Problem 33 - Essential Calculus

Question

In Exercises 33-36, find the length of the curve over the given interval. $$ \begin{array}{ll} \text { Polar Equation } & \text { Interval } \\ \hline r=a & 0 \leq \theta \leq 2 \pi \end{array} $$

Step-by-Step Solution

Answer

The length of the curve over the given interval is \(2a\pi\)

Step 1: Recall the formula for the arc length in polar coordinates.

The arc length (L) in polar coordinates is calculated using the formula \( L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta \), where r is the polar function, and \( \alpha \) and \( \beta \) are the limits of the interval. Here, \( r = a \) (a constant) so the derivative \( \frac{dr}{d\theta} \) will be zero.

Step 2: Substitute the given values into the formula.

When you substitute the given values into the formula, you will have: \( L = \int_{0}^{2\pi} \sqrt{a^2 + \left(0\right)^2} d\theta = \int_{0}^{2\pi} a d\theta \). This turns out to be an integral of a constant, which is a straightforward problem.

Step 3: Compute the integral.

The integral of a constant 'a' over the interval from 0 to \( 2\pi \) is \( a * (\theta \mid_{0}^{2\pi}) = a * (2\pi - 0) = 2a\pi \). Therefore, the length of the curve over the given interval is \( 2a\pi \)

Key Concepts

Polar Equations

Polar equations describe curves on a plane using polar coordinates. In this system, each point on a curve is determined by an angle \( \theta \) and a radius \( r \), forming a relation \( r = f(\theta) \). This is different from Cartesian coordinates where points are described using \( x \) and \( y \).

• Radius \( r \): The distance from the origin (center of the coordinate system) to a point on the curve.
• Angle \( \theta \): The angle measured from the positive x-axis to the line connecting the origin to the point.

For example, in the given problem, the polar equation is \( r = a \), where \( a \) is a constant. This implies that the radius is constant and the curve is a circle centered at the origin with radius \( a \). Calculating arc lengths in polar coordinates often involves working with such relationships to understand the shape and extent of curves.

Integrals in Calculus

Integrals in calculus are fundamental tools for analyzing and measuring various kinds of quantities. In the context of arc length, integrals help in determining the length of a curve defined by a polar equation.
An integral can be visualized as the summation of infinitely small lengths, areas, or volumes. When you integrate a function, you're effectively calculating the total quantity under the curve that it describes, between two points.

• Definite Integrals: Used for finding the exact value of area, length, or volume within a specific interval. They have limits of integration like \( \alpha \) and \( \beta \).
• Indefinite Integrals: Generally used for finding the antiderivative of a function, which helps in solving problems involving accumulations.

In this exercise, a definite integral is used: \( L = \int_{0}^{2\pi} a \, d\theta \). This integral sums up the tiny contributions of length from the curve spanning the interval \( 0 \) to \( 2\pi \). The process involves recognizing the length's components and assembling them through integration.

Curve Length Calculation

Calculating the curve's length in polar coordinates involves understanding the relationship between radius, angle, and the changes in these values over a specified interval. The arc length formula for polar coordinates is crucial here:
\[ L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta \]
To find the arc length, we'll need:

• Understanding the function \( r = f(\theta) \): How the radius changes with the angle.
• The derivative \( \frac{dr}{d\theta} \): Reflects how quickly \( r \) changes with respect to \( \theta \). In cases where \( r \) is constant, this derivative is zero.

In our problem, \( r \) is constant: \( r = a \). Therefore, \( \frac{dr}{d\theta} = 0 \). Thus, the formula simplifies to \( L = \int_{0}^{2\pi} a \, d\theta \). This results in \( 2a\pi \), representing the circular path length around the origin, confirming the calculation of the curve's length efficiently and accurately. Such processes in calculus transform complex geometric investigations into manageable computations.