Question

The position of a particle at time \(t\) is given by \(\mathbf{r}=i \cos t+i \sin t+k t\). Show that both the speed and the magnitude of the acceleration are constant. Describe the motion.

Short answer

The speed is \sqrt{2} and the magnitude of the acceleration is 1. The particle moves in a helical path.

Step-by-step solution

Define the Position Vector

Given the position vector \(\( \mathbf{r} = i \cos t + i \sin t + k t \)\), separate the components: \( x(t) = \cos t \), \( y(t) = \sin t \), and \( z(t) = t \).

Find the Velocity Vector

Differentiate the position vector with respect to time \( t \): \(\frac{d \mathbf{r}}{dt} = \frac{d}{dt}(i \cos t + i \sin t + k t) \). This results in \(-i \sin t + j \cos t + k\).

Determine the Speed

Calculate the speed, which is the magnitude of the velocity vector: \( \text{Speed} = \lVert\mathbf{v}\rVert = \sqrt{(-i \sin t)^2 + (j \cos t)^2 + k^2} = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{1 + 1} = \sqrt{2} \).

Determine the Acceleration Vector

Differentiate the velocity vector with respect to time \( t \) to get the acceleration vector: \( \frac{d \mathbf{v}}{dt} = \frac{d}{dt}(-i \sin t + j \cos t + k) \). This results in \(-i \cos t - j \sin t\).

Calculate the Magnitude of the Acceleration

Calculate the magnitude of the acceleration vector: \( \text{Magnitude of acceleration} = \sqrt{(-i \cos t)^2 + (-j \sin t)^2} = \sqrt{\cos^2 t + \sin^2 t} = \sqrt{1} = 1 \).

Describe the Motion

The position vector shows the particle moving in a helical path. The circular motion in the \( x \) and \( y \) plane, combined with the constant upward motion along the \( z \)-axis represents a helix.

Key concepts

velocity vector

The velocity vector of a particle gives us the direction and speed at which the particle is moving at any moment.
The velocity vector is the first derivative of the position vector with respect to time. Given the position vector \( \mathbf{r} = i \cos t + j \sin t + k t\), the velocity vector can be found by differentiating this with respect to time \(t\).
This results in \(\mathbf{v} = -i \sin t + j \cos t + k\).
This tells us how fast and in which direction the particle is moving:

• \( \mathbf{-i \sin t}\): Speed in the x-direction.
• \( \mathbf{j \cos t}\): Speed in the y-direction.
• \( \mathbf{k}\): Constant speed in the z-direction.

To get the magnitude of the velocity vector, also known as speed, we use the formula:\[ \text{Speed} = \sqrt{(-i \sin t)^2 + (j \cos t)^2 + k^2} = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2} \]
This confirms the particle moves with a constant speed, giving us vital insight into the motion of the particle.

acceleration vector

The acceleration vector of a particle tells us how the velocity of the particle changes over time.
It’s the second derivative of the position vector with respect to time or the first derivative of the velocity vector. By differentiating \( \mathbf{v} = -i \sin t + j \cos t + k\ \, with respect to time \ t\) again, we get the acceleration vector:
\[ \mathbf{a} = -i \cos t - j \sin t\];
This tells us how the speed and direction of the particle change:

• \( \mathbf{-i \cos t}\): Change in x-direction speed.
• \( \mathbf{-j \sin t}\): Change in y-direction speed.
• No term for z-direction: No change in speed in the z-direction (constant climb).

To find the magnitude of the acceleration vector (acceleration), use the following formula:
\[ \text{Acceleration} = \sqrt{(-i \cos t)^2 + (-j \sin t)^2} = \sqrt{\cos^2 t + \sin^2 t} = \sqrt{1} = 1\];
This tells us the magnitude of acceleration remains constant, so the particle’s rate of speeding up (or slowing down) is consistent.

helical motion

Helical motion describes a particle moving in a spiral along a helical path.
In this problem, the position vector \(ewline \mathbf{r} = i \cos t + j \sin t + k t \ewline \) gives us a clear picture of this helical motion.
Here’s why:

• In the \ x \ and \ y \ planes: The components \ i \cos(t) \ewline \ewline \ and \ \ j \sin(t) \ewline \ewline denote circular motion. As \ t \ changes, the particle traces out a circle in the \ x \ and \ y \ planes.
• In the \ z \ plane:The \ k t term shows linear motion. As \ t \ changes, the particle moves at a constant speed upward along the \ z \ axis.

Combining these, at every moment, the particle traces out a point on a circle in the \ ( x, y) \ planet and moves linearly (vertically) along \ z \ axis, forming a helix.
This spiral, helix, spirals upwards steadily over time, demonstrating the three-dimensional aspect of motion.

constant speed

Constant speed means the magnitude of the velocity vector does not change.
To confirm this, look at the velocity vector \( \mathbf{v} = -i \sin t + j \cos t + k\);
The formula for speed, the magnitude of the velocity vector, is:\[ \text {Speed} = \sqrt{(-i \sin t)^2 + (j \cos t)^2 + k^2} = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}\];
Since this results in a constant number \ \sqrt{2}, it confirms the particle moves at a constant speed.
This means all changes in direction do not affect how fast the particle moves. This is important in understanding overall motion.