Problem 15 - Thomas Calculus

Question

In Exercises \(13-18, \mathbf{r}(t)\) 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. $$ \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(3 \sin t) \mathbf{j}+4 t \mathbf{k}, \quad t=\pi / 2 $$

Step-by-Step Solution

Answer

Velocity: \(-2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k}\) and speed: \(2\sqrt{5}\); direction: \(\frac{-1}{\sqrt{5}} \mathbf{i} + \frac{2}{\sqrt{5}} \mathbf{k}\).

Step 1: Find the Velocity Vector

To find the velocity vector \( \mathbf{v}(t) \), we need to calculate the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). Given \( \mathbf{r}(t) = (2\cos t) \mathbf{i} + (3\sin t) \mathbf{j} + 4t \mathbf{k} \), we compute each component separately: - The derivative of \( 2\cos t \) with respect to \( t \) is \( -2\sin t \).- The derivative of \( 3\sin t \) with respect to \( t \) is \( 3\cos t \).- The derivative of \( 4t \) with respect to \( t \) is \( 4 \).Thus, \( \mathbf{v}(t) = -2\sin t \mathbf{i} + 3\cos t \mathbf{j} + 4\mathbf{k} \).

Step 2: Find the Acceleration Vector

To find the acceleration vector \( \mathbf{a}(t) \), we need to calculate the derivative of the velocity vector \( \mathbf{v}(t) = -2\sin t \mathbf{i} + 3\cos t \mathbf{j} + 4\mathbf{k} \). - The derivative of \( -2\sin t \) is \( -2\cos t \). - The derivative of \( 3\cos t \) is \( -3\sin t \). - The derivative of the constant \( 4 \) is \( 0 \).Thus, \( \mathbf{a}(t) = -2\cos t \mathbf{i} - 3\sin t \mathbf{j} \).

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

Substitute \( t = \frac{\pi}{2} \) into the velocity and acceleration vectors:- For \( \mathbf{v}(t) = -2\sin t \mathbf{i} + 3\cos t \mathbf{j} + 4\mathbf{k} \), we have: \[ \mathbf{v}\left(\frac{\pi}{2}\right) = -2\sin\left(\frac{\pi}{2}\right) \mathbf{i} + 3\cos\left(\frac{\pi}{2}\right) \mathbf{j} + 4\mathbf{k} = -2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} \]- For \( \mathbf{a}(t) = -2\cos t \mathbf{i} - 3\sin t \mathbf{j} \), we have: \[ \mathbf{a}\left(\frac{\pi}{2}\right) = -2\cos\left(\frac{\pi}{2}\right) \mathbf{i} - 3\sin\left(\frac{\pi}{2}\right) \mathbf{j} = 0 \mathbf{i} - 3 \mathbf{j} \]

Step 4: Calculate Particle's Speed

The particle's speed is the magnitude of the velocity vector at \( t = \frac{\pi}{2} \). For \( \mathbf{v}\left(\frac{\pi}{2}\right) = -2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} \), the speed is:\[ \text{Speed} = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5} \]

Step 5: Determine Direction of Motion

The direction of motion is given by the unit vector of the velocity at \( t = \frac{\pi}{2} \). This is done by dividing \( \mathbf{v}\left(\frac{\pi}{2}\right) \) by its magnitude (speed): \[ \text{Direction} = \frac{\mathbf{v}\left(\frac{\pi}{2}\right)}{\|\mathbf{v}\left(\frac{\pi}{2}\right)\|} = \frac{-2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k}}{2\sqrt{5}} = \frac{-2}{2\sqrt{5}} \mathbf{i} + \frac{0}{2\sqrt{5}} \mathbf{j} + \frac{4}{2\sqrt{5}} \mathbf{k} = \frac{-1}{\sqrt{5}} \mathbf{i} + 0 \mathbf{j} + \frac{2}{\sqrt{5}} \mathbf{k} \]

Step 6: Express Velocity as Product of Speed and Direction

Express the velocity \( \mathbf{v}\left(\frac{\pi}{2}\right) \) as the product of the speed and the unit direction vector:\[ \mathbf{v}\left(\frac{\pi}{2}\right) = 2\sqrt{5}\left( \frac{-1}{\sqrt{5}} \mathbf{i} + 0 \mathbf{j} + \frac{2}{\sqrt{5}} \mathbf{k}\right) \] This shows that \( \mathbf{v}\left(\frac{\pi}{2}\right) \) is indeed the product of its speed \( 2\sqrt{5} \) and its direction vector.

Key Concepts

Position Vector

The concept of the position vector is crucial when discussing the motion of particles in space. The position vector, often denoted as \( \mathbf{r}(t) \), represents the location of a particle at a specific time \( t \). It consists of components along the standard basis vectors \( \mathbf{i}, \mathbf{j}, \text{and } \mathbf{k} \), which align with the x, y, and z coordinates, respectively. In the given exercise, the position vector \( \mathbf{r}(t) = (2\cos t) \mathbf{i} + (3\sin t) \mathbf{j} + 4t \mathbf{k} \) describes a particle's path as it moves through space. Here's a breakdown of its components:

• The term \( 2\cos(t) \mathbf{i} \) describes the x-coordinate as a function of cosine.
• \( 3\sin(t) \mathbf{j} \) provides the y-coordinate depending on sine.
• The term \( 4t \mathbf{k} \) gives a linear increase with time, representing the z-coordinate.

This vector shows how the particle's position is modified as time progresses, which is why taking its derivative gives us the velocity vector.

Velocity Vector

The velocity vector indicates how the position of a particle changes over time. It's derived by differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \). In our exercise, the velocity vector \( \mathbf{v}(t) = -2\sin t \mathbf{i} + 3\cos t \mathbf{j} + 4\mathbf{k} \) shows the rate of change in each direction:

• \( -2\sin t \mathbf{i} \) highlights how the x-position varies.
• \( 3\cos t \mathbf{j} \) indicates the change in the y-position.
• The constant \( 4 \mathbf{k} \) represents a steady increase along the z-axis.

Evaluating this at \( t=\pi/2 \), we obtain \( \mathbf{v}(\pi/2) = -2\mathbf{i} + 0\mathbf{j} + 4\mathbf{k} \), meaning the particle moves with velocity components of \(-2\) in the x-direction and \(4\) in the z-direction at that moment.

Acceleration Vector

The acceleration vector reflects how the velocity of a particle changes over time. By differentiating the velocity vector, \( \mathbf{v}(t) = -2\sin t \mathbf{i} + 3\cos t \mathbf{j} + 4\mathbf{k} \), we gather insights into this change:

• \( -2\cos t \mathbf{i} \) shows a periodic change in acceleration along the x-axis.
• \( -3\sin t \mathbf{j} \) offers a similar pattern for the y-axis.
• Since there is no \( t \) in \( 4\mathbf{k} \), we see no change in the z-axis component (it's zero).

When we evaluate this at \( t = \pi/2 \), we find \( \mathbf{a}(\pi/2) = 0\mathbf{i} - 3\mathbf{j} \), suggesting there's no acceleration along the x-axis, while the y-axis has an acceleration of \(-3\). This reflects the nature of the motion, as the particle moves through space.

Magnitude of Velocity

Magnitude of velocity, commonly referred to as speed, informs us about how fast the particle is moving regardless of its direction. We find this by calculating the length of the velocity vector. Given \( \mathbf{v}(t) = -2\mathbf{i} + 0\mathbf{j} + 4\mathbf{k} \) at \( t = \pi/2 \), we calculate the magnitude as follows:\[ \text{Speed} = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5} \]This tells us that the particle is moving with a speed of \( 2\sqrt{5} \). Speed can solely change based on velocity changes, without consideration for direction changes. For more insight, direction is found by normalizing the velocity vector, effectively finding a unit vector, which in this case was \( \frac{-1}{\sqrt{5}}\mathbf{i} + 0\mathbf{j} + \frac{2}{\sqrt{5}}\mathbf{k} \).