Problem 39 - Calculus Early Transcendentals

Question

Find the mass of a thin wire shaped in the form of the helix \(x=3 \cos t, y=3 \sin t, z=4 t(0 \leq t \leq \pi / 2)\) if the density function is \(\delta=k x /\left(1+y^{2}\right)(k>0)\)

Step-by-Step Solution

Answer

The mass of the wire is \( \frac{5\pi k}{4} \).

Step 1: Understand the problem

We need to find the mass of a thin wire shaped according to the given parametric equations for a helix. The mass can be calculated as the integral of the density function multiplied by the differential arc length.

Step 2: Set up the integral expression

The mass of the wire is given by the integral \( M = \int_C \delta \, ds \), where \( \delta \) is the density function \( \frac{kx}{1+y^2} \). The differential arc length \( ds \) can be expressed as \( \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \ dt \).

Step 3: Calculate the derivatives

First, calculate \( \frac{dx}{dt} = -3 \sin t \), \( \frac{dy}{dt} = 3 \cos t \), and \( \frac{dz}{dt} = 4 \).

Step 4: Find the differential arc length

Compute \( ds = \sqrt{(-3 \sin t)^2 + (3 \cos t)^2 + 4^2} \ dt = \sqrt{9 \sin^2 t + 9 \cos^2 t + 16} \ dt = \sqrt{9 + 16} \ dt = 5 \ dt \), using \( \sin^2 t + \cos^2 t = 1 \).

Step 5: Substitute into the mass integral

Substitute the density function and \( ds \) into the mass integral to get \( M = \int_0^{\pi/2} \left( \frac{k(3 \cos t)}{1 + (3 \sin t)^2} \right) 5 \, dt \).

Step 6: Simplify the integrand

The integrand simplifies to \( 15k \int_0^{\pi/2} \frac{\cos t}{1 + 9 \sin^2 t} \, dt \).

Step 7: Solve the integral

To solve the integral, perform a substitution \( u = \sin t \), such that \( du = \cos t \, dt \). Then the integral becomes \( 15k \int_0^1 \frac{1}{1+9u^2} \, du \), which evaluates to \( 15k \times \frac{1}{3}\left[ \tan^{-1}(3u) \right]_0^1 \).

Step 8: Evaluate the definite integral

Upon evaluating, substitute the limits to get \( 15k \times \frac{1}{3} \left( \tan^{-1}(3 \times 1) - \tan^{-1}(3 \times 0) \right) = 15k \times \frac{\pi}{12} \).

Step 9: Simplify the final expression

Simplify the result to obtain the final mass = \( \frac{5\pi k}{4} \).

Key Concepts

Helix

A helix is a 3-dimensional curve that spirals around an axis with a consistent radius. Imagine a spring or a corkscrew, and that's essentially the shape of a helix. A helix can be described using parametric equations. These equations define the x, y, and z coordinates as a function of parameter 't', which typically represents time or angle:

• For the exercise given, the helix is described by:
• \( x = 3 \cos t \)
• \( y = 3 \sin t \)
• \( z = 4t \)

This helix curls around the z-axis and extends vertically. The parameters such as the radius and vertical rise help determine the specific form of the helix.
The helix is limited to the part from \( t = 0 \) to \( t = \pi/2 \), capturing this segment of the spiral.

Density Function

In physics and mathematical modeling, a density function describes how mass is distributed along an object, such as a wire in this case. For this wire, the density function is given by \[ \delta = \frac{k x}{1 + y^2} \]where:

• \( k \) is a constant that affects the density;
• \( x \) and \( y \) coordinate values influence how density varies along the helix.

The function distinguishes how mass is located along different points in the helix based on its x-coordinate, with heavier concentration where x is larger and influenced by the term \( 1 + y^2 \). This results in a non-uniform density along the wire.

Parametric Equations

Parametric equations provide a powerful means to describe geometric entities in space through parameters, often simplifying complex relationships. They define a curve not as a single equation involving x and y, but by expressing each coordinate as a function of a parameter t.
For the helix discussed:

• \( x = 3 \cos t \)
• \( y = 3 \sin t \)
• \( z = 4t \)

Each equation describes how that particular coordinate varies as the parameter t changes. This representation is particularly convenient for curves that loop back on themselves like a helix, as it captures the motion and position in a straightforward, continuous format. Plus, using these types of equations often simplifies the process of calculating properties like length or mass associated with the curve.

Arc Length Integral

The arc length integral is a fundamental concept used to determine the length of a curve described by parametric equations over a specified interval. In our problem, it forms part of the process to calculate the mass of the helical wire. The integral used here is:\[\int_C \delta \ ds\]where \( ds \), the differential arc length, is crucial because it describes a small segment of the curve over which to integrate.
Here, \( ds \) is computed using the derivatives of the parametric equations:

• \( \frac{dx}{dt} = -3 \sin t, \quad \frac{dy}{dt} = 3 \cos t, \quad \frac{dz}{dt} = 4 \)
• Combining these gives \( ds = \sqrt{(-3 \sin t)^2 + (3 \cos t)^2 + 4^2} \, dt = 5 \, dt \)

This represents the infinitesimal change in the curve's length as 't' changes, and is used to calculate the total mass by integrating with respect to the density function over the interval from 0 to \( \pi/2 \). This approach efficiently combines geometry and calculus.