Problem 11 - Thomas Calculus

Question

\({r}(t)\) is the position of a particle in space at time \(t .\) Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of \(t\) . Write the particle's velocity at that time as the product of its speed and direction. \begin{equation} \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(3 \sin t) \mathbf{j}+4 t \mathbf{k}, \quad t=\pi / 2 \end{equation}

Step-by-Step Solution

Answer

The velocity is \(-2 \mathbf{i} + 4 \mathbf{k}\) and acceleration is \(-3 \mathbf{j}\) at \(t=\frac{\pi}{2}\). Speed is \(2\sqrt{5}\) and direction is \(-\frac{1}{\sqrt{5}} \mathbf{i} + \frac{2}{\sqrt{5}} \mathbf{k}\).

Step 1: Differentiate Position Vector to Find Velocity Vector

The velocity vector \( \mathbf{v}(t) \) of a particle is the derivative of its position vector \( \mathbf{r}(t) \) with respect to time \( t \). Thus, we need to differentiate each component of \( \mathbf{r}(t) \):\( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + 4t \mathbf{k} \).Differentiating:- \( \frac{d}{dt} (2 \cos t) = -2 \sin t \),- \( \frac{d}{dt} (3 \sin t) = 3 \cos t \),- \( \frac{d}{dt} (4t) = 4 \).Thus, the velocity vector is:\[ \mathbf{v}(t) = (-2 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 4 \mathbf{k}. \]

Step 2: Differentiate Velocity Vector to Find Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Differentiating each component of \( \mathbf{v}(t) \):\( \mathbf{v}(t) = (-2 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 4 \mathbf{k} \).Differentiating:- \( \frac{d}{dt} (-2 \sin t) = -2 \cos t \),- \( \frac{d}{dt} (3 \cos t) = -3 \sin t \),- \( \frac{d}{dt} (4) = 0 \).Thus, the acceleration vector is:\[ \mathbf{a}(t) = (-2 \cos t) \mathbf{i} - (3 \sin t) \mathbf{j}. \]

Step 3: Evaluate Velocity and Acceleration at t = \( \frac{\pi}{2} \)

To find the velocity and acceleration at \( t = \frac{\pi}{2} \), substitute \( t = \frac{\pi}{2} \) into the expressions for \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \).For velocity:- \( -2 \sin(\frac{\pi}{2}) = -2 \times 1 = -2 \),- \( 3 \cos(\frac{\pi}{2}) = 3 \times 0 = 0 \).\[ \mathbf{v}(\frac{\pi}{2}) = -2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} = -2 \mathbf{i} + 4 \mathbf{k}. \]For acceleration:- \( -2 \cos(\frac{\pi}{2}) = -2 \times 0 = 0 \),- \( -3 \sin(\frac{\pi}{2}) = -3 \times 1 = -3 \).\[ \mathbf{a}(\frac{\pi}{2}) = 0 \mathbf{i} - 3 \mathbf{j}. \]

Step 4: Calculate Speed and Direction at \( t = \frac{\pi}{2} \)

The speed of the particle is the magnitude of the velocity vector at \( t = \frac{\pi}{2} \):\[ \| \mathbf{v}(\frac{\pi}{2}) \| = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}. \]The direction is given by the unit vector of \( \mathbf{v}(\frac{\pi}{2}) \):\[ \text{Direction} = \frac{\mathbf{v}(\frac{\pi}{2})}{\| \mathbf{v}(\frac{\pi}{2}) \|} = \frac{-2 \mathbf{i} + 4 \mathbf{k}}{2 \sqrt{5}} = -\frac{1}{\sqrt{5}} \mathbf{i} + \frac{2}{\sqrt{5}} \mathbf{k}. \]

Step 5: Express Velocity as Product of Speed and Direction

The velocity vector at \( t = \frac{\pi}{2} \) can be expressed as the product of its speed and direction of motion:\[ \mathbf{v}(\frac{\pi}{2}) = 2 \sqrt{5} \left( -\frac{1}{\sqrt{5}} \mathbf{i} + \frac{2}{\sqrt{5}} \mathbf{k} \right). \]This shows that the velocity is composed of a speed of \( 2\sqrt{5} \) and a unit direction vector.

Key Concepts

Particle Motion

In vector calculus, understanding the motion of a particle involves analyzing how its position changes over time. When we describe a particle's position in space, we often use a position vector, typically denoted as \( \mathbf{r}(t) \), which represents the location of the particle at any given time \( t \). This vector is usually broken down into its components along the Cartesian coordinate axes, denoted by \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), representing the x, y, and z directions, respectively.

Tracking how these components change as time progresses allows us to understand the path the particle follows in space, its speed, and its direction at any moment. Imagine this as tracing the line left behind by the particle as it moves, visualizing its trajectory. Analyzing particle motion is fundamental in physics and other sciences, as it helps us predict future positions and understand the forces acting upon it.

Velocity Vector

The velocity vector is key in defining the motion of a particle and it gives both the speed and direction of a particle at any particular moment. The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). It is calculated by differentiating each component of the position vector, considering how quickly each part changes.

In practice, this differentiation tells us how fast the particle moves along each axis of our coordinate system. For instance, if the position vector is given as \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + 4t \mathbf{k} \), then the velocity vector is \( \mathbf{v}(t) = (-2 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 4 \mathbf{k} \).

Thus, the velocity vector encapsulates all necessary information about how fast and in what direction the particle moves at any time \( t \). It serves as a crucial tool in analyzing motion.

Acceleration Vector

Acceleration is just as important as velocity, as it explains how the velocity of a particle changes over time. The acceleration vector \( \mathbf{a}(t) \) is found by differentiating the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Each component is differentiated again to show changes in the velocity's speed and direction.

Understanding the acceleration vector allows us to see whether the particle is speeding up, slowing down, or changing directions. For example, from the velocity vector \( \mathbf{v}(t) = (-2 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 4 \mathbf{k} \), the acceleration becomes \( \mathbf{a}(t) = (-2 \cos t) \mathbf{i} - (3 \sin t) \mathbf{j} \).

This acceleration vector provides insights into the forces or influences acting on the particle, helping us understand the dynamics of its motion more fully.

Magnitude of a Vector

The magnitude of a vector represents its length and serves as a measure of its size, entirely separate from its direction. For velocity, the magnitude is what we call 'speed'. Speed indicates how fast a particle is moving, irrespective of its direction. To calculate it, we use the formula for the magnitude of a vector. It involves taking the square root of the sum of the squares of its components:

• For a vector like \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude is found as \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \).

In our example, the velocity vector at \( t = \frac{\pi}{2} \) is \( -2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} \), giving a speed of \( \sqrt{20} = 2\sqrt{5} \).

Knowing the magnitude aids in understanding the overall motion and dynamics of the particle without worrying about direction, letting us know how fast it actually moves in space.