91Ó°ÊÓ

A helical path is like a spring, twisting and turning in a three-dimensional space. It spirals around a central axis, like the familiar shape of a slinky. Understanding the path is crucial to finding the mass of an object shaped this way. For the given problem, the path is described by the vector function:

• \(\mathbf{r}(t)=3 \cos t \mathbf{i} + 3 \sin t \mathbf{j} + 2t \mathbf{k}\)

This equation incorporates trigonometric functions for the x and y components, creating a circular motion in the horizontal plane. The z-component, \(2t\), extends linearly, lifting the spiral upwards as it revolves around its axis. The helix makes two full circles, or rotations, which completes the image of a spring-like path in space. These elements combined help us understand how to represent the object's path for further calculations.

A density function describes how mass is distributed along an object. In this case, we are given a simple density function that depends on the z-coordinate: \(\rho(x, y, z) = z\). This means that the object's density varies with its height in the vertical direction.
For the helical path, where \(z = 2t\), the density function becomes a function of \(t\), the parameter of the path. With this information, we can express the density as:

• \(\rho(t) = 2t\)

This linear relationship implies that as the parameter \(t\) increases, the density along the spring’s path increases proportionally. Understanding this transition from a spatial coordinate to a parameter-based density function is essential for accurately calculating the mass along the path.

The arc length differential, denoted as \(ds\), is a small segment of the curve's length. It's an important way to measure how long a curve or path is and helps us later calculate the mass or other physical properties related to the path. For a parametric path in three-dimensional space, the arc length differential is computed with:

• \(ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2} \, dt\)

In our specific example, the derivative components \(\frac{dx}{dt} = -3\sin t\), \(\frac{dy}{dt} = 3\cos t\), and \(\frac{dz}{dt} = 2\) are calculated from the path function.
Plugging these into the arc length differential formula gives:

• \(ds = dt\sqrt{(3 \sin t)^2 + (-3 \cos t)^2 + 2^2} = dt\sqrt{10}\)

This result simplifies our calculations and is crucial for defining how we integrate along the curve.

Definite integrals help us sum up small quantities over a specified interval, giving a total result. To find the mass of the spring, we perform a definite integral over the arc length, accounting for density variations along the path. Mass as an integral over the arc length is given by:

• \(Mass = \int_{0}^{4\pi} \rho(t) \, ds\)

The density \(\rho(t) = 2t\sqrt{10}\) and the arc length differential \(ds = dt\sqrt{10}\) are substituted into this integral, resulting in:

• \(Mass = \int_{0}^{4\pi} 2t\sqrt{10} \, dt = \sqrt{10} \left[ \frac{2t^2}{2} \right]_0^{4\pi}\)

The calculus steps simplify to provide the total mass:

• \(Mass = 8\pi^2\sqrt{10}\)

This integral effectively sums up the continuously varying mass along the helical path from \(t=0\) to \(t=4\pi\), providing a complete view of the mass distribution.