Question

Use a calculator to find the approximate length of the following curves. The limaçon \(r=2-4 \sin \theta\)

Short answer

Answer: The approximate length of the curve is 9.848.

Step-by-step solution

Determine the range of values for θ

To determine the range of values for \(\theta\), observe the curve \(r = 2 - 4 \sin \theta\): - When \(\sin \theta = 0\), we have \(r=2\). This happens at \(\theta = 0, \pi\). - When \(\sin \theta = 1\) or \(-1\), we have \(r = -2\) and \(r=6\), which happens at \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\) respectively. Thus, the complete curve is drawn as \(\theta\) ranges from \(0\) to \(2\pi\). So, \(a = 0\) and \(b = 2\pi\).

Compute the derivative of r with respect to θ

Now we compute the derivative of \(r\) with respect to \(\theta\). Since \(r = 2 - 4 \sin \theta\), we can write: \(\frac{dr}{d\theta} = -4 \cos \theta\)

Substitute everything into the polar curve length formula and evaluate the integral

Substituting the given function and its derivative into the polar curve length formula, we get: \( L = \int_{0}^{2\pi} \sqrt{ (2 - 4 \sin \theta)^2 + (-4 \cos \theta)^2 } d\theta \) After simplifying and using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have: \( L = \int_{0}^{2\pi} \sqrt{ 16 \sin^2 \theta - 16 \sin \theta \cos \theta + 16 \cos^2 \theta} d\theta \) Factor out the common factor of 16: \( L = \int_{0}^{2\pi} \sqrt{16 (\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)} d\theta \) Now, evaluate the integral using a calculator: Approximating the definite integral, we get \( L \approx 9.848 \) So, the approximate length of the curve \(r = 2 - 4 \sin \theta\) is 9.848.

Key concepts

Polar Coordinates

Polar coordinates offer a unique way to represent points on a plane using a combination of angles and distances. Unlike the Cartesian coordinate system (which uses x and y axes), polar coordinates define a point by its distance from a reference point (usually the origin) and an angle from a reference direction (such as the positive x-axis).
Polar coordinates are quite useful for representing circular and spiral shapes, which are relatively complex to express using regular coordinates. The feature of polar coordinates makes them ideal for describing the Limaçon curve like the one given in the exercise, defined by the polar equation \(r = 2 - 4 \sin \theta\).

• The variable \(r\) describes the distance from the origin to the curve at a specific angle.
• The angle \(\theta\) indicates the direction of this distance, measured in radians.

This system helps simplify the process of working with curves that exhibit radial symmetry or periodic behavior, making them more approachable for calculus operations like finding length.

Curve Length

In calculus, finding the length of a curve involves summing up the tiny little segments that make up the entire curve. This concept extends naturally to polar curves like the Limaçon, which can exhibit loops or complex shapes. The formula for finding the length of a curve in polar coordinates is:\[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]Where:

• \(L\) is the curve length.
• \(\frac{dr}{d\theta}\) is the derivative of \(r\) with respect to \(\theta\), showing how the distance changes as the angle \(\theta\) changes.
• \(r\) represents the radial distance at a given angle \(\theta\).

By breaking down the curve into infinitesimally small linear segments and summing their lengths from \( \theta = a \) to \( \theta = b \), we effectively gain the total length of the curve.
This particular exercise specifically considered \( \theta \) ranging from 0 to \( 2\pi \), capturing the complete shape of the Limaçon.

Calculus Integration

Calculus integration is a powerful mathematical tool used to find areas, volumes, central points, and lengths of curves. In our context of finding the Limaçon curve length, integration was employed to sum up all small parts along the curve's path.
The integral used to find the length of the Limaçon involved a square root, which poses additional challenges. This is because dealing with square roots and trigonometric functions inside an integral can sometimes require clever tricks or numerical methods to evaluate accurately. In the exercise example, a calculator was used to estimate:\[ L = \int_{0}^{2\pi} \sqrt{ (2 - 4 \sin \theta)^2 + (-4 \cos \theta)^2 } \, d\theta \]This represents the entire computation needed to trace out the complete loop of the Limaçon, spanning angles from 0 to \(2\pi\).

• The integration range \((0, 2\pi)\) ensures that the entire curve length is calculated.
• Using a calculator helps handle complex functions like trigonometric identities that are tough to simplify manually.

Understanding integration in this context involves recognizing how it helps sum infinitesimals to get accurate and comprehensive measurements of the curve.