Problem 17 - Calculus Early Transcendentals

Question

\(17-18\) (a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. (b) Use a computer to graph the path of the particle. $$\mathbf{a}(t)=2 t \mathbf{i}+\sin t \mathbf{j}+\cos 2 t \mathbf{k}, \quad \mathbf{v}(0)=\mathbf{i}, \quad \mathbf{r}(0)=\mathbf{j}$$

Step-by-Step Solution

Answer

The position vector is \( \mathbf{r}(t) = \frac{t^3}{3}\mathbf{i} + (1-\sin t)\mathbf{j} + \left(\frac{1}{4} - \frac{1}{4}\cos 2t\right)\mathbf{k} \). Graphing this will show the particle's path.

Step 1: Integrate the Acceleration to Find Velocity

Since acceleration is the derivative of velocity, integrate the given acceleration vector \( \mathbf{a}(t) = 2t\mathbf{i} + \sin t\mathbf{j} + \cos 2t\mathbf{k} \) to find the velocity vector \( \mathbf{v}(t) \). Upon integration, we obtain:\[\mathbf{v}(t) = \int (2t\mathbf{i} + \sin t\mathbf{j} + \cos 2t\mathbf{k})\, dt = (t^2 + C_1)\mathbf{i} - \cos t\mathbf{j} + \frac{1}{2}\sin 2t\mathbf{k} + \mathbf{C}_2.\]

Step 2: Apply Initial Velocity Condition

Using the initial velocity condition \( \mathbf{v}(0) = \mathbf{i} \), substitute \( t = 0 \) into the integrated velocity expression:\[ (t^2 + C_1)\mathbf{i} - \cos t\mathbf{j} + \frac{1}{2}\sin 2t\mathbf{k} + \mathbf{C}_2 = \mathbf{i}. \]We find \( C_1 = 0 \) and \( \mathbf{C}_2 = \mathbf{0} \). Thus, \( \mathbf{v}(t) = t^2\mathbf{i} - \cos t\mathbf{j} + \frac{1}{2}\sin 2t\mathbf{k} \).

Step 3: Integrate Velocity to Find Position

Now integrate the velocity \( \mathbf{v}(t) = t^2\mathbf{i} - \cos t\mathbf{j} + \frac{1}{2}\sin 2t\mathbf{k} \) to get the position vector \( \mathbf{r}(t) \):\[\mathbf{r}(t) = \int (t^2\mathbf{i} - \cos t\mathbf{j} + \frac{1}{2}\sin 2t\mathbf{k})\, dt = \frac{t^3}{3}\mathbf{i} - \sin t\mathbf{j} - \frac{1}{4}\cos 2t\mathbf{k} + \mathbf{C}_3.\]

Step 4: Apply Initial Position Condition

Using the initial position \( \mathbf{r}(0) = \mathbf{j} \), substitute \( t = 0 \) in the integrated position expression:\[\frac{0^3}{3}\mathbf{i} - \sin(0)\mathbf{j} - \frac{1}{4}\cos(2 \cdot 0)\mathbf{k} + \mathbf{C}_3 = \mathbf{j} \]This gives \( \mathbf{C}_3 = \mathbf{j} + \frac{1}{4}\mathbf{k} \). Therefore, \( \mathbf{r}(t) = \frac{t^3}{3}\mathbf{i} + (1-\sin t)\mathbf{j} + \left(\frac{1}{4} - \frac{1}{4}\cos 2t\right)\mathbf{k} \).

Step 5: Plot the Path of the Particle Using a Computer

Use graphing software to plot the path described by the position vector \( \mathbf{r}(t) = \frac{t^3}{3}\mathbf{i} + (1-\sin t)\mathbf{j} + \left(\frac{1}{4} - \frac{1}{4}\cos 2t\right)\mathbf{k} \). This will show the trajectory of the particle in three-dimensional space.

Key Concepts

Position Vector

The position vector of a particle in calculus refers to the vector that gives the location of the particle in space at any point in time, typically denoted as \( \mathbf{r}(t) \). This vector is crucial because it allows us to pinpoint exactly where the particle is within a given coordinate system. The process of determining the position vector involves starting from known initial conditions and integrating the velocity vector, which in turn is derived from the acceleration vector.

For example, if we have a particle whose motion is described by certain initial velocity and position conditions, the position vector allows us to express its trajectory. By integrating the velocity function, we obtain the position vector, which in this exercise was derived as:

• \( \mathbf{r}(t) = \frac{t^3}{3}\mathbf{i} + (1-\sin t)\mathbf{j} + \left(\frac{1}{4} - \frac{1}{4}\cos 2t\right)\mathbf{k} \)

The position vector is thus a fundamental construct in determining and visualizing the path traced by the particle.

Acceleration

Acceleration describes how the velocity of a particle changes over time. It is the derivative of the velocity vector and provides information about the rate of change of speed and direction of the particle.

In this context, the acceleration vector \( \mathbf{a}(t) = 2t\mathbf{i} + \sin t\mathbf{j} + \cos 2t\mathbf{k} \) encompasses components in three dimensions, each capable of affecting the particle's motion differently:

• The \( 2t \mathbf{i} \) component suggests that the acceleration in the \( x \)-direction increases linearly with time.
• The \( \sin t \mathbf{j} \) and \( \cos 2t \mathbf{k} \) components introduce periodic changes in the \( y \)-direction and \( z \)-direction, causing the particle to possibly oscillate or curve.

This varying acceleration influences the velocity and subsequently, the position of the particle, emphasizing the interconnectedness of these vector components in determining motion.

Integration

Integration is a fundamental calculus operation used to find a function when its derivative is known. In the context of particle motion, integration serves to determine the velocity and position vectors from the known acceleration.

Starting with the acceleration vector \( \mathbf{a}(t) \), integrating it grants us the velocity vector \( \mathbf{v}(t) \). The procedure involves:

• Integrating each component of the acceleration vector separately: \[ \mathbf{v}(t) = \int (2t\mathbf{i} + \sin t\mathbf{j} + \cos 2t\mathbf{k}) \ dt \]
• Applying initial conditions to determine constants of integration, such as \( C_1 \) and \( \mathbf{C}_2 \).

This process is repeated, integrating the velocity vector to find the position vector, again using initial conditions to resolve constants of integration. Integration, in essence, is how we reconstruct the path of the particle from its changing rate of velocity.

Velocity

Velocity in calculus is the rate of change of the position vector with respect to time. It's a vector quantity, meaning it has both magnitude and direction. After integrating the acceleration vector, we find the velocity vector \( \mathbf{v}(t) \), which outlines how swiftly and in what direction the particle moves at any given instant.

For the exercise, starting from the acceleration, the velocity vector is determined through:

• Integrating the acceleration: \( \mathbf{v}(t) = t^2\mathbf{i} - \cos t\mathbf{j} + \frac{1}{2}\sin 2t\mathbf{k} \)
• Utilizing the initial condition \( \mathbf{v}(0) = \mathbf{i} \) to solve for constants.

This vector quantitatively describes the particle's speed along its path and how this speed evolves over time. Understanding velocity helps predict future positions and adjust the integration to account for initial conditions.

Graphing in Calculus

Graphing in calculus involves plotting mathematical functions and expressions to visually interpret their behavior over a period of time or within a specified range. When grappling with particle motion, graphing the position vector can provide a clear image of the trajectory.

Upon deriving the position vector, computational tools or graphing software are employed to plot the path it describes. This involves

• Evaluating the position vector \( \mathbf{r}(t) \) at various values of \( t \)
• Representing these points in three-dimensional space to show the particle's trajectory

Such visualizations aid in comprehending how the particle moves through its environment, revealing the intricacies of its path, like curves and oscillations driven by its acceleration. Graphing offers an intuitive look at the complex relationships outlined by calculus, making these relationships accessible and understandable.