Problem 3 - Essential Calculus Early Transcendentals

Question

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: Differentiate the position function to find velocity

The velocity of a particle is the derivative of its position function with respect to time, i.e., \( \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) \). Given \( \mathbf{r}(t) = 3 \cos t \, \mathbf{i} + 2 \sin t \, \mathbf{j} \), we differentiate each component:- \( \frac{d}{dt} (3 \cos t) = -3 \sin t \)- \( \frac{d}{dt} (2 \sin t) = 2 \cos t \)Thus, the velocity vector is \( \mathbf{v}(t) = -3 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \).

Step 2: Evaluate velocity at specific time

To find the velocity at \( t = \frac{\pi}{3} \), substitute the value into the velocity function:\[ \mathbf{v}\left(\frac{\pi}{3}\right) = -3 \sin\left(\frac{\pi}{3}\right) \, \mathbf{i} + 2 \cos\left(\frac{\pi}{3}\right) \, \mathbf{j} \]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: Differentiate velocity function to find acceleration

The acceleration of a particle is the derivative of its velocity function, i.e., \( \mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) \). Differentiate each component of the velocity function:- \( \frac{d}{dt} (-3 \sin t) = -3 \cos t \)- \( \frac{d}{dt} (2 \cos t) = -2 \sin t \)Thus, the acceleration vector is \( \mathbf{a}(t) = -3 \cos t \, \mathbf{i} - 2 \sin t \, \mathbf{j} \).

Step 4: Evaluate acceleration at specific time

Substitute \( t = \frac{\pi}{3} \) into the acceleration function:\[ \mathbf{a}\left(\frac{\pi}{3}\right) = -3 \cos\left(\frac{\pi}{3}\right) \, \mathbf{i} - 2 \sin\left(\frac{\pi}{3}\right) \, \mathbf{j} \]Using \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \), we have:\[ \mathbf{a}\left(\frac{\pi}{3}\right) = -\frac{3}{2} \, \mathbf{i} - \sqrt{3} \, \mathbf{j} \]

Step 5: Calculate speed of the particle

Speed is the magnitude of the velocity vector. For \( \mathbf{v}(t) = -3 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \), the speed is:\[ \| \mathbf{v}(t) \| = \sqrt{(-3 \sin t)^2 + (2 \cos t)^2} \]Plug \( t = \frac{\pi}{3} \) into the speed formula:\[ \| \mathbf{v}\left(\frac{\pi}{3}\right) \| = \sqrt{\left(-\frac{3\sqrt{3}}{2}\right)^2 + 1^2} = \sqrt{\frac{27}{4} + 1} = \sqrt{\frac{31}{4}} = \frac{\sqrt{31}}{2} \]

Step 6: Sketch particle path and vectors

The path described by \( \mathbf{r}(t)=3 \cos t \, \mathbf{i}+2 \sin t \, \mathbf{j} \) is an ellipse with a semi-major axis of 3 along the x-axis and a semi-minor axis of 2 along the y-axis. At \( t = \frac{\pi}{3} \), plot the point corresponding to \( \mathbf{r}\left(\frac{\pi}{3}\right) = \left( 3 \cos\left(\frac{\pi}{3}\right), 2 \sin\left(\frac{\pi}{3}\right) \right) = \left( \frac{3}{2}, \sqrt{3} \right) \).Draw the velocity vector starting from this point in the direction \( (-\frac{3\sqrt{3}}{2}, 1) \) and the acceleration vector in the direction \( (-\frac{3}{2}, -\sqrt{3}) \).

Key Concepts

Velocity Vector

The velocity vector of a particle gives us insight into how the position of the particle changes over time. In this problem, the particle's position function is given by \( \mathbf{r}(t) = 3 \cos t \, \mathbf{i} + 2 \sin t \, \mathbf{j} \). To find the velocity vector, we differentiate the position function with respect to time \( t \). This involves taking the derivative of each component separately.

• For the component \( 3 \cos t \mathbf{i} \), the derivative is \( -3 \sin t \mathbf{i} \).
• For the component \( 2 \sin t \mathbf{j} \), the derivative is \( 2 \cos t \mathbf{j} \).

Thus, the velocity vector is given by \( \mathbf{v}(t) = -3 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \). This vector shows us how fast and in what direction the particle is moving at any time \( t \). For a specific time, such as \( t = \frac{\pi}{3} \), the velocity vector \( \mathbf{v}\left( \frac{\pi}{3} \right) \) provides the exact movement characteristics at that moment.

Acceleration Vector

The acceleration vector tells us how the velocity of the particle changes over time. Conceptually, it's the derivative of the velocity vector. Starting from the velocity vector \( \mathbf{v}(t) = -3 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} \), we differentiate each component to get:

• The derivative of \( -3 \sin t \, \mathbf{i} \) is \( -3 \cos t \, \mathbf{i} \).
• The derivative of \( 2 \cos t \, \mathbf{j} \) is \( -2 \sin t \, \mathbf{j} \).

Thus, the acceleration vector is \( \mathbf{a}(t) = -3 \cos t \, \mathbf{i} - 2 \sin t \, \mathbf{j} \). It illustrates how the particle's velocity is changing over time. Evaluating this acceleration at a specific point, such as \( t = \frac{\pi}{3} \), gives \( \mathbf{a}\left( \frac{\pi}{3} \right) \), providing the change rate of velocity at that moment.

Particle Path Sketching

To visualize how a particle moves, we sketch its path. For the given position function \( \mathbf{r}(t) = 3 \cos t \, \mathbf{i} + 2 \sin t \, \mathbf{j} \), the path is an ellipse. This is because \( \cos t \) and \( \sin t \) describe a circular motion, and the coefficients (3 and 2) stretch the circle into an ellipse.
Key features of this path include:

• A semi-major axis of length 3 along the x-axis, which is the longest diameter of the ellipse.
• A semi-minor axis of length 2 along the y-axis.

At any given time \( t \), such as \( t = \frac{\pi}{3} \), we can calculate the exact position on this path using \( \mathbf{r}(t) \). These calculations help us determine specific points through which the particle moves, including those where velocity and acceleration vectors are plotted.

Position Function Differentiation

The process of differentiation lies at the heart of finding how things change. In this exercise, differentiating the position function helps to uncover both the velocity and the acceleration of a particle. Differentiation transforms the position function \( \mathbf{r}(t) \), given by \( 3 \cos t \mathbf{i} + 2 \sin t \mathbf{j} \), into the velocity vector first.
Differentiation consists of:

• Computing the derivative of the \( 3 \cos t \mathbf{i} \) to get \( -3 \sin t \mathbf{i} \).
• Computing the derivative of the \( 2 \sin t \mathbf{j} \) to get \( 2 \cos t \mathbf{j} \).

These derivatives form the velocity vector \( \mathbf{v}(t) \). Further differentiation of \( \mathbf{v}(t) \) provides the acceleration vector. In essence, each step of differentiation unveils the next layer of motion attributes — speed, direction, changes, everything needed to understand the particle's journey.