Problem 13 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

\( \mathbf{v}(t) = e^t(-\sin t + \cos t) \mathbf{i} + e^t(\cos t + \sin t) \mathbf{j} + e^t(t + 1) \mathbf{k} \); \( \mathbf{a}(t) = e^t(-2\sin t) \mathbf{i} + e^t(2\cos t) \mathbf{j} + e^t(t + 2) \mathbf{k} \); Speed: \( \sqrt{2e^{2t} + e^{2t}(t+1)^2} \).

Step 1: Differentiate Position to Find Velocity

The first step is to find the velocity of the particle by differentiating the position function \( \mathbf{r}(t) = e^t(\cos t \mathbf{i} + \sin t \mathbf{j} + t \mathbf{k}) \). Use the product rule in differentiation. The velocity \( \mathbf{v}(t) \) is given by:\[\mathbf{v}(t) = \frac{d}{dt}\left( e^t \cos t \right) \mathbf{i} + \frac{d}{dt}\left( e^t \sin t \right) \mathbf{j} + \frac{d}{dt}\left( e^t t \right) \mathbf{k}\]Calculating each component, we get:\[\mathbf{v}(t) = e^t(-\sin t + \cos t) \mathbf{i} + e^t(\cos t + \sin t) \mathbf{j} + (e^t(t + 1)) \mathbf{k}\]

Step 2: Differentiate Velocity to Find Acceleration

To find the acceleration, differentiate the velocity function \( \mathbf{v}(t) \). Use the product rule again for each component:\[\mathbf{a}(t) = \frac{d}{dt}\left( e^{t}(-\sin t + \cos t) \right) \mathbf{i} + \frac{d}{dt}\left( e^{t}(\cos t + \sin t) \right) \mathbf{j} + \frac{d}{dt}\left( e^{t}(t + 1) \right) \mathbf{k}\]Calculating the derivatives gives:\[\mathbf{a}(t) = e^t(-2\sin t) \mathbf{i} + e^t(2\cos t) \mathbf{j} + (e^t(t + 2)) \mathbf{k}\]

Step 3: Calculate Speed of the Particle

The speed of the particle is the magnitude of the velocity vector \( \mathbf{v}(t) \). It is calculated as:\[||\mathbf{v}(t)|| = \sqrt{(e^t(-\sin t + \cos t))^2 + (e^t(\cos t + \sin t))^2 + (e^t(t + 1))^2}\]Simplifying each term inside the square root gives:\[||\mathbf{v}(t)|| = \sqrt{2e^{2t} + e^{2t}(t+1)^2}\]

Key Concepts

Velocity Calculation

In vector calculus, determining the velocity of a particle involves differentiating its position vector with respect to time. The given position function for this exercise is:
\( \mathbf{r}(t) = e^t(\cos t \mathbf{i} + \sin t \mathbf{j} + t \mathbf{k}) \).
To find the velocity, noted as \( \mathbf{v}(t) \), we differentiate \( \mathbf{r}(t) \). This means calculating the derivative of each component with respect to \( t \). The product rule, which we'll discuss later, comes in handy here because each term is a product of functions depending on \( t \).
By performing these operations, we obtain:

• \( \mathbf{v}(t) = e^t(-\sin t + \cos t) \mathbf{i} \)
• \( + e^t(\cos t + \sin t) \mathbf{j} \)
• \( + (e^t(t + 1)) \mathbf{k} \)

This vector \( \mathbf{v}(t) \) represents the change in position with time, commonly known as velocity, and involves how fast each component (\( \mathbf{i}, \mathbf{j}, \mathbf{k} \)) changes.

Acceleration Differentiation

To comprehend acceleration in vector calculus, we need to take the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time, \( t \). This yields the acceleration vector \( \mathbf{a}(t) \).
In our specific example, the velocity function is
\( \mathbf{v}(t) = e^t(-\sin t + \cos t) \mathbf{i} + e^t(\cos t + \sin t) \mathbf{j} + (e^t(t + 1)) \mathbf{k} \).
Applying the differentiation again by utilizing the product rule for each component, we find

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

Acceleration measures changes in velocity over time and reveals how the particle's speed and direction are adjusted as it moves, giving us a full picture of its dynamic behavior.

Speed Magnitude

In vector calculus, the speed of a particle is defined as the magnitude of its velocity vector. The velocity vector \( \mathbf{v}(t) \) obtained in the previous steps can be decomposed into its components, and its magnitude calculated to find speed.
To compute the speed, we use:
\[ ||\mathbf{v}(t)|| = \sqrt{(e^t(-\sin t + \cos t))^2 + (e^t(\cos t + \sin t))^2 + (e^t(t + 1))^2} \]Upon simplification, the expression becomes:
\[ ||\mathbf{v}(t)|| = \sqrt{2e^{2t} + e^{2t}(t+1)^2} \]
This equation gives us the scalar speed, a crucial measurement showing how fast the particle moves regardless of direction. Speed focuses solely on the rate of travel and is a non-negative value.

Product Rule in Differentiation

The product rule is an essential tool in vector calculus, especially when differentiating functions involving products of two or more differentiable functions. The rule states that if \( u(t) \) and \( v(t) \) are functions of \( t \), then the derivative of their product \( u(t)v(t) \) is given by
\( (uv)' = u'v + uv' \).
This rule helps us handle components of vector functions that involve such products, as was the case in both the velocity and acceleration calculations.

• For \( e^t \cos t \), differentiate both \( e^t \) and \( \cos t \).
• Apply the product rule: derivative of first times the second plus the first times derivative of the second.

Repeating this for each component was vital in accurately finding both \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \). Understanding the product rule is vital, as it enables differentiation when more complex functions come into play.