Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the velocity \(\mathbf{v}\), acceleration \(\mathbf{a}\), and speed \(s\) at the indicated time \(t=t_{1}\). $$ \mathbf{r}(t)=t \sin \pi t \mathbf{i}+t \cos \pi t \mathbf{j}+e^{-t} \mathbf{k} ; t_{1}=2 $$

Step-by-Step Solution

Answer

Velocity: \(2\pi \mathbf{i} + \mathbf{j} - e^{-2} \mathbf{k}\); Acceleration: \(\pi \mathbf{i} - 2\pi^2 \mathbf{j} + e^{-2} \mathbf{k}\); Speed: \(\sqrt{4\pi^2 + 1 + e^{-4}}\).

Step 1: Compute the Velocity Vector

The velocity vector \( \mathbf{v} \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Compute the derivative: \[ \mathbf{v}(t) = \frac{d}{dt} \left[ t \sin \pi t \mathbf{i} + t \cos \pi t \mathbf{j} + e^{-t} \mathbf{k} \right] \]Using the product rule and chain rule:\[ \mathbf{v}(t) = \left( \sin \pi t + t \pi \cos \pi t \right) \mathbf{i} + \left( \cos \pi t - t \pi \sin \pi t \right) \mathbf{j} - e^{-t} \mathbf{k}\] Evaluate at \( t_1 = 2 \): \[ \mathbf{v}(2) = \left( \sin 2\pi + 2\pi \cos 2\pi \right) \mathbf{i} + \left( \cos 2\pi - 2\pi \sin 2\pi \right) \mathbf{j} - e^{-2} \mathbf{k} \] Since \( \sin 2\pi = 0 \) and \( \cos 2\pi = 1 \): \[ \mathbf{v}(2) = 2\pi \mathbf{i} + \mathbf{j} - e^{-2} \mathbf{k} \]

Step 2: Compute the Acceleration Vector

The acceleration vector \( \mathbf{a} \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Compute the derivative: \[ \mathbf{v}(t) = \left( \sin \pi t + t \pi \cos \pi t \right) \mathbf{i} + \left( \cos \pi t - t \pi \sin \pi t \right) \mathbf{j} - e^{-t} \mathbf{k}\] Therefore:\[ \mathbf{a}(t) = \left( \pi \cos \pi t - \pi^2 t \sin \pi t \right) \mathbf{i} + \left( -\pi \sin \pi t - \pi^2 t \cos \pi t \right) \mathbf{j} + e^{-t} \mathbf{k} \]Evaluate at \( t_1 = 2 \): \[ \mathbf{a}(2) = \left( \pi \cos 2\pi - 2\pi^2 \sin 2\pi \right) \mathbf{i} + \left( -\pi \sin 2\pi - 2\pi^2 \cos 2\pi \right) \mathbf{j} + e^{-2} \mathbf{k} \]Since \( \sin 2\pi = 0 \) and \( \cos 2\pi = 1 \):\[ \mathbf{a}(2) = \pi \mathbf{i} - 2\pi^2 \mathbf{j} + e^{-2} \mathbf{k} \]

Step 3: Compute the Speed

The speed \( s \) is the magnitude of the velocity vector \( \mathbf{v}(2) \). The velocity vector is:\[ \mathbf{v}(2) = 2\pi \mathbf{i} + \mathbf{j} - e^{-2} \mathbf{k} \] Calculate the magnitude:\[ s = \| \mathbf{v}(2) \| = \sqrt{(2\pi)^2 + 1^2 + (-e^{-2})^2} \] \[ s = \sqrt{4\pi^2 + 1 + e^{-4}} \]

Key Concepts

Velocity Vector

To understand the concept of a velocity vector, imagine following the path of an object as it moves in space over time. The velocity vector \( \mathbf{v}(t) \) gives a precise description of both the speed and direction of the object's motion at any point along its path. In mathematical terms, the velocity vector is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \), which tells us how the position changes instantaneously.

For our example, given the position \( \mathbf{r}(t) = t \sin \pi t \mathbf{i} + t \cos \pi t \mathbf{j} + e^{-t} \mathbf{k} \), calculating the derivative gives us the velocity vector: \[ \mathbf{v}(t) = (\sin \pi t + t \pi \cos \pi t) \mathbf{i} + (\cos \pi t - t \pi \sin \pi t) \mathbf{j} - e^{-t} \mathbf{k} \]

At \( t = 2 \), it evaluates to \( \mathbf{v}(2) = 2\pi \mathbf{i} + \mathbf{j} - e^{-2} \mathbf{k} \). This means at the instant \( t = 2 \), the object is moving mainly in the \( \mathbf{i} \)-direction with a bit in the \( \mathbf{j} \)-direction while decaying exponentially in the \( \mathbf{k} \)-direction.

Acceleration Vector

The acceleration vector provides insight into how the velocity of an object is changing with time. It's the derivative of the velocity vector \( \mathbf{v}(t) \), indicating the rate of change of velocity. If the velocity vector represents an object's motion, acceleration explains how that motion is speeding up, slowing down, or changing direction.

In our exercise, the velocity vector was derived as \( \mathbf{v}(t) = (\sin \pi t + t \pi \cos \pi t) \mathbf{i} + (\cos \pi t - t \pi \sin \pi t) \mathbf{j} - e^{-t} \mathbf{k} \). Differentiating the velocity gives us the acceleration:

• \( \mathbf{a}(t) = (\pi \cos \pi t - \pi^2 t \sin \pi t) \mathbf{i} \)
• \( + (-\pi \sin \pi t - \pi^2 t \cos \pi t) \mathbf{j} \)
• \( + e^{-t} \mathbf{k} \)

Evaluating at \( t = 2 \) results in \( \mathbf{a}(2) = \pi \mathbf{i} - 2\pi^2 \mathbf{j} + e^{-2} \mathbf{k} \).

This shows that at this moment, there's a positive force in the \( \mathbf{i} \)-direction, pulling back strongly in the \( \mathbf{j} \)-direction, and a tiny boost in the \( \mathbf{k} \)-direction.

Magnitude of Velocity

The magnitude of the velocity vector, also known simply as "speed," tells us how fast an object is moving regardless of direction. It's calculated by taking the square root of the sum of the squares of the velocity components.

For the given vector, \( \mathbf{v}(2) = 2\pi \mathbf{i} + \mathbf{j} - e^{-2} \mathbf{k} \), the magnitude is computed as:

• Square each component: \( (2\pi)^2 , 1^2 , (-e^{-2})^2 \)
• Add them up: \( 4\pi^2 + 1 + e^{-4} \)
• Take the square root: \( s = \sqrt{4\pi^2 + 1 + e^{-4}} \)

This value gives us the speed at \( t = 2 \). It represents the total pace at which the object is moving without regards to its direction. Calculating the magnitude helps differentiate between how fast and how far different components of velocity contribute to the motion.