Problem 12 - 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)=(\sec t) \mathbf{i}+(\tan t) \mathbf{j}+\frac{4}{3} t \mathbf{k}, \quad t=\pi / 6 \end{equation}

Step-by-Step Solution

Answer

At \( t = \frac{\pi}{6} \), the velocity is \( 2\sqrt{7} \) times its direction vector.

Step 1: Find the Velocity Vector

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

Step 2: Find the Acceleration Vector

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

Step 3: Evaluate Velocity at t = π/6

Substitute \( t = \frac{\pi}{6} \) into the velocity vector:\[\mathbf{v}\left(\frac{\pi}{6}\right) = (\sec(\frac{\pi}{6}) \cdot \tan(\frac{\pi}{6})) \mathbf{i} + (\sec^2(\frac{\pi}{6})) \mathbf{j} + \frac{4}{3} \mathbf{k}\]Knowing that \( \sec(\frac{\pi}{6}) = 2 \) and \( \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \), the vector becomes:\[\mathbf{v}\left(\frac{\pi}{6}\right) = \left(2 \cdot \frac{1}{\sqrt{3}}\right) \mathbf{i} + (2^2) \mathbf{j} + \frac{4}{3} \mathbf{k} = \left(\frac{2}{\sqrt{3}}\right) \mathbf{i} + 4 \mathbf{j} + \frac{4}{3} \mathbf{k}\]

Step 4: Compute the Speed at t = π/6

The speed of the particle is the magnitude of the velocity vector:\[\|\mathbf{v}\left(\frac{\pi}{6}\right)\| = \sqrt{\left(\frac{2}{\sqrt{3}}\right)^2 + 4^2 + \left(\frac{4}{3}\right)^2}\]Calculating this, we get:\[\|\mathbf{v}\left(\frac{\pi}{6}\right)\| = \sqrt{\frac{4}{3} + 16 + \frac{16}{9}}\]\[\|\mathbf{v}\left(\frac{\pi}{6}\right)\| = \sqrt{\frac{84}{9}} = 2\sqrt{7}\]

Step 5: Find the Direction of Motion at t = π/6

The direction of motion is given by the unit vector in the direction of the velocity vector:\[\mathbf{d}\left(\frac{\pi}{6}\right) = \frac{\mathbf{v}\left(\frac{\pi}{6}\right)}{\|\mathbf{v}\left(\frac{\pi}{6}\right)\|}\] Plugging in known values:\[\mathbf{d}\left(\frac{\pi}{6}\right) = \frac{\left(\frac{2}{\sqrt{3}}\right) \mathbf{i} + 4 \mathbf{j} + \frac{4}{3}\mathbf{k}}{2\sqrt{7}}\]\[\mathbf{d}\left(\frac{\pi}{6}\right) = \left(\frac{1}{\sqrt{7\cdot 3}}\right) \mathbf{i} + \left(\frac{2}{\sqrt{7}}\right) \mathbf{j} + \left(\frac{2}{3\sqrt{7}}\right) \mathbf{k}\]

Step 6: Express Velocity as Product of Speed and Direction

The velocity vector at \( t = \frac{\pi}{6} \) is given as the product of the speed and direction:\[\mathbf{v}\left(\frac{\pi}{6}\right) = 2\sqrt{7} \cdot \left(\frac{1}{\sqrt{21}} \mathbf{i} + \frac{2}{\sqrt{7}} \mathbf{j} + \frac{2}{3\sqrt{7}} \mathbf{k}\right)\]This confirms that the velocity is the product of its speed and direction vectors at the specified time.

Key Concepts

Velocity vector

In calculus, the velocity vector represents the rate of change of the position of a particle with respect to time. It is derived by differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Simply put, it gives you information about how fast and in which direction a particle is moving at any given moment. For the position vector given by \( \mathbf{r}(t)= (\sec t) \mathbf{i} + (\tan t) \mathbf{j} + \frac{4}{3} t \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) is calculated by applying differentiation to each component:

• The rate of change of \( \sec t \) is \( \sec t \tan t \)
• For \( \tan t \), it's \( \sec^2 t \)
• For \( \frac{4}{3}t \), it is simply \( \frac{4}{3} \)

The resulting velocity vector is \( \mathbf{v}(t) = (\sec t \tan t) \mathbf{i} + (\sec^2 t) \mathbf{j} + \frac{4}{3} \mathbf{k} \). This vector helps us to understand not just the speed but also the precise direction of the particle's movement across x, y, and z axes.

Acceleration vector

The acceleration vector is central to understanding how a particle's velocity changes over time. It is computed by differentiating the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). This provides us with the second derivative of the position vector \( \mathbf{r}(t) \).

• The derivative of \( \sec t \tan t \) is \( \sec^3 t + \sec t \tan^2 t \)
• For \( \sec^2 t \) it is \( 2\sec^2 t \tan t \)
• Since \( \frac{4}{3} \) is constant, its derivative is 0

The acceleration vector thus becomes \( \mathbf{a}(t) = (\sec^3 t + \sec t \cdot \tan^2 t) \mathbf{i} + (2 \sec^2 t \cdot \tan t) \mathbf{j} \). The acceleration vector defines not only how quickly the speed of the particle is changing but also the new direction into which the particle is accelerating.

Particle motion

Particle motion is the analysis of a particle's path, encompassing both velocity and acceleration. In the context of our example, particle motion reflects how the particle's position changes over time, derived from its velocity and acceleration vectors.

• Velocity tells us how position changes with time
• Acceleration informs us how velocity changes

By calculating both vectors, we can fully describe the path of a particle through space and determine its exact location and direction at any given time. This forms the basis of predicting future states and behaviors of any moving object.

Speed and direction

Speed and direction are key components in understanding how a particle moves along its path. Speed is a scalar quantity representing the magnitude of the velocity vector, which indicates how fast the particle is moving regardless of the direction.
The direction, on the other hand, is expressed as a unit vector showing where the particle is pointing at any given moment. To find the speed, compute the magnitude of the velocity vector:\[ \|\mathbf{v}(t)\| = \sqrt{(v_x)^2 + (v_y)^2 + (v_z)^2} \]The direction is found by normalizing the velocity vector to form a unit vector:\[ \mathbf{d}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \]Thus, speed and direction together provide a complete picture of the particle's motion at any particular time, with speed telling how fast and direction showing where.

Differentiation

Differentiation is a fundamental technique in calculus used to find the rate at which one quantity changes with respect to another. In problems involving velocity and acceleration, differentiation allows us to find these rates of change:

• By differentiating the position vector, we obtain the velocity vector
• A further differentiation of the velocity vector yields the acceleration vector

This process involves applying rules of differentiation to each component of a vector individually. For example, when differentiating trigonometric functions like \( \sec t \) or polynomial terms like \( \frac{4}{3}t \), we utilize specific differentiation formulas. Mastery of differentiation empowers us to solve complex motion problems, comprehending how a particle moves and accelerates in multi-dimensional space.