Problem 12 - Thomas Calculus

Question

In Exercises \(9-14, \mathrm{r}(t)\) 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. $$ \mathbf{r}(t)=(\sec t) \mathbf{i}+(\tan t) \mathbf{j}+\frac{4}{3} t \mathbf{k}, \quad t=\pi / 6 $$

Step-by-Step Solution

Answer

Velocity at \( t = \pi/6 \) is \( \mathbf{v}(\pi/6) = \frac{2}{3} \mathbf{i} + \frac{4}{3} \mathbf{j} + \frac{4}{3} \mathbf{k} \); speed is 2; direction is \( \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k} \).

Step 1: Find the Velocity Vector

To find the velocity vector \( \mathbf{v}(t) \), we need to differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \). The position vector is given by \( \mathbf{r}(t) = \sec(t) \mathbf{i} + \tan(t) \mathbf{j} + \frac{4}{3} t \mathbf{k} \). So we find: \[ \mathbf{v}(t) = \frac{d}{dt}[\sec(t) \mathbf{i} + \tan(t) \mathbf{j} + \frac{4}{3} t \mathbf{k}] = \sec(t) \tan(t) \mathbf{i} + \sec^2(t) \mathbf{j} + \frac{4}{3} \mathbf{k}. \]

Step 2: Find the Acceleration Vector

To find the acceleration vector \( \mathbf{a}(t) \), differentiate the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Use the result from Step 1: \[ \mathbf{a}(t) = \frac{d}{dt}[\sec(t) \tan(t) \mathbf{i} + \sec^2(t) \mathbf{j} + \frac{4}{3} \mathbf{k}] = (\sec^3(t) + \sec(t) \tan^2(t)) \mathbf{i} + 2\sec(t)\tan(t) \mathbf{j}. \]

Step 3: Evaluate Velocity at \( t = \pi/6 \)

Substitute \( t = \pi/6 \) into \( \mathbf{v}(t) \) to find the velocity at this specific time. \[ \sec(\pi/6) = \frac{2}{\sqrt{3}}, \quad \tan(\pi/6) = \frac{1}{\sqrt{3}}. \] Thus, \[ \mathbf{v}(\pi/6) = \left(\frac{2}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}\right) \mathbf{i} + \left(\frac{2}{\sqrt{3}}\right)^2 \mathbf{j} + \frac{4}{3} \mathbf{k} = \frac{2}{3} \mathbf{i} + \frac{4}{3} \mathbf{j} + \frac{4}{3} \mathbf{k}. \]

Step 4: Calculate Particle's Speed

The speed is the magnitude of the velocity vector, calculated as: \[ \|\mathbf{v}(\pi/6)\| = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{36}{9}} = 2. \]

Step 5: Find Direction of Motion

The direction of motion is the normalized velocity vector at \( t = \pi/6 \). This means dividing the velocity vector by its magnitude: \[ \mathbf{u} = \frac{\mathbf{v}(\pi/6)}{2} = \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}. \]

Step 6: Express Velocity as Speed and Direction

Finally, express the velocity as the product of its speed and direction (unit vector): \[ \mathbf{v}(\pi/6) = 2 \times \left(\frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}\right). \]

Key Concepts

Velocity Vector

When discussing motion in calculus, understanding the velocity vector is crucial. The velocity vector represents the rate at which an object changes its position in space, and it is the derivative of the position vector with respect to time. For example, if a particle's position vector is given by \( \mathbf{r}(t) = \sec(t) \mathbf{i} + \tan(t) \mathbf{j} + \frac{4}{3} t \mathbf{k} \), then the velocity vector, denoted as \( \mathbf{v}(t) \), is obtained by differentiating each component:

• The \( \mathbf{i} \) component becomes \( \sec(t) \tan(t) \).
• The \( \mathbf{j} \) component becomes \( \sec^2(t) \).
• The \( \mathbf{k} \) component is constant and becomes \( \frac{4}{3} \).

Thus, the velocity vector is \( \mathbf{v}(t) = \sec(t) \tan(t) \mathbf{i} + \sec^2(t) \mathbf{j} + \frac{4}{3} \mathbf{k} \). Differentiation helps us see how the particle's position changes over time, and the direction of \( \mathbf{v}(t) \) gives us insight into which direction the particle is moving.

Acceleration Vector

In calculus, the acceleration vector provides information about the rate at which the velocity of a particle is changing. It is essentially the derivative of the velocity vector with respect to time. If the velocity vector is \( \mathbf{v}(t) = \sec(t) \tan(t) \mathbf{i} + \sec^2(t) \mathbf{j} + \frac{4}{3} \mathbf{k} \), the acceleration vector \( \mathbf{a}(t) \) is obtained by differentiating each of these components:

• The differentiation of \( \sec(t) \tan(t) \) is \( \sec^3(t) + \sec(t) \tan^2(t) \).
• The differentiation of \( \sec^2(t) \) is \( 2\sec(t)\tan(t) \).
• The constant term \( \frac{4}{3} \) disappears because its derivative is zero.

Therefore, the acceleration vector is \( \mathbf{a}(t) = (\sec^3(t) + \sec(t) \tan^2(t)) \mathbf{i} + 2\sec(t)\tan(t) \mathbf{j} \). Knowing the acceleration allows us to understand how the speed of the particle is changing, whether it's speeding up, slowing down, or changing direction.

Speed

Speed is an important scalar quantity in the study of motion, and it's defined as the magnitude of the velocity vector. It tells us how fast an object is moving regardless of its direction. To calculate speed, we simply take the square root of the sum of the squares of the components of the velocity vector.In our previous example, the velocity vector at \( t = \pi/6 \) is \( \mathbf{v}(\pi/6) = \frac{2}{3} \mathbf{i} + \frac{4}{3} \mathbf{j} + \frac{4}{3} \mathbf{k} \). The speed is given by:\[\|\mathbf{v}(\pi/6)\| = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \sqrt{\frac{36}{9}} = 2.\]This calculation reveals that the particle's speed at \( t = \pi/6 \) is 2. By understanding speed, we gain insight into the amount of distance an object covers per time unit, which is a fundamental aspect of its motion.

Direction of Motion

The direction of motion in a three-dimensional space can be understood by looking at the normalized velocity vector, often referred to as a unit vector. This vector shows the direction in which the particle is heading, but it has a magnitude of 1, meaning it conveys only directional information, and not speed.To determine the direction of motion at a specific time, such as \( t = \pi/6 \), we take the velocity vector \( \mathbf{v}(\pi/6) = \frac{2}{3} \mathbf{i} + \frac{4}{3} \mathbf{j} + \frac{4}{3} \mathbf{k} \) and divide each component by the magnitude of the velocity vector (which is the speed, 2 in this scenario):\[\mathbf{u} = \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}.\]This unit vector, \( \mathbf{u} \), then indicates the precise direction of motion. Understanding direction alone helps in knowing the path an object follows, separate from its speed. It indicates the trajectory the particle is following at that particular moment in time.