Problem 221 - Calculus Volume 3

Question

Find the length of the curve over the given interval. \(r=8+8 \cos \theta\) on the interval \(0 \leq \theta \leq \pi\)

Step-by-Step Solution

Answer

The length of the curve is 32 units.

Step 1: Understand the Polar Curve Equation

We have the polar curve equation given as \( r = 8 + 8 \cos \theta \). This equation represents a limaçon with an inner loop.

Step 2: Use Polar Coordinates Length Formula

The length of a curve in polar coordinates is given by the formula:\[L = \int_{\theta_1}^{\theta_2} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta\]Here, \( \theta_1 = 0 \) and \( \theta_2 = \pi \).

Step 3: Calculate the Derivative \( \frac{dr}{d\theta} \)

The derivative of \( r \) with respect to \( \theta \) is:\[ \frac{dr}{d\theta} = -8 \sin \theta \]

Step 4: Substitute Values Into the Length Formula

Substitute \( r = 8 + 8 \cos \theta \) and \( \frac{dr}{d\theta} = -8 \sin \theta \) into the length formula:\[ L = \int_{0}^{\pi} \sqrt{ (-8 \sin \theta)^2 + (8 + 8 \cos \theta)^2 } \, d\theta \]

Step 5: Simplify the Expression Inside the Square Root

Simplifying inside the square root:\[L = \int_{0}^{\pi} \sqrt{ 64 \sin^2 \theta + 64 + 128 \cos \theta + 64 \cos^2 \theta } \, d\theta\] Apply the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):\[L = \int_{0}^{\pi} \sqrt{ 128 + 128 \cos \theta } \, d\theta\]

Step 6: Factor and Simplify Further

Factor out 128:\[L = \int_{0}^{\pi} \sqrt{ 128(1 + \cos \theta) } \, d\theta = \int_{0}^{\pi} \sqrt{ 256 \cos^2 \left(\frac{\theta}{2}\right) } \, d\theta\] This can be simplified using \( 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2} \):\[L = \int_{0}^{\pi} 16 \cos \frac{\theta}{2} \, d\theta \]

Step 7: Evaluate the Integral

Evaluate the integral:\[L = \int_{0}^{\pi} 16 \cos \frac{\theta}{2} \, d\theta \]Make the substitution \( u = \frac{\theta}{2} \), \( du = \frac{1}{2} d\theta \):\[ L = 32 \int_{0}^{\frac{\pi}{2}} \cos u \, du = 32 \left[ \sin u \right]_{0}^{\frac{\pi}{2}}\]\[L = 32(\sin \frac{\pi}{2} - \sin 0 ) = 32(1 - 0)\]

Step 8: Final Step: Calculate the Final Length

The final length of the curve is:\[L = 32\]

Key Concepts

Limaçon

A limaçon is a type of polar curve represented by the equation \[ r = a + b \cos \theta \] or \[ r = a + b \sin \theta. \]In our exercise, we encountered the limaçon \( r = 8 + 8 \cos \theta \), which exhibits unique features based on the values of \( a \) and \( b \).

• When \( a = b \), as in our case (\( a = b = 8 \)), the limaçon forms a cardioid shape.
• Here, it has an inner loop, revealing its attractive resemblance to a heart shape, especially over the interval \( 0 \leq \theta \leq \pi \).

Limaçons have unique symmetrical properties and can have loops or dimpled appearances, depending on the ratio of \( a \) to \( b \).Understanding these properties helps in visualizing not only the shape but also in setting the correct intervals for integration or other calculations.

Curve Length

Finding the length of a curve formulated in polar coordinates can be an enriching lesson in calculus. The general formula to find the length, or arc length, of a curve \( r(\theta) \) in polar coordinates is given by:\[L = \int_{\theta_1}^{\theta_2} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta\]This formula combines the square of the derivative \( \frac{dr}{d\theta} \) with the square of the polar radius \( r \).

• Each segment of the curve is calculated, mashed, and summed through integration across the specified limits of angle \( \theta \).
• The square root ensures that minor changes and discrepancies in curve direction are accounted for, providing a smooth and continuous length result.

This process is foundational in understanding how curve lengths translate from Cartesian coordinates, a common starting point for new learners, to polar coordinates.

Polar Curve Integration

Integrating polar curves to find important attributes like arc length combines calculus and trigonometry, offering several noteworthy challenges. In the current problem, we integrated over \( 0 \leq \theta \leq \pi \) for the limaçon curve. This process required us to first compute the derivative \( \frac{dr}{d\theta} = -8 \sin \theta \) and then substitute into the length formula.To make integration simpler, we transformed the integrand using trigonometric identities:

• Replace \( 1 + \cos \theta \) with \( 2 \cos^2 \left(\frac{\theta}{2}\right) \).
• This facilitated the integral into a more manageable form \( 16 \int_{0}^{\pi} \cos \frac{\theta}{2} \, d\theta \).

These transformations are vital in making polar integrations solvable and interpretable, as they often reduce complex trigonometric expressions into simpler terms.

Trigonometric Identities

Trigonometric identities help in simplifying calculations of polar curves, especially when seeking integrals or derivatives. In this problem, we utilized the following identities:

• Basic identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
• Half-angle identity: \( 1 + \cos \theta = 2 \cos^2 \left( \frac{\theta}{2} \right) \)

These identities allow us to effectively transform our equations, simplifying both computations and integrals needed to find the arc length of the limaçon.