Problem 13 - 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)=(2 \ln (t+1)) \mathbf{i}+t^{2} \mathbf{j}+\frac{t^{2}}{2} \mathbf{k}, \quad t=1 \end{equation}

Step-by-Step Solution

Answer

Velocity: \(\mathbf{i} + 2 \mathbf{j} + \mathbf{k}\), Acceleration: \(-\frac{1}{2} \mathbf{i} + 2 \mathbf{j} + \mathbf{k}\), Speed: \(\sqrt{6}\), Direction: \(\frac{1}{\sqrt{6}} \mathbf{i} + \frac{2}{\sqrt{6}} \mathbf{j} + \frac{1}{\sqrt{6}} \mathbf{k}\).

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Differentiate each component: \(\frac{d}{dt}(2\ln(t+1)) \mathbf{i} = \frac{2}{t+1} \mathbf{i} \); \( \frac{d}{dt}(t^2) \mathbf{j} = 2t \mathbf{j} \); \( \frac{d}{dt}(\frac{t^2}{2}) \mathbf{k} = t \mathbf{k} \). Thus, \( \mathbf{v}(t) = \frac{2}{t+1} \mathbf{i} + 2t \mathbf{j} + t \mathbf{k} \).

Step 2: Find the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Differentiate each component: \(\frac{d}{dt}(\frac{2}{t+1}) \mathbf{i} = -\frac{2}{(t+1)^2} \mathbf{i} \); \( \frac{d}{dt}(2t) \mathbf{j} = 2 \mathbf{j} \); \( \frac{d}{dt}(t) \mathbf{k} = 1 \mathbf{k} \). Thus, \( \mathbf{a}(t) = -\frac{2}{(t+1)^2} \mathbf{i} + 2 \mathbf{j} + \mathbf{k} \).

Step 3: Evaluate Velocity at \(t = 1\)

Substitute \( t = 1 \) into the velocity vector \( \mathbf{v}(t) = \frac{2}{t+1} \mathbf{i} + 2t \mathbf{j} + t \mathbf{k} \) to find \( \mathbf{v}(1) = \frac{2}{2} \mathbf{i} + 2 \cdot 1 \mathbf{j} + 1 \mathbf{k} = \mathbf{i} + 2 \mathbf{j} + \mathbf{k} \).

Step 4: Find the Speed at \(t = 1\)

The speed is the magnitude of the velocity vector \( \mathbf{v}(1) \). Calculate it as \( ||\mathbf{v}(1)|| = \sqrt{(1)^2 + (2)^2 + (1)^2} = \sqrt{6} \).

Step 5: Determine Direction of Motion at \(t = 1\)

The direction of motion is the unit vector of \( \mathbf{v}(1) \). Find it by dividing each component of \( \mathbf{v}(1) \) by its magnitude: \( \mathbf{e} = \frac{1}{\sqrt{6}} \mathbf{i} + \frac{2}{\sqrt{6}} \mathbf{j} + \frac{1}{\sqrt{6}} \mathbf{k} \).

Step 6: Express Velocity as Speed and Direction

The velocity \( \mathbf{v}(1) \) can be expressed as the product of speed and direction: \( \mathbf{v}(1) = \sqrt{6}( \frac{1}{\sqrt{6}} \mathbf{i} + \frac{2}{\sqrt{6}} \mathbf{j} + \frac{1}{\sqrt{6}} \mathbf{k} ) = \sqrt{6} \mathbf{e} \).

Key Concepts

Velocity Vector

The velocity vector is a crucial concept in particle motion studies. It represents how fast the particle's position changes over time and in which direction this change occurs. To determine the velocity vector from the position function, you need to take the derivative of each component of the position vector with respect to time. In our given function \[ \mathbf{r}(t) = (2 \ln (t+1)) \mathbf{i} + t^2 \mathbf{j} + \frac{t^2}{2} \mathbf{k}, \]we differentiate each component:

• For the \( \mathbf{i} \)-component, derivative of \( 2 \ln (t+1) \) gives \( \frac{2}{t+1} \).
• For the \( \mathbf{j} \)-component, derivative of \( t^2 \) gives \( 2t \).
• For the \( \mathbf{k} \)-component, derivative of \( \frac{t^2}{2} \) gives \( t \).

Thus, the velocity vector \( \mathbf{v}(t) \) is \[ \mathbf{v}(t) = \frac{2}{t+1} \mathbf{i} + 2t \mathbf{j} + t \mathbf{k}. \]This vector helps pinpoint exactly how the particle is moving at any time \( t \).

Acceleration Vector

The acceleration vector indicates how the velocity of a particle changes over time, which means it is the derivative of the velocity vector. By finding this, you can understand how the motion speed or direction changes as time progresses.For the velocity function already found:\[ \mathbf{v}(t) = \frac{2}{t+1} \mathbf{i} + 2t \mathbf{j} + t \mathbf{k}, \]differentiate each component:

• The \( \mathbf{i} \)-component's derivative, \( \frac{2}{t+1} \), becomes \(-\frac{2}{(t+1)^2} \).
• The \( \mathbf{j} \)-component's derivative, \( 2t \), becomes \( 2 \).
• The \( \mathbf{k} \)-component's derivative, \( t \), becomes \( 1 \).

The resulting acceleration vector \( \mathbf{a}(t) \) is:\[ \mathbf{a}(t) = -\frac{2}{(t+1)^2} \mathbf{i} + 2 \mathbf{j} + \mathbf{k}. \]This vector reveals any changes in the particle's motion pattern, making it a vital tool in dynamic analysis.

Direction of Motion

The direction of motion of a particle is represented by a unit vector. This is derived from the velocity vector, providing insight into the precise path of the particle without regard to how quickly it is moving. At any specific point in time, dividing the velocity vector by its magnitude results in a unit vector.At \( t = 1 \), the velocity vector \( \mathbf{v}(1) = \mathbf{i} + 2\mathbf{j} + \mathbf{k} \). To find the direction:First, calculate the magnitude of \( \mathbf{v}(1) \):\[ ||\mathbf{v}(1)|| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}. \]Then, divide each component by this magnitude:

• The \( \mathbf{i} \)-component: \( \frac{1}{\sqrt{6}} \)
• The \( \mathbf{j} \)-component: \( \frac{2}{\sqrt{6}} \)
• The \( \mathbf{k} \)-component: \( \frac{1}{\sqrt{6}} \)

Thus, the direction of motion is captured by:\[ \mathbf{e} = \frac{1}{\sqrt{6}} \mathbf{i} + \frac{2}{\sqrt{6}} \mathbf{j} + \frac{1}{\sqrt{6}} \mathbf{k}. \]Knowing this allows you to visualize which direction the particle is heading in, irrespective of its speed.

Speed Calculation

Speed is the scalar measure of how fast an object is moving, determined by the magnitude of the velocity vector. Unlike velocity, speed has no direction. It's a crucial metric in evaluating the rate of movement.To calculate speed at \( t = 1 \) based on the previously obtained velocity vector \( \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} + \mathbf{k} \), use the formula:\[ \text{Speed} = ||\mathbf{v}(1)|| = \sqrt{(1)^2 + (2)^2 + (1)^2}. \]This results in:\[ \text{Speed} = \sqrt{6}. \]Thus, the speed of the particle at \( t = 1 \) is \( \sqrt{6} \). A clear understanding of speed is essential for determining how rapidly the particle covers distance at any specific moment.