Problem 93 - Calculus Volume 3

Question

A spring is made of a thin wire twisted into the shape of a circular helix \(x=2 \cos t, y=2 \sin t, z=t .\) Find the mass of two turns of the spring if the wire has constant mass density.

Step-by-Step Solution

Answer

The mass of the spring is \( \rho \times 4\pi \sqrt{5} \).

Step 1: Understanding the Helix Equations

The helix is expressed parametrically as \( x = 2 \cos t \), \( y = 2 \sin t \), and \( z = t \). This describes a circular helix with a radius of \( 2 \) in the \( xy \)-plane and a height increase along the \( z \)-axis as \( t \) increases.

Step 2: Determine the Value of \( t \) for Two Turns

Since the helix completes one turn when \( t \) goes from 0 to \( 2\pi \), two turns corresponds to \( t \) going from 0 to \( 4\pi \).

Step 3: Parametric Representation of the Curve

The parameter \( t \) traces the path of the helix along \( x(t), y(t), \) and \( z(t) \). We need to compute the arc length of this path over one cycle, \( t \in [0, 4\pi]\).

Step 4: Arc Length of the Helix

The arc length of a parametric curve given by \( (x(t), y(t), z(t)) \) is calculated using:\[L = \int_{t_1}^{t_2} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt\]For this helix, the derivatives are:\( \frac{dx}{dt} = -2 \sin t \), \( \frac{dy}{dt} = 2 \cos t \), \( \frac{dz}{dt} = 1 \).

Step 5: Calculate the Arc Length

Substitute the derivatives into the arc length formula:\[L = \int_{0}^{4\pi} \sqrt{(-2 \sin t)^2 + (2 \cos t)^2 + 1^2} \, dt = \int_{0}^{4\pi} \sqrt{4 \sin^2 t + 4 \cos^2 t + 1} \, dt\]Use the identity \( \sin^2 t + \cos^2 t = 1 \) to simplify:\[L = \int_{0}^{4\pi} \sqrt{5} \, dt = \sqrt{5} \cdot [t]_{0}^{4\pi} = \sqrt{5} \cdot 4\pi\]

Step 6: Compute the Mass of the Spring

If the wire has constant mass density \( \rho \), the mass of the spring is proportional to its length:\[ \text{Mass} = \rho \times L = \rho \times 4\pi \sqrt{5} \]

Step 7: Conclusion: Compute the Final Mass

The final mass of two turns of the spring is given by multiplying the constant mass density \( \rho \) by the calculated arc length \( 4\pi \sqrt{5} \).

Key Concepts

Parametric Equations

Parametric equations are mathematical expressions where a set of variables are expressed as functions of one or more independent parameters. In this exercise, the parametric equations define the position of points along the curve of the helix. Specifically:

• For the x-coordinate: \( x = 2 \cos t \)
• For the y-coordinate: \( y = 2 \sin t \)
• For the z-coordinate: \( z = t \)

These parametric equations describe a curve in three-dimensional space. The parameter \( t \) takes on values along the curve, influencing its position in 3D space over time. Each of these equations captures a different aspect of the helical shape:

• The expressions for \( x \) and \( y \) suggest circular motion in the \( xy \)-plane with radius 2.
• The z-coordinate increases linearly, indicating a constant upward motion.

Together, these aspects form the familiar coiled shape of a helix.

Arc Length Calculation

Knowing how to calculate the arc length of a parametric curve is crucial for understanding physical properties such as mass along the curve. The arc length \( L \) of a parametric curve described by \( (x(t), y(t), z(t)) \) can be determined using the following formula:\[L = \int_{t_1}^{t_2} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt\]For the given helical spring:

• \( \frac{dx}{dt} \) becomes \( -2 \sin t \)
• \( \frac{dy}{dt} \) becomes \( 2 \cos t \)
• \( \frac{dz}{dt} \) becomes \( 1 \)

By substituting these derivatives into the formula, and simplifying using the identity \( \sin^2 t + \cos^2 t = 1 \), the arc length for two turns (from \( t = 0 \) to \( t = 4\pi \)) measures \( 4\pi \sqrt{5} \). This calculation is essential because the length of the curve directly impacts the mass of the helical spring.

Mass Density

Mass density is a fundamental concept in physics which measures how much mass is contained in a given length or volume of material. In this exercise, the wire forming the helical spring possesses constant mass density, denoted by \( \rho \). With a constant mass density, the mass of the spring is simply the product of the density and the total length of the wire.To find the mass of the spring:

• First, calculate the arc length of the helix, which we've already found to be \( 4\pi \sqrt{5} \).
• Then, multiply this length by the mass density \( \rho \): \( \text{Mass} = \rho \times 4\pi \sqrt{5} \).

This approach makes it seamless to understand how the geometry of the spring and its material properties combine to define its overall mass. It's an elegant demonstration of how geometric and physical properties intermingle in real-world applications.