Question

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

Short answer

Answer: The approximate length of the curve is 18.425.

Step-by-step solution

Find \(\frac{dr}{d\theta}\)

We have the polar equation as \(r = 4 - 2\cos\theta\). Differentiating \(r\) with respect to \(\theta\) to find \(\frac{dr}{d\theta}\): $$ \frac{dr}{d\theta}= \frac{d}{d\theta}(4 - 2\cos\theta) = 2\sin\theta $$ Now we have \(r\) and \(\frac{dr}{d\theta}\).

Determine the limits of integration

To find the limits of integration, we need to find the range of \(\theta\) for which the polar equation traces the complete curve. For the limaçon \(r = 4 - 2\cos\theta\), this happens over the interval \(0 \leq \theta < 2\pi\). Thus, the limits of integration are \(\alpha = 0\) and \(\beta = 2\pi\).

Calculate the length of the limaçon using a calculator

Using the formula, we can substitute the values \(r\), \(\frac{dr}{d\theta}\), and the limits of integration: $$ L = \int_0^{2\pi}\sqrt{(4-2\cos\theta)^2+(2\sin\theta)^2}d\theta $$ Now, use a calculator with an integral function to find the approximate length of the limaçon. The calculator yields: $$ L \approx 18.425 $$ So, the approximate length of the given limaçon is \(18.425\).

Key concepts

Polar Coordinates

Polar coordinates offer a unique approach to defining the position of a point in the plane using a radius and an angle. Unlike Cartesian coordinates, which use x and y to denote a position, polar coordinates utilize two different values:

• Radius (\(r\)): The distance from the origin to the point.
• Angle (\(\theta\)): The angle measured counterclockwise from the positive x-axis to the line connecting the origin to the point.

These coordinates are especially handy when dealing with curves and shapes that are naturally circular or spiral, such as the limaçon discussed in the given exercise.

A limaçon, like the one described, is easily expressed in polar form and can exhibit various shapes such as loops or dimples, all dependent on the parameters in the equation. Understanding how to convert from polar to Cartesian coordinates, and vice versa, can significantly enhance your ability to interpret such curves.

Curve Length

The length of a curve in polar coordinates is an intriguing topic as it introduces a distinct method for solving problems compared to Cartesian coordinates. The formula used for this, \(L = \int_\alpha^\beta \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \, d\theta\), provides the means to calculate the length of a curve defined by a polar equation over a given interval.

This formula integrates the square root of the sum of squares which consists of:

• \(r^2\)
• \(\left(\frac{dr}{d\theta}\right)^2\)

Each component captures the contributions of changes in both the radius and the angle, reflecting how the curve stretches and compresses along its path.

Understanding curve length in polar coordinates not only helps in calculating specific examples like the limaçon but also provides insight into the geometric interpretations of polar graphs in general.

Differentiation

Differentiation in the context of polar coordinates focuses on finding how the radius \(r\) changes with respect to the angle \(\theta\). This is a crucial step when determining features like slopes and curve lengths.

In the problem provided, we begin by differentiating \(r = 4 - 2\cos\theta\) with respect to \(\theta\). This gives us \(\frac{dr}{d\theta} = 2\sin\theta\).

• The differentiation involves using trigonometric identities.
• This process reveals how the radial distance changes as the angle increments – important for evaluating the nature of a curve.

Differentiation in such cases aids in formulating the exact equation needed for further calculations like finding the total curve length.

Integration

In calculus, integration is employed to determine areas, volumes, or lengths, particularly where a sum over continuous data is necessary. When applied in polar coordinates, as with the given exercise, integration facilitates finding the length of a curve.

For the problem at hand, the integral \(L = \int_0^{2\pi}\sqrt{(4-2\cos\theta)^2+(2\sin\theta)^2}d\theta\) is computed over the interval \(0\leq \theta < 2\pi\), ensuring that the entire curve is accounted for.

The integral comprises the very components we obtained through differentiation and appears as a somewhat complex expression due to the involvement of trigonometric functions. However, it simplifies the calculation by reducing the equation to evaluating the integral over the specified range, ultimately yielding a numerical approximation which is essential when using a calculator.

Although sometimes challenging, mastering integration in polar coordinates enhances one’s ability to solve real-world problems.