91Ó°ÊÓ

In this exercise, the concept of a density function is crucial for determining the mass of a parametrized curve, specifically a wire. The density function, denoted by \( \rho(\theta) \), describes how mass is distributed along the wire for each unit of arc length. In our case, the density function is given by \( \rho(\theta) = \frac{2\theta}{\pi} + 1 \). This means that at any given point on the curve (or wire), the density is determined by the angle \( \theta \).

• The density function \( \rho(\theta) \) changes with the angle \( \theta \), indicating that different sections of the wire have different masses per unit length.
• The function starts at 1 when \( \theta = 0 \) and increases linearly with \( \theta \), reaching its maximum at \( 0 \leq \theta \leq \pi \).

This variable nature of the density function is what gives the wire its total mass when integrated over the entire curve length. Understanding the density function allows us to set up an integral that accounts for varying mass across the wire, ultimately helping us to calculate total mass.

To find the mass of a curve, the arc length element \( ds \) is a key component. Essentially, it represents a small segment of the arc length along the curve, which is used to calculate mass or other integral-based quantities. In our exercise, the arc length element is derived from the parametrization of the curve \( \mathbf{r}(\theta) = \langle\cos \theta, \sin \theta\rangle \).

• For a parametrized curve, \( ds \) can be expressed by taking the derivative of \( \mathbf{r}(\theta) \) with respect to \( \theta \) and finding its magnitude.
• In this case, the derivative is \( \frac{d\mathbf{r}}{d\theta} = \langle-\sin \theta, \cos \theta\rangle \).
• Calculating the magnitude, we find it is always 1 for this specific curve, simplifying \( ds \) to just \( d\theta \).

Therefore, the arc length element \( ds = d\theta \) indicates that each infinitesimal segment of arc length is directly equal to a change in \( \theta \). This elegant simplification makes it straightforward to integrate over the curve.

Integration in calculus is a fundamental tool used to compute quantities like area under a curve or, in this case, the mass of a parametrized curve (wire). Here, integration is applied to sum up contributions of mass along the wire, considering its density function \( \rho(\theta) \) and arc length element \( ds \).

• Our task is to set up and evaluate the integral \( M = \int_{0}^{\pi} \rho(\theta)\,ds \), where \( \rho(\theta) = \frac{2\theta}{\pi} + 1 \) and \( ds = d\theta \).
• This integral represents a continuous sum of mass units across the wire length from \( \theta = 0 \) to \( \theta = \pi \).

The integral is broken into two components:

• \( \int_{0}^{\pi} \frac{2\theta}{\pi} d\theta \) evaluates the contribution from the variable part of the density function.
• \( \int_{0}^{\pi} d\theta \) computes the contribution from the constant part of the density function.

Evaluating these, we sum the results to find the mass \( M = 2\pi \). Integration here elegantly sums up the mass distributed along this circular path, capturing both the constant and linearly varying elements of the density.