector \( -\frac{3\sqrt{3}}{2} \mathbf{i} + 1 \mathbf{j} \) and the acceleration vector \( -\frac{3}{2} \mathbf{i} - \sqrt{3} \mathbf{j} \) originating from this position.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Velocity
When we think about a moving particle, its velocity tells us how fast it’s moving and in what direction. In terms of parametric equations, the velocity is represented as a vector. This vector is the derivative of the position vector with respect to time. For a given position vector, \( \mathbf{r}(t) = 3 \cos t \mathbf{i} + 2 \sin t \mathbf{j} \), the velocity vector is calculated by differentiating each component with respect to time \( t \).
- Differentiate \( 3 \cos t \) to get \( -3 \sin t \).
- Differentiate \( 2 \sin t \) to get \( 2 \cos t \).
Acceleration
Acceleration indicates how the velocity of a particle is changing over time. In the context of parametric equations, it is the derivative of the velocity vector. For the velocity vector \( \mathbf{v}(t) = -3 \sin t \mathbf{i} + 2 \cos t \mathbf{j} \), we find the acceleration vector by differentiating each component again.
- Differentiate \( -3 \sin t \) to get \( -3 \cos t \).
- Differentiate \( 2 \cos t \) to get \( -2 \sin t \).
Particle Path
The path of a particle offers a visual representation of its trajectory over time. In the example given by the position function \( \mathbf{r}(t) = 3 \cos t \mathbf{i} + 2 \sin t \mathbf{j} \), the path is an ellipse. This means the particle moves in a loop-like motion, traced out by the parametric equations.
- The term \( 3 \cos t \) suggests an x-component of the path with a maximum stretch of 3 units, which is the semi-major axis.
- The term \( 2 \sin t \) indicates a y-component stretch of 2 units, establishing the semi-minor axis.