Question

For the following exercises, find the arc length of the curve over the given interval. \(r=6 \cos \theta, 0 \leq \theta \leq 2 \pi .\) Check your answer by geometry.

Short answer

The arc length is \(12\pi\).

Step-by-step solution

Understand the Curve

The polar equation given is \( r = 6 \cos \theta \). This represents a circle centered at \( (3, 0) \) with a radius of 3 in the Cartesian coordinate system.

Set Up the Arc Length Formula for Polar Curves

The formula to find the arc length \( L \) of a polar curve \( r = f(\theta) \) from \( \theta = a \) to \( \theta = b \) is:\[ L = \int_a^b \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]

Derive the Expression for \( \frac{dr}{d\theta} \)

Determine \( \frac{dr}{d\theta} \) by differentiating \( r = 6 \cos \theta \):\[ \frac{dr}{d\theta} = -6 \sin \theta \]

Substitute into the Arc Length Formula

Substitute \( r = 6 \cos \theta \) and \( \frac{dr}{d\theta} = -6 \sin \theta \) into the arc length formula:\[ L = \int_0^{2\pi} \sqrt{ (-6 \sin \theta)^2 + (6 \cos \theta)^2 } \, d\theta \]Simplify the expression inside the square root:\[ L = \int_0^{2\pi} \sqrt{ 36 \sin^2 \theta + 36 \cos^2 \theta } \, d\theta \]

Simplify the Integral Expression

Notice that \( \sin^2 \theta + \cos^2 \theta = 1 \).Thus:\[ L = \int_0^{2\pi} \sqrt{ 36 (\sin^2 \theta + \cos^2 \theta) } \, d\theta \]This simplifies to:\[ L = \int_0^{2\pi} 6 \, d\theta \]

Evaluate the Integral

Evaluate the integral:\[ L = 6 \int_0^{2\pi} \, d\theta = 6 \left[ \theta \right]_0^{2\pi} \]This results in:\[ L = 6(2\pi - 0) = 12\pi \]

Verify Answer with Geometry

The curve \( r = 6 \cos \theta \) is a circle with a diameter of 6 (radius \(3\)), thus its circumference is \( 2\pi \times 3 = 6\pi \). As the interval covers the circle twice (from 0 to \(2\pi\)), the total arc length is indeed \(12\pi\).

Key concepts

Polar Coordinates

In mathematics, polar coordinates offer an elegant way to describe locations and shapes on a plane. Unlike rectangular coordinates which use horizontal and vertical distances (x, y), polar coordinates express points based on the distance from a central point and the angle from a reference direction. Specifically, a point in polar coordinates is represented as \((r, \theta)\):

• \(r\) is the radial distance from the origin (though it can be negative, which reflects across the origin).
• \(\theta\) is the angle from the positive x-axis, measured in radians.

For example, the polar equation \(r = 6 \cos \theta\) describes a circle in polar coordinates. Here, \(r\) varies with \(\theta\) dependent on the cosine function. The curve forms a circle where the radius decreases as \(\theta\) changes, reaching zero at the center, and grows again as \(\theta\) continues. When translating to Cartesian coordinates, this circle is centered at \((3, 0)\) with a radius of 3. Polar coordinates are particularly useful in problems involving symmetry and curves that pivot around a point, such as circles or spirals.

Arc Length Formula

The arc length formula is a tool used to calculate the distance along a curve. In polar coordinates, determining the arc length requires a special formula. For a curve described by \(r = f(\theta)\), the arc length \(L\) over an interval \([a, b]\) is given by:\[ L = \int_a^b \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \] This formula integrates the expression that arises from the Pythagorean Theorem applied to infinitesimal segments along the curve:

• \(r^2\) accounts for the radial distance squared.
• \(\left( \frac{dr}{d\theta} \right)^2\) captures the change in \(r\) as \(\theta\) changes, squared.

Using this formula allows us to integrate over the specified interval, providing the complete arc length of the curve. In the example given, this calculation for \(r = 6 \cos \theta\) accurately determines the path length as \(12\pi\), validating that these relationships hold true for periodic and symmetrical curves.

Differentiation

Differentiation is a fundamental concept in calculus, focusing on finding the rate at which one quantity changes with respect to another. In the context of arc length in polar coordinates, differentiation is crucial for deriving \(\frac{dr}{d\theta}\), which tells us how the radial distance \(r\) changes as the angle \(\theta\) varies. For the functional form \(r = 6 \cos \theta\), we apply differentiation to find: \[ \frac{dr}{d\theta} = -6 \sin \theta \] This represents the slope of \(r\) with respect to \(\theta\), reflecting the oscillatory nature of the cosine function — as the curve rotates, distances shrink and grow.

• The negative sign indicates that \(r\) decreases when \(\theta\) increases across certain intervals.
• Sine and cosine functions derive each other's behavior in differentiation due to their trigonometric properties.

Understanding this derivative is necessary to use the arc length formula. It illustrates how calculus allows us to explore the nuanced dynamics of curves, revealing distances that are not immediately apparent from the equations alone.