Problem 14 - Calculus Early Transcendentals

Question

\(9-14\) Find the velocity, acceleration, and speed of a particle with the given position function. $$\mathbf{r}(t)=t \sin t \mathbf{i}+t \cos t \mathbf{j}+t^{2} \mathbf{k}$$

Step-by-Step Solution

Answer

The velocity is \( \mathbf{v}(t) = (\sin t + t \cos t) \mathbf{i} + (\cos t - t \sin t) \mathbf{j} + 2t \mathbf{k} \), acceleration is \( \mathbf{a}(t) = (2 \cos t - t \sin t) \mathbf{i} + (-2 \sin t - t \cos t) \mathbf{j} + 2 \mathbf{k} \), and speed is \( \sqrt{1 + 5t^2} \)."

Step 1: Find the Velocity Vector

The velocity vector is the derivative of the position vector \( \mathbf{r}(t) \). We find the derivative of each component with respect to \( t \):\[ \mathbf{v}(t) = \frac{d}{dt}(t \sin t) \mathbf{i} + \frac{d}{dt}(t \cos t) \mathbf{j} + \frac{d}{dt}(t^2) \mathbf{k} \].Using product rule for \( t \sin t \) and \( t \cos t \):- \( \frac{d}{dt}(t \sin t) = \sin t + t \cos t \)- \( \frac{d}{dt}(t \cos t) = \cos t - t \sin t \)- \( \frac{d}{dt}(t^2) = 2t \)Thus, the velocity vector is:\[ \mathbf{v}(t) = (\sin t + t \cos t) \mathbf{i} + (\cos t - t \sin t) \mathbf{j} + 2t \mathbf{k} \]

Step 2: Find the Acceleration Vector

The acceleration vector is the derivative of the velocity vector \( \mathbf{v}(t) \). We differentiate each component of \( \mathbf{v}(t) \):\[ \mathbf{a}(t) = \frac{d}{dt}(\sin t + t \cos t) \mathbf{i} + \frac{d}{dt}(\cos t - t \sin t) \mathbf{j} + \frac{d}{dt}(2t) \mathbf{k} \].Applying the product rule again:- \( \frac{d}{dt}(\sin t + t \cos t) = \cos t + \cos t - t \sin t = 2 \cos t - t \sin t \)- \( \frac{d}{dt}(\cos t - t \sin t) = -\sin t - \sin t - t \cos t = -2 \sin t - t \cos t \)- \( \frac{d}{dt}(2t) = 2 \)Hence, the acceleration vector is:\[ \mathbf{a}(t) = (2 \cos t - t \sin t) \mathbf{i} + (-2 \sin t - t \cos t) \mathbf{j} + 2 \mathbf{k} \]

Step 3: Calculate the Speed of the Particle

The speed of a particle is the magnitude of the velocity vector \( \mathbf{v}(t) \). We compute:\[ | \mathbf{v}(t) = \sqrt{(\sin t + t \cos t)^2 + (\cos t - t \sin t)^2 + (2t)^2} \ = \sqrt{(\sin^2 t + 2t \sin t \cos t + t^2 \cos^2 t) + (\cos^2 t - 2t \cos t \sin t + t^2 \sin^2 t) + 4t^2 } \ = \sqrt{(\sin^2 t + \cos^2 t) + t^2(\cos^2 t + \sin^2 t) + 4t^2 } \ = \sqrt{1 + t^2 + 4t^2 } \ = \sqrt{1 + 5t^2 } \]

Key Concepts

Position Function

The position function \(\mathbf{r}(t) = t \sin t \, \mathbf{i} + t \cos t \, \mathbf{j} + t^{2} \, \mathbf{k}\) describes the location of a particle in space at time \( t \). \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) represent the unit vectors in the directions of the \( x, y, \) and \( z \) axes respectively. As \( t \) changes, the particle moves along a path defined by this vector function.
Position functions are crucial for determining the trajectory of moving objects. When motion is described in three-dimensional space, each component function (\( t \sin t \), \( t \cos t \), and \( t^2 \)) describes movement along each axis. Here are some key points to understand:

• Within the position function, each term is a piece of the particle’s movement along an axis.
• The term referring to \( t \sin t \) affects movement in the \( x \) direction.
• The \( t \cos t \) term impacts the \( y \) direction.
• The \( t^2 \) term controls the vertical \( z \)-axis movement.

The function enables us to pinpoint exactly where the particle is at any given moment. This foundational element is what we differentiate to find velocity and acceleration.

Derivative

In calculus, the derivative is used to determine the rate of change of a function. When applied to the position function, it helps us find the velocity of the particle. By differentiating the position vector \( \mathbf{r}(t) \), we obtain the velocity vector \( \mathbf{v}(t) \).
For a function \( f(t) = t \sin t \), its derivative is calculated using the product rule: \( \frac{d}{dt}(t \sin t) = \sin t + t \cos t \). The same rule applies for \( t \cos t \), resulting in \( \cos t - t \sin t \), and for \( t^2 \), resulting in \( 2t \). These calculations result in the velocity vector:

• \( \mathbf{v}(t) = (\sin t + t \cos t) \, \mathbf{i} + (\cos t - t \sin t) \, \mathbf{j} + 2t \, \mathbf{k} \)

Knowing how to differentiate is vital because:

• The derivative tells us how fast the particle is moving in each direction.
• It indicates changes in speed and direction along its path.

Mastering derivatives provides you with a deeper insight into how dynamic systems behave over time.

Magnitude of Velocity

The magnitude of velocity, often known as speed, is a scalar quantity. It provides the absolute value of the velocity vector \( \mathbf{v}(t) \), showing how fast the particle is travelling, without regard to direction. To find this, we calculate the square root of the sum of each component of the velocity squared:
\[|\mathbf{v}(t)| = \sqrt{(\sin t + t \cos t)^2 + (\cos t - t \sin t)^2 + (2t)^2}\]
Upon simplification, we integrate trigonometric identities like \( \sin^2 t + \cos^2 t = 1 \) and find:

• \[ |\mathbf{v}(t)| = \sqrt{1 + 5t^2} \]

Highlights of the magnitude of velocity include:

• It shows how speed changes over time; if \( t \) increases, speed changes as \( \sqrt{1 + 5t^2} \) changes.
• Unlike velocity, speed provides no directional information.

Understanding this concept helps clarify how quickly the particle moves at any point along its path.