Problem 49 - Calculus III

Question

A particle travels along the path of an ellipse with the equation \(\mathbf{r}(t)=\cos t \mathbf{i}+2 \sin t \mathbf{j}+0 \mathbf{k}\). Find the following:Velocity of the particle

Step-by-Step Solution

Answer

The velocity of the particle is \( \mathbf{v}(t) = -\sin t \mathbf{i} + 2\cos t \mathbf{j} \).

Step 1: Identify the Velocity Expression

The velocity of a particle is given by the derivative of its position vector with respect to time, that is, the derivative of \( \mathbf{r}(t) \) with respect to \( t \). Hence, the velocity \( \mathbf{v}(t) = \frac{d}{dt}(\mathbf{r}(t)) \).

Step 2: Differentiate the Position Vector

To find \( \mathbf{v}(t) \), differentiate each component of \( \mathbf{r}(t) \). The position vector is \( \mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k} \). Differentiating, we find:- \( \frac{d}{dt}(\cos t) = -\sin t \)- \( \frac{d}{dt}(2\sin t) = 2\cos t \)- \( \frac{d}{dt}(0) = 0 \). So, \( \mathbf{v}(t) = -\sin t \mathbf{i} + 2\cos t \mathbf{j} + 0 \mathbf{k} \).Thus, \( \mathbf{v}(t) = -\sin t \mathbf{i} + 2\cos t \mathbf{j} \).

Step 3: Consolidate the Velocity Vector

Combine the differentiated components into a single velocity vector expression. The velocity vector is:\[ \mathbf{v}(t) = -\sin t \mathbf{i} + 2\cos t \mathbf{j} \].

Key Concepts

Particle Motion

When we talk about particle motion, we're considering how a particle moves through space over time. This often involves analyzing the path it takes and its position at different times. In this exercise, the particle follows the path of an ellipse, and its position is described by the vector function \( \mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k} \).
This function tells us the location of the particle at any time \( t \). The components \( \cos t \) and \( 2 \sin t \) represent the particle's coordinates in the \( i \) and \( j \) directions respectively.
- **Understanding Motion**: To fully grasp how the particle moves, it's crucial to break down its path into these directional components. - **Elliptical Path**: Given that the components consist of trigonometric functions, the particle traces out an ellipse in the \( xy \)-plane.Understanding particle motion involves mastering the path and trajectory, allowing for insights into how variables change along this path.

Velocity Vector

The velocity vector is crucial for understanding how fast and in which direction a particle moves at any given time. When dealing with vector calculus, this vector is derived by differentiating the position vector function with respect to time.
The velocity formula is \( \mathbf{v}(t) = \frac{d}{dt}(\mathbf{r}(t)) \). In the context of our problem, where \( \mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k} \), we need to find the derivative of each component to form the full velocity vector:- The \( i \)-component results in \( -\sin t \)- The \( j \)-component results in \( 2\cos t \)Putting it all together, the velocity vector is \( \mathbf{v}(t) = -\sin t \mathbf{i} + 2 \cos t \mathbf{j} \). This tells us not only how fast the particle is moving but also the direction of movement at any moment.

Differentiation

Differentiation is a key tool in calculus used to find rates of change, such as velocity. It's the process of finding a derivative, which provides information about how a function changes with respect to its variables.
For our task, we differentiated the position vector \( \mathbf{r}(t) = \cos t \mathbf{i} + 2 \sin t \mathbf{j} + 0 \mathbf{k} \).
Each component of the vector was differentiated separately:

• For \( \cos t \), differentiation yielded \( -\sin t \)
• For \( 2\sin t \), differentiation yielded \( 2\cos t \)
• The constant 0 remained unchanged

These individual derivatives were then combined to form the particle's velocity vector. Differentiation not only helps in finding velocities but also in understanding other dynamic changes, like acceleration and curvature, in various applications.