Question

Find the displacement, distance traveled, average velocity and average speed of the described object on the given interval. An object with position function \(\vec{r}(t)=\langle 2 \cos t, 2 \sin t, 3 t\rangle\), where distances are measured in feet and time is in seconds, on \([0,2 \pi]\).

Short answer

Displacement: \(\langle 0, 0, 6\pi \rangle\); Distance: \(2\pi\sqrt{13}\) feet; Average velocity: \(\langle 0, 0, 3 \rangle\) ft/s; Average speed: \(\sqrt{13}\) ft/s.

Step-by-step solution

Determine Displacement

Displacement is the change in position from the initial to the final point. We find \(\vec{r}(0)\) and \(\vec{r}(2\pi)\) first. Then, \(\text{Displacement} = \vec{r}(2\pi) - \vec{r}(0)\).1. Calculate \(\vec{r}(0) = \langle 2\cos(0), 2\sin(0), 3\cdot0 \rangle = \langle 2, 0, 0 \rangle\).2. Calculate \(\vec{r}(2\pi) = \langle 2\cos(2\pi), 2\sin(2\pi), 3\cdot2\pi \rangle = \langle 2, 0, 6\pi \rangle\).3. \(\text{Displacement} = \langle 2, 0, 6\pi \rangle - \langle 2, 0, 0 \rangle = \langle 0, 0, 6\pi \rangle\).

Calculate Distance Traveled

The distance traveled involves finding the arc length of the path defined by \(\vec{r}(t)\), which is the integral of the speed function \(\|\vec{r}'(t)\|\) over the interval \([0, 2\pi]\).1. Compute the derivative \(\vec{r}'(t) = \langle -2 \sin t, 2 \cos t, 3 \rangle\).2. Calculate the magnitude \(\|\vec{r}'(t)\| = \sqrt{(-2 \sin t)^2 + (2 \cos t)^2 + 3^2} = \sqrt{4\sin^2 t + 4\cos^2 t + 9} = \sqrt{4 + 9} = \sqrt{13}\).3. The distance is \(\int_{0}^{2\pi} \|\vec{r}'(t)\|\, dt = \int_{0}^{2\pi} \sqrt{13}\, dt = \sqrt{13} \cdot 2\pi = 2\pi\sqrt{13} \text{ feet}\).

Calculate Average Velocity

Average velocity is the displacement divided by the total time.1. The displacement vector is \(\langle 0, 0, 6\pi \rangle\).2. Time elapsed is \([0, 2\pi]\), which is \(2\pi\) seconds.3. \(\text{Average velocity} = \frac{\langle 0, 0, 6\pi \rangle}{2\pi} = \langle 0, 0, 3 \rangle\) feet/second.

Calculate Average Speed

Average speed is the total distance traveled divided by the total time.1. Total distance traveled is \(2\pi\sqrt{13}\) feet (from Step 2).2. Time interval is \(2\pi\) seconds.3. \(\text{Average speed} = \frac{2\pi\sqrt{13}}{2\pi} = \sqrt{13}\) feet/second.

Key concepts

Displacement

Displacement is one of the primary concepts in vector calculus and can be understood as the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction.
When solving problems involving displacement, you focus on the initial and final positions of the object. It's calculated by subtracting the initial position vector from the final position vector.
In this exercise, the object's position is given by the vector function \( \vec{r}(t) = \langle 2 \cos t, 2 \sin t, 3t \rangle \), where \(t\) is time in seconds.
The initial time \(t = 0\) and final time \(t = 2\pi\), provides the position vectors \(\vec{r}(0) = \langle 2, 0, 0 \rangle\) and \(\vec{r}(2\pi) = \langle 2, 0, 6\pi \rangle\).
Displacement, therefore, is \(\vec{r}(2\pi) - \vec{r}(0) = \langle 0, 0, 6\pi \rangle\), indicating a change along the z-axis by \(6\pi\) feet, with no change in the x or y directions.

Distance Traveled

The distance traveled by an object is different from displacement, as it considers the entire path taken, not just the start and end points. This concept reflects the actual length of the path traveled by an object, and it is a scalar quantity, only having magnitude, not direction.
To calculate the distance traveled, you need to integrate the magnitude of the velocity vector, \(\|\vec{r}'(t)\|\), over the time interval.
This involves the computation of the derivative of the position function, yielding \(\vec{r}'(t) = \langle -2 \sin t, 2 \cos t, 3 \rangle\).
Its magnitude is \(\|\vec{r}'(t)\| = \sqrt{4 + 9} = \sqrt{13}\).
The integral of the speed over \([0, 2\pi]\) gives the length of the path: \( \int_{0}^{2\pi} \sqrt{13} \, dt = 2\pi\sqrt{13}\) feet, describing the total distance covered by the object.

Average Velocity

Average velocity highlights a way to measure how fast, and in which direction, an object moves from one point to another. It is a vector, summarizing an object's change in position over a specified time. Calculating average velocity involves dividing the displacement by the total time taken.
In the discussed exercise, displacement is the vector \(\langle 0, 0, 6\pi \rangle\), obtained over a time interval of \(2\pi\) seconds.
Using the formula: \( \text{Average velocity} = \frac{\text{Displacement}}{\text{Total Time}} \), the average velocity is \(\frac{\langle 0, 0, 6\pi \rangle}{2\pi} = \langle 0, 0, 3 \rangle\) feet per second.
This calculation tells us that while the object loops around in the xy-plane, its net movement is a straightforward climb along the z-axis at 3 feet per second.

Average Speed

Average speed is different from average velocity, as it only considers the path's total distance divided by the total time, regardless of direction. It's specifically a scalar quantity, focusing solely on how much ground an object covers within a time frame, not on the start and end points.
In this context, by knowing the total distance traveled (\(2\pi\sqrt{13}\) feet) and the duration (\(2\pi\) seconds), you calculate the average speed.
Apply the formula: \( \text{Average speed} = \frac{\text{Total Distance}}{\text{Total Time}} \).
The result is \(\frac{2\pi\sqrt{13}}{2\pi} = \sqrt{13}\) feet per second.
This gives a clear view that the object travels, on average, at \(\sqrt{13}\) feet per second along its path, irrespective of its direction at various intervals.