Problem 18 - Thomas Calculus

Question

In Exercises \(13-18, \mathbf{r}(t)\) 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. $$ \mathbf{r}(t)=e^{-t} \mathbf{i}+(2 \cos 3 t) \mathbf{j}+(2 \sin 3 t) \mathbf{k}, \quad t=0 $$

Step-by-Step Solution

Answer

Velocity: \( \mathbf{v}(0) = -1 \mathbf{i} + 6 \mathbf{k} \), Speed: \( \sqrt{37} \), Direction: \( \frac{-1}{\sqrt{37}} \mathbf{i} + \frac{6}{\sqrt{37}} \mathbf{k} \).

Step 1: Find the Velocity Vector

To find the velocity vector, we need to take the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). \[ \mathbf{v}(t) = \frac{d}{dt}\left( e^{-t} \mathbf{i} + 2 \cos 3t \mathbf{j} + 2 \sin 3t \mathbf{k} \right) \]This results in:\[ \mathbf{v}(t) = -e^{-t} \mathbf{i} - 6 \sin 3t \mathbf{j} + 6 \cos 3t \mathbf{k} \]

Step 2: Find the Acceleration Vector

The acceleration vector is found by taking the derivative of the velocity vector \( \mathbf{v}(t) \):\[ \mathbf{a}(t) = \frac{d}{dt}\left( -e^{-t} \mathbf{i} - 6 \sin 3t \mathbf{j} + 6 \cos 3t \mathbf{k} \right) \]This results in:\[ \mathbf{a}(t) = e^{-t} \mathbf{i} - 18 \cos 3t \mathbf{j} - 18 \sin 3t \mathbf{k} \]

Step 3: Evaluate the Velocity at \( t = 0 \)

Substitute \( t = 0 \) into the velocity vector \( \mathbf{v}(t) \):\[ \mathbf{v}(0) = -e^{0} \mathbf{i} - 6 \sin(0) \mathbf{j} + 6 \cos(0) \mathbf{k} \]This simplifies to:\[ \mathbf{v}(0) = -1 \mathbf{i} + 6 \mathbf{k} \]

Step 4: Calculate the Speed at \( t = 0 \)

The speed of the particle is the magnitude of the velocity vector at \( t = 0 \):\[ \text{Speed} = \| \mathbf{v}(0) \| = \sqrt{(-1)^2 + 6^2} \]\[ \text{Speed} = \sqrt{1 + 36} = \sqrt{37} \]

Step 5: Determine the Direction of Motion at \( t = 0 \)

The direction of motion is the unit vector of the velocity vector \( \mathbf{v}(0) \): \[ \text{Direction} = \frac{\mathbf{v}(0)}{\| \mathbf{v}(0) \|} = \frac{-1 \mathbf{i} + 6 \mathbf{k}}{\sqrt{37}} \]

Step 6: Write the Velocity as Product of Speed and Direction

Express the velocity \( \mathbf{v}(0) \) as the product of its speed and direction:\[ \mathbf{v}(0) = \sqrt{37} \left( \frac{-1}{\sqrt{37}} \mathbf{i} + \frac{6}{\sqrt{37}} \mathbf{k} \right) \]This verifies that the velocity vector can be decomposed into speed and direction.

Key Concepts

Velocity Vector

The velocity vector represents the speed and direction of a particle's movement at any point in time. It is derived from the position vector, which gives the location of a particle in space with components like \( x \), \( y \), and \( z \). To find the velocity vector \( \mathbf{v}(t) \), we take the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). By using differentiation rules, the derivative of each component of the position vector is calculated:

• The derivative of \( e^{-t} \) yields \( -e^{-t} \).
• The derivative of \( 2 \cos 3t \) gives \( -6 \sin 3t \) because of the chain rule.
• The derivative of \( 2 \sin 3t \) results in \( 6 \cos 3t \), again using the chain rule.

With these components, the velocity vector \( \mathbf{v}(t) \) is:\[\mathbf{v}(t) = -e^{-t} \mathbf{i} - 6 \sin 3t \mathbf{j} + 6 \cos 3t \mathbf{k}.\] This vector tells us not only how fast the particle moves but also the direction of its velocity at every point in time.

Acceleration Vector

Acceleration is the rate at which the velocity of a particle changes with time. To find the acceleration vector \( \mathbf{a}(t) \), we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Differentiation rules apply here as well:

• For \( -e^{-t} \), the derivative is \( e^{-t} \).
• The derivative of \(-6\sin 3t\) is \(-18\cos 3t\), using the chain rule.
• The derivative of \(6\cos 3t\) results in \(-18\sin 3t\).

Thus, the acceleration vector \( \mathbf{a}(t) \) is:\[\mathbf{a}(t) = e^{-t} \mathbf{i} - 18 \cos 3t \mathbf{j} - 18 \sin 3t \mathbf{k}.\]This vector helps us understand how the particle's velocity is changing at each instant, indicating any accelerations or decelerations.

Speed and Direction

The speed of a particle at any moment is determined by the magnitude of its velocity vector. To compute this magnitude at \( t = 0 \), we use:\[\text{Speed} = \| \mathbf{v}(0) \| = \sqrt{(-1)^2 + 6^2} = \sqrt{37}.\]Speed is a scalar quantity and tells us how fast the particle is moving regardless of direction. The direction of motion is indicated by a unit vector, meaning it has a length of 1 but points in the same direction as the velocity vector. This is obtained by dividing each component of the velocity vector by its magnitude:\[\text{Direction} = \frac{\mathbf{v}(0)}{\| \mathbf{v}(0) \|} = \frac{-1 \mathbf{i} + 6 \mathbf{k}}{\sqrt{37}}.\]This unit vector points in the direction of the particle's movement, giving us insight into where it is headed.

Derivative of Vector Functions

Taking derivatives of vector functions allows us to determine key aspects of a particle's motion, like velocity and acceleration. The process involves finding derivatives of each individual component of the vector function:

• Start with the position vector \( \mathbf{r}(t) \), consisting of functions \( f(t) \), \( g(t) \), and \( h(t) \).
• The velocity vector is given by \( \mathbf{v}(t) = \frac{d}{dt}(f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}) \).
• The acceleration vector is \( \mathbf{a}(t) = \frac{d}{dt}(\mathbf{v}(t)) \), which provides details on changes in velocity.

Differentiating is essential because it translates a position function into a velocity vector, thus enabling the calculation of speed and the analysis of directional movement. It builds foundational knowledge for predicting particle motion.