Problem 340 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

The arc length of the curve is \(12\pi\).

Step 1: Set up the arc length formula for polar coordinates

The formula for finding the arc length of a curve in polar coordinates is \[L = \int_{\theta_1}^{\theta_2} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta\]In this exercise, we have \(r = 6 \cos \theta\) and \(0 \leq \theta \leq 2\pi\).

Step 2: Differentiate \(r\) with respect to \(\theta\)

To find \(\frac{dr}{d\theta}\), differentiate \(r = 6 \cos \theta\) with respect to \(\theta\):\[\frac{dr}{d\theta} = -6 \sin \theta\].

Step 3: Substitute into the arc length formula

Substitute \(r = 6 \cos \theta\) and \(\frac{dr}{d\theta} = -6 \sin \theta\) into the arc length formula:\[L = \int_{0}^{2\pi} \sqrt{ (-6 \sin \theta)^2 + (6 \cos \theta)^2 } \, d\theta\].

Step 4: Simplify the expression inside the integral

Calculate the expression inside the square root:\[(-6 \sin \theta)^2 + (6 \cos \theta)^2 = 36 \sin^2 \theta + 36 \cos^2 \theta = 36(\sin^2 \theta + \cos^2 \theta) = 36\].The expression simplifies to 36 because \( \sin^2 \theta + \cos^2 \theta = 1\).

Step 5: Evaluate the integral

The integral becomes:\[L = \int_{0}^{2\pi} \sqrt{36} \, d\theta = \int_{0}^{2\pi} 6 \, d\theta\].Evaluating this, we get:\[L = 6\left[\theta\right]_{0}^{2\pi} = 6(2\pi - 0) = 12\pi\].

Step 6: Verify with geometry

Since \(r = 6\cos\theta\) is a circle with radius 3 centered at \((3, 0)\), its full circumference is \(2\pi \times 3 = 6\pi\), but from geometrical inspection, the curve goes through twice, doubling the arc length, leading to \(12\pi\), which confirms the calculated result.

Key Concepts

Polar Coordinates

Polar coordinates are essential when dealing with curves that aren't easily expressed using Cartesian coordinates. Unlike in the Cartesian system, where points are determined by a pair of perpendicular axis values - the x and y values - the polar coordinate system establishes a point's position using a radius and an angle. The radius, denoted as \( r \), measures the distance from a central point, typically the origin, and the angle, \( \theta \), indicates the direction from the reference positive x-axis.Using polar coordinates enables the description of circular and spiral shapes in a much simpler form. For example, the equation \( r = 6 \cos \theta \) represents a circle when plotted, with a specific orientation and position. In this exercise, understanding this circle in polar form helps set up the equation needed to calculate the arc length.

Arc Length Formula

Calculating the arc length of a curve in polar coordinates involves a special formula. For a given polar function \( r(\theta) \), the arc length formula is: \[ L = \int_{\theta_1}^{\theta_2} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]This formula stems from the Pythagorean theorem and integrates the infinitesimal segments of the curve from the starting angle \( \theta_1 \) to the ending angle \( \theta_2 \). In effect, it combines both the rate at which \( r \) changes with respect to \( \theta \) and the magnitude of \( r \) itself.In our exercise, by substituting \( r(\theta) = 6 \cos \theta \), the formula enables us to explore these infinitesimal elements and eventually sum them up to evaluate the full length of the curve, resulting in a computed result of \( 12\pi \), confirmed through geometry.

Differentiation in Calculus

Differentiation is a crucial tool in calculus for understanding rate changes. In this context, we differentiate to find the rate of change of the radius in polar coordinates. Given the curve equation \( r = 6 \cos \theta \), differentiating with respect to \( \theta \) gives \( \frac{dr}{d\theta} = -6 \sin \theta \). This derivative tells us how the distance from the origin varies as the angle changes, a vital component in calculating arc length.By using this derivative in the arc length formula, we effectively include how the curve bends and shifts, integrating it over the specified interval from \( 0 \) to \( 2\pi \). This highlights the utility of differentiation in uncovering essential dynamics of a curve's geometry.

Integration of Trigonometric Functions

Integration is the process of finding areas under curves, and it plays a fundamental role in calculating arc length. Trigonometric functions, common in polar curves, often require specific techniques for integration.In this exercise, the integration involves the expression \( \sqrt{36} \), resulting directly from simplifying \((-6 \sin \theta)^2 + (6 \cos \theta)^2\). The trig identity \( \sin^2 \theta + \cos^2 \theta = 1 \) simplifies this before integration. As you integrate \( 6 \) over the interval \( 0 \) to \( 2\pi \), you account for the arc's full scope. Integration not only aggregates the segments but also factors in the circle's radius during its complete traversal, validated geometrically in the final result.