Problem 107 - Calculus 3

Question

Find the length of one turn of the helix given by \(\mathbf{r}(t)=\frac{1}{2} \cos t \mathbf{i}+\frac{1}{2} \sin t \mathbf{j}+\sqrt{\frac{3}{4}} t \mathbf{k}\)

Step-by-Step Solution

Answer

The length of one turn of the helix is \(2\pi\).

Step 1: Determine the parameter range for one complete turn

For a helix, one complete turn corresponds to the parameter going from 0 to \(2\pi\). This is because a complete circular rotation in terms of \(\cos t\) and \(\sin t\) occurs within this interval.

Step 2: Calculate the derivative of the position vector

Take the derivative of \(\mathbf{r}(t)= \frac{1}{2}\cos t \mathbf{i} + \frac{1}{2}\sin t \mathbf{j} + \sqrt{\frac{3}{4}}t \mathbf{k} \) with respect to \(t\): \(\mathbf{r}'(t) = -\frac{1}{2}\sin t \mathbf{i} + \frac{1}{2}\cos t \mathbf{j} + \sqrt{\frac{3}{4}} \mathbf{k}\).

Step 3: Find the magnitude of the derivative

The magnitude of \(\mathbf{r}'(t)\) is calculated using the formula \(|\mathbf{r}'(t)| = \sqrt{\left(-\frac{1}{2}\sin t\right)^2 + \left(\frac{1}{2}\cos t\right)^2 + \left(\sqrt{\frac{3}{4}}\right)^2}\). Simplifying this, we get \( |\mathbf{r}'(t)| = \sqrt{ \frac{1}{4}(\sin^2 t + \cos^2 t) + \frac{3}{4} } = \sqrt{1}\) since \(\sin^2 t + \cos^2 t = 1\).Therefore, \(|\mathbf{r}'(t)| = 1\).

Step 4: Integrate the magnitude over one period

The arc length \(L\) of one turn is given by the integral of the magnitude of the derivative over one period: \( L = \int_0^{2\pi} 1 \, dt = [t]_0^{2\pi} = 2\pi \).

Key Concepts

Parametric Equations

Parametric equations are a useful tool in mathematics and physics, especially in representing curves in space. Unlike standard functions that express one variable in terms of another, parametric equations use one or more parameters to describe a set of related quantities. For a helix, these equations become a vital way to describe the three-dimensional curve in terms of a single parameter, usually time \(t\).

In this problem, the position vector \(\mathbf{r}(t)\) is given in terms of \(t\) as:

• \( \frac{1}{2} \cos t \mathbf{i} \) for the x-coordinate,
• \( \frac{1}{2} \sin t \mathbf{j} \) for the y-coordinate,
• \( \sqrt{\frac{3}{4}}t \mathbf{k} \) for the z-coordinate.

This allows us to visualize how the helix progresses along each axis as the parameter \(t\) changes.

Parametric equations are particularly powerful because they allow for easy manipulations and derivations, like finding the arc length of curves through differentiation. This makes them a cornerstone concept in studying motion, trajectories, and other spatial phenomena.

Arc Length

Arc length is the measure of the distance along a curved line, which is different from the straight-line distance between two endpoints. For parametrically defined curves, such as a helix, determining the arc length involves integrating the speed function over a specified interval of the parameter.

In calculating the arc length of the helix, we first found the derivative of the position vector \(\mathbf{r}'(t)\). Next, we computed its magnitude \(|\mathbf{r}'(t)|\), which represents the speed of the moving point on the curve. In this case, the magnitude simplified to a constant, \(1\).

Finally, to find the length of one turn of the helix, you integrate the constant magnitude from \(0\) to \(2\pi\), corresponding to one complete turn around the circular path. This gives an arc length of \(2\pi\), indicating how far the point travels in one full revolution along the helix.

Vector Calculus

Vector calculus is an essential branch of mathematics that deals with vector fields and operations on vectors. In this exercise, vector calculus helps in determining the arc length of a helix, which involves finding the derivative and magnitude of vector functions.

Taking the derivative of the position vector \(\mathbf{r}(t)\) is a vector calculus operation that gives us the velocity \(\mathbf{r}'(t)\) of the point on the helix. This derivative provides the components of the rate of change in each spatial direction. By computing the magnitude of this derivative, we measure the rate of travel, or speed, along the curve.

Vector calculus is not just about computing derivatives; it also involves integrating vector functions to find quantities such as arc lengths or areas. Understanding these fundamental operations is crucial for analyzing motion and forces in physics and engineering.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in describing periodic phenomena, such as circular and helical paths. They are crucial in parametric equations to represent circular motion.

In the exercise, the helix's x and y coordinates involve the trigonometric functions \(\cos t\) and \(\sin t\). The periodic nature of these functions ensures that the path repeats every \(2\pi\) units, corresponding to a full circle.

These functions help express how the helix oscillates in the x-y plane while simultaneously rising along the z-axis (z given by \(\sqrt{\frac{3}{4}}t\)). The relation \(\sin^2 t + \cos^2 t = 1\) simplifies the magnitude of the derivative during calculations, reflecting their power in simplifying and solving complex expressions as seen in the arc length derivation.

Understanding trigonometric functions is therefore vital in handling periodic motion problems, offering insight into angles, lengths, and periodicity.