Question

Consider the trajectory given by the position function $$\mathbf{r}(t)=\left\langle 50 e^{-t} \cos t, 50 e^{-t} \sin t, 5\left(1-e^{-t}\right)\right), \quad \text { for } t \geq 0$$ a. Find the initial point \((t=0)\) and the "terminal" point \(\left(\lim _{t \rightarrow \infty} \mathbf{r}(t)\right)\) of the trajectory. b. At what point on the trajectory is the speed the greatest? c. Graph the trajectory.

Short answer

Question: Find the initial and terminal points, the point with the greatest speed, and graph the trajectory of the position function \( \mathbf{r}(t) = \langle 50e^{-t} \cos t, 50e^{-t} \sin t, 5(1 - e^{-t})\rangle \). Solution: a. Initial point: \((50, 0, 0)\) and Terminal point: \((0, 0, 5)\) b. Point with the greatest speed: \((50, 0, 0)\) c. Use a 3D graphing tool to plot the trajectory of the position function \(\mathbf{r}(t)\).

Step-by-step solution

a. Initial and Terminal Points

Evaluate the position function \(\mathbf{r}(t)\) at \(t=0\): $$ \mathbf{r}(0) = \left\langle 50 e^{-0} \cos 0, 50 e^{-0} \sin 0, 5\left(1-e^{-0}\right)\right\rangle = \left\langle 50, 0, 0 \right\rangle $$ The initial point is \((50, 0, 0)\). Now find the limit of the position function as \(t\rightarrow\infty\): $$ \lim_{t\rightarrow\infty}\mathbf{r}(t) = \lim_{t\rightarrow\infty} \left\langle 50 e^{-t} \cos t, 50 e^{-t} \sin t, 5(1 - e^{-t})\right\rangle = \left\langle 0, 0, 5 \right\rangle $$ The terminal point is \((0, 0, 5)\).

b. Point with Greatest Speed

Differentiate the position function with respect to \(t\) to find the velocity function: $$ \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \left\langle (-50 e^{-t}\cos t - 50 e^{-t}\sin t), (-50 e^{-t}\sin t + 50 e^{-t}\cos t), 5 e^{-t} \right\rangle $$ Now find the speed function by finding the magnitude of the velocity function: $$ |\mathbf{v}(t)| = \sqrt{(-50 e^{-t}\cos t - 50 e^{-t}\sin t)^2 + (-50 e^{-t}\sin t + 50 e^{-t}\cos t)^2 + (5 e^{-t})^2} $$ Simplify the expression: $$ |\mathbf{v}(t)| = 50e^{-t}\sqrt{2\cos^2t+2\sin^2t+0.01} $$ To find the critical points, find the first derivative of the speed function, and set it equal to zero or analyze where it is undefined: $$ \frac{d|\mathbf{v}(t)|}{dt} = 0 $$ Since the simplified expression doesn't have any critical points that could lead to local maxima or minima, the speed is greatest at the initial point. Therefore, we don't need to find the first derivative of the speed function. The point on the trajectory where the speed is the greatest is \((50, 0, 0)\), the initial point.

c. Graph the Trajectory

To graph the trajectory, plot the position function \(\mathbf{r}(t)\) for the domain \(t\geq0\). The graph will start at the initial point \((50, 0, 0)\) and moves towards the terminal point \((0, 0, 5)\) in a spiral fashion while gradually increasing in the \(z\) direction. The graph has a parametric form, so use a 3D graphing tool to plot the function. Recap: a. Initial point: \((50, 0, 0)\) and Terminal point: \((0, 0, 5)\) b. Point with greatest speed: \((50, 0, 0)\) c. Graph the trajectory using a 3D graphing tool for the function \(\mathbf{r}(t)\).

Key concepts

Velocity Function

In the context of parametric equations, especially for trajectories in 3D space, the velocity function plays a crucial role in understanding how an object moves. The velocity function, denoted as \( \mathbf{v}(t) \), is the derivative of the position function \( \mathbf{r}(t) \) with respect to time, \( t \). This function provides us with information on both the direction and the speed at which an object is moving at any given time.

To find the velocity function, we differentiate each component of the position vector. In this specific problem, the position function is given by:

• \( \mathbf{r}(t) = \langle 50 e^{-t} \cos t, 50 e^{-t} \sin t, 5(1 - e^{-t}) \rangle \)

Differentiating this gives the velocity function as:

• \( \mathbf{v}(t) = \langle -50 e^{-t} (\cos t + \sin t), -50 e^{-t} (\sin t - \cos t), 5 e^{-t} \rangle \)

This function tells us how fast and in which direction the object is moving at any time \( t \). The velocity vector components represent changes in the \( x \), \( y \), and \( z \) dimensions.

Initial and Terminal Points

Understanding the initial and terminal points of a trajectory helps in visualizing the path an object takes in space, which is essential for many physical problems. The initial point corresponds to the position of the object at \( t = 0 \), while the terminal point is the position as \( t \rightarrow \infty \).

For our trajectory:

• The initial point is found by evaluating the position function at \( t = 0 \):
  \( \mathbf{r}(0) = \langle 50, 0, 0 \rangle \).
• The terminal point is determined by taking the limit as \( t \) approaches infinity:
  \( \lim_{t \rightarrow \infty} \mathbf{r}(t) = \langle 0, 0, 5 \rangle \).

These points highlight the start and end of the motion in three-dimensional space, providing key observations into where the object begins and where it eventually settles in its motion.

Speed Analysis

Speed analysis of a trajectory involves understanding how fast an object moves along its path. Speed is the magnitude of the velocity vector, calculated using the formula \( |\mathbf{v}(t)| = \sqrt{v_x^2 + v_y^2 + v_z^2} \). This provides a scalar value indicating how fast the object is moving without regard to direction.

In our example:

• The simplified speed function is \( |\mathbf{v}(t)| = 50e^{-t}\sqrt{2} \sqrt{1 + 0.005} = 50 e^{-t} \sqrt{2.01} \).

Analyzing this function shows that the speed decreases over time due to the exponential factor \( e^{-t} \). Initially, at \( t = 0 \), the speed is highest, and it diminishes as \( t \) increases.

Since no critical points were detected that result in speed greater than the initial, it's concluded that the greatest speed is precisely at the starting point \( t = 0 \). This helps to determine where along the trajectory the object moves most rapidly.

3D Graphing

Visualizing trajectories using 3D graphing is an excellent way to comprehend an object's motion within three dimensions. By plotting the parametric equations of the position function, we gain a comprehensive view of how the object moves over time.

In our scenario:

• 3D graphing allows us to plot \( \mathbf{r}(t) = \langle 50 e^{-t} \cos t, 50 e^{-t} \sin t, 5(1 - e^{-t}) \rangle \) and see its development through space.
• The trajectory begins at the initial point \( (50, 0, 0) \), spiraling inward while climbing towards the terminal point \( (0, 0, 5) \).
• The path taken is spiral-like in nature, with the \( z \)-component incrementing gradually.

Using a 3D graphing tool enriches our understanding by displaying the trajectory visually, making it easier to connect mathematical concepts with tangible geometrical interpretations.