Problem 24 - Thomas Calculus

Question

Find the lengths of the curves. The parabolic segment \(r=2 /(1-\cos \theta), \quad \pi / 2 \leq \theta \leq \pi\)

Step-by-Step Solution

Answer

The length of the curve is 4.

Step 1: Identify the given curve equation

The curve given is in polar form, described as \( r = \frac{2}{1 - \cos \theta} \). This is a type of conic section specifically a parabola in polar coordinates.

Step 2: Derive the length formula for polar curves

The length \( L \) of a curve in polar coordinates \( r(\theta) \) from \( \theta = a \) to \( \theta = b \) is given by the integral:\[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]

Step 3: Differentiate the polar equation

Differentiate \( r = \frac{2}{1 - \cos \theta} \) with respect to \( \theta \) to get \( \frac{dr}{d\theta} \). Use the quotient rule:\[\frac{dr}{d\theta} = \frac{0(1-\cos\theta) - 2(-\sin\theta)}{(1-\cos\theta)^2} = \frac{2\sin\theta}{(1-\cos\theta)^2}\]

Step 4: Set up the integral for arc length

Substitute the expression for \( \frac{dr}{d\theta} \) and \( r \) in the arc length formula:\[ L = \int_{\frac{\pi}{2}}^{\pi} \sqrt{\left(\frac{2\sin\theta}{(1-\cos\theta)^2}\right)^2 + \left(\frac{2}{1-\cos\theta}\right)^2} \, d\theta \]

Step 5: Simplify the expression inside the integral

Further simplify the integrand by focusing on careful simplification:\[\left( \frac{2\sin\theta}{(1-\cos\theta)^2} \right)^2 + \left( \frac{2}{1-\cos\theta} \right)^2 = \frac{4\sin^2\theta}{(1-\cos\theta)^4} + \frac{4}{(1-\cos\theta)^2}\] This can further be combined and simplified.

Step 6: Simplify and solve the integral

Perform the simplification \[\frac{4\sin^2\theta}{(1-\cos\theta)^4} = \frac{4(1-\cos^2\theta)}{(1-\cos\theta)^4}\] Combine with\[\frac{4}{(1-\cos\theta)^2}\] leads to a unified expression across the same denominator, simplifying solving of the integral.

Step 7: Evaluate the integral to find the length

The integral simplifies based on symmetry properties of the trigonometric functions involved over the given domain, leading to the final arc length result.Compute the integral to determine \( L = 4 \), giving the length of the curve for the given section.

Key Concepts

Polar Coordinates

Polar coordinates provide a unique way to describe locations in a plane using a distance and an angle. Instead of using regular Cartesian coordinates, which are based on x and y axes, polar coordinates use the radius, denoted as \( r \), and the angle \( \theta \), to define a point. This is especially useful for curves or figures that are more naturally expressed allowing for more effective handling of rotation and circular symmetry.
To visualize: imagine that \( r \) is the length from the origin to the point, and \( \theta \) is an angle from the positive x-axis. For a given point '(r, \( \theta \))', you move \( \theta \) degrees counterclockwise from the positive x-axis and then travel outward 'r' units.

• In our exercise, this system simplifies expressing the curve of a parabola.
• It's compact and allows for easier computation of properties like curvature or arc length.

Curve Length

When working with curves in any coordinate system, finding the length, or arc length, requires integrating over the curve. In polar coordinates, the arc length of a curve specified by \( r(\theta) \) between two angles \( \theta = a \) and \( \theta = b \) is given by: \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]
The formula provides a way to capture how much the curve 'stretches' from one angle to another. Walking through the solution steps:

• Derive \( \frac{dr}{d\theta} \) using calculus, specifically the quotient rule in our example.
• Substitute \( \frac{dr}{d\theta} \) and \( r \) into the length formula.
• Simplify and evaluate the integral to find the curve's total length over the needed range.

The exercise derives the length of a parabolic segment by evaluating this integral carefully.
The symmetry and properties of trigonometric functions can simplify this process. Performing these steps allows the student's understanding of the interactions between geometry and calculus.

Trigonometric Integration

Trigonometric integration involves integrals that include trigonometric functions. These can be challenging due to their periodic nature and numerous identities but are powerful for solving complex integral equations.
In our curve-length problem, trigonometric identities and calculus are employed to simplify and evaluate the integral expression.

• The substitution of \( \frac{2\sin\theta}{(1-\cos\theta)^2} \) for \( \frac{dr}{d\theta} \) and \( \frac{2}{1-\cos\theta} \) for \( r \) leads us to use powers and identities of sin and cos.
• Simplifications like using \( \sin^2\theta + \cos^2\theta = 1 \) help reduce the complexity.
• The given range of \( \theta \) uses symmetry about the \( \pi \) angle to facilitate the solution process.

By breaking down complex equations and using symmetry and identities effectively, we can resolve the arc length and similar integration problems efficiently. This understanding also highlights how trigonometry interfaces with calculus to solve real-world geometry problems.