Problem 5 - Calculus Early Transcendentals

Question

\(3-8\) 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 \(t .\) $$\mathbf{r}(t)=3 \cos t \mathbf{i}+2 \sin t \mathbf{j}, \quad t=\pi / 3$$

Step-by-Step Solution

Answer

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 1: 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:\[ \mathbf{v}(t) = \frac{d}{dt} (3 \cos t) \mathbf{i} + \frac{d}{dt} (2 \sin t) \mathbf{j} = -3 \sin t \mathbf{i} + 2 \cos t \mathbf{j}. \]

Step 2: 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:\[ \mathbf{v}\left(\frac{\pi}{3}\right) = -\frac{3\sqrt{3}}{2} \mathbf{i} + 1 \mathbf{j}. \]

Step 3: 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:\[ \mathbf{a}(t) = \frac{d}{dt}(-3 \sin t \mathbf{i} + 2 \cos t \mathbf{j}) = -3 \cos t \mathbf{i} - 2 \sin t \mathbf{j}. \]

Step 4: 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:\[ \mathbf{a}\left(\frac{\pi}{3}\right) = -\frac{3}{2} \mathbf{i} - \sqrt{3} \mathbf{j}. \]

Step 5: Find the speed of the particle

The speed of the particle is the magnitude of the velocity vector. Calculate the magnitude of \( \mathbf{v}\left(\frac{\pi}{3}\right) = -\frac{3\sqrt{3}}{2} \mathbf{i} + 1 \mathbf{j} \):\[ \text{Speed} = \sqrt{\left(-\frac{3\sqrt{3}}{2}\right)^2 + 1^2} = \sqrt{\frac{27}{4} + 1} = \sqrt{\frac{31}{4}} = \frac{\sqrt{31}}{2}. \]

Step 6: 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.

Key Concepts

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 \( 2 \cos t \).

This results in the velocity vector \( \mathbf{v}(t) = -3 \sin t \mathbf{i} + 2 \cos t \mathbf{j} \). Evaluating this at \( t = \frac{\pi}{3} \), we find \( \mathbf{v}\left(\frac{\pi}{3}\right) = -\frac{3\sqrt{3}}{2} \mathbf{i} + 1 \mathbf{j} \). In essence, the velocity vector tells us the speed and direction of the particle's motion at a specific instant in time.

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 \( -2 \sin t \).

The resulting acceleration vector is \( \mathbf{a}(t) = -3 \cos t \mathbf{i} - 2 \sin t \mathbf{j} \). At \( t = \frac{\pi}{3} \), the evaluated vector is \( \mathbf{a}\left(\frac{\pi}{3}\right) = -\frac{3}{2} \mathbf{i} - \sqrt{3} \mathbf{j} \). This provides insight into how the speed and direction of the particle are changing at a given moment. While velocity measures movement, acceleration measures changes in that movement.

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.

At \( t = \frac{\pi}{3} \), the particle's position is \( \mathbf{r}\left(\frac{\pi}{3}\right) = \frac{3}{2} \mathbf{i} + \sqrt{3} \mathbf{j} \), marking its location on the ellipse. Drawing the velocity and acceleration vectors from this point helps interpret not just how the particle moves, but clear visual guidance on its changing motion. Through vectors, we grasp both speed direction and how they are modifying over time.