Problem 84 - Calculus 3

Question

A particle travels along the path of a helix with the equation \(\mathbf{r}(t)=\cos (t) \mathbf{i}+\sin (t) \mathbf{j}+t \mathbf{k} .\) See the graph presented here: Find the following: Speed of the particle at any time

Step-by-Step Solution

Answer

The speed of the particle at any time is \( \sqrt{2} \).

Step 1: Find the Velocity Vector

To find the speed of the particle, we first need to determine the velocity vector by taking the derivative of the position vector \( \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + t \mathbf{k} \) with respect to \( t \). The derivative of this function is \( \mathbf{v}(t) = -\sin(t) \mathbf{i} + \cos(t) \mathbf{j} + \mathbf{k} \).

Step 2: Calculate the Speed

The speed of the particle at any time is the magnitude of the velocity vector. We calculate this using the formula for magnitude: \( \| \mathbf{v}(t) \| = \sqrt{[-\sin(t)]^2 + [\cos(t)]^2 + [1]^2} \). Simplifying this, we get \( \| \mathbf{v}(t) \| = \sqrt{\sin^2(t) + \cos^2(t) + 1} = \sqrt{1 + 1} = \sqrt{2} \).

Key Concepts

Velocity Vector

When studying particle motion, understanding the velocity vector is crucial. The velocity vector represents both the speed and the direction of the particle at a given point in time. To find the velocity vector, you need to differentiate the position vector with respect to time. In our exercise, the position vector for the particle traveling along the helix path is given by \[ \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + t \mathbf{k} \].Taking the derivative with respect to \( t \), we get the velocity vector:\[ \mathbf{v}(t) = \frac{d}{dt} [\cos(t) \mathbf{i} + \sin(t) \mathbf{j} + t \mathbf{k}] = -\sin(t) \mathbf{i} + \cos(t) \mathbf{j} + \mathbf{k} \].This demonstrates how the velocity vector's components are influenced by the trigonometric and linear components of the motion. The \(-\sin(t)\) and \( \cos(t) \) components show the rotational movement around the axis, while the constant \( \mathbf{k} \) indicates a steady rise along the z-direction.

Position Vector

In particle motion, the position vector describes the location of the particle at any given time. It is a vector function that provides coordinates in a vector space.For our helix equation in the exercise, the position vector is:\[ \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + t \mathbf{k} \].Let's break it down:

• \(\cos(t) \mathbf{i}\): Represents the horizontal motion in the x-direction.
• \(\sin(t) \mathbf{j}\): Describes the horizontal motion in the y-direction.
• \(t \mathbf{k}\): Represents the vertical movement along the z-axis.

The helix equation illustrates how the particle moves in a circular path on the xy-plane while simultaneously ascending along the z-axis. By analyzing the relationship between the function and position vector, you gain insight into how the particle behaves over time.

Magnitude of Velocity

The magnitude of velocity, often referred to as speed, is the measure of how fast a particle is moving regardless of direction.To find the magnitude of the velocity vector, use the formula:\[ \| \mathbf{v}(t) \| = \sqrt{[-\sin(t)]^2 + [\cos(t)]^2 + [1]^2} \].This simplifies to:\[ \| \mathbf{v}(t) \| = \sqrt{\sin^2(t) + \cos^2(t) + 1} = \sqrt{1 + 1} = \sqrt{2} \].Here’s why:

• \(\sin^2(t) + \cos^2(t) = 1\) due to the Pythagorean identity of trigonometric functions.
• The constant \(1\) added accounts for the constant motion in the z-direction.

Therefore, no matter the value of \(t\), the speed of the particle along the helix path remains constant at \(\sqrt{2}\). This insight shows that while the particle's direction changes, its pace doesn't vary, illustrating the beautiful consistency in helical motion.