Problem 25 - Thomas Calculus

Question

Find the lengths of the curves in Exercises \(21-28 .\) The parabolic segment \(r=6 /(1+\cos \theta), \quad 0 \leq \theta \leq \pi / 2\)

Step-by-Step Solution

Answer

The length of the parabolic segment is obtained using numerical integration techniques due to complexity in special forms involved.

Step 1: Understand the Problem

We need to find the length of the polar curve given by \( r = \frac{6}{1+\cos\theta} \) over the interval \( 0 \leq \theta \leq \frac{\pi}{2} \). This represents a portion of a conic section, specifically a parabola segment.

Step 2: Recall the Arc Length Formula for Polar Curves

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

Step 3: Differentiate r with respect to \( \theta \)

Given \( r = \frac{6}{1+\cos\theta} \), we find the derivative \( \frac{dr}{d\theta} \). Using the quotient rule:\[\frac{dr}{d\theta} = \frac{d}{d\theta} \left( \frac{6}{1+\cos\theta} \right) = \frac{-6\sin\theta}{(1+\cos\theta)^2}\]

Step 4: Substitute into the Arc Length Formula

Substitute \( r \) and \( \frac{dr}{d\theta} \) into the arc length formula:\[L = \int_{0}^{\frac{\pi}{2}} \sqrt{\left( \frac{-6\sin\theta}{(1+\cos\theta)^2} \right)^2 + \left( \frac{6}{1+\cos\theta} \right)^2} \, d\theta\]

Step 5: Simplify the Integrand

Simplify the expression inside the square root:\[L = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{36\sin^2\theta}{(1+\cos\theta)^4} + \frac{36}{(1+\cos\theta)^2}} \, d\theta\]Factor the integrand:\[L = \int_{0}^{\frac{\pi}{2}} \frac{6}{(1+\cos\theta)^2} \sqrt{\sin^2\theta + (1+\cos\theta)^2} \, d\theta\]The inner expression simplifies to \( \sin^2\theta + 1 + 2\cos\theta + \cos^2\theta \). Use the identity \( \sin^2\theta + \cos^2\theta = 1 \) to further simplify:\[\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta = 2 + 2\cos\theta\]

Step 6: Simplification of the Expression

Further simplify \( \sqrt{2 + 2\cos\theta} \) to \( \sqrt{2(1 + \cos\theta)} = \sqrt{2} \sqrt{1 + \cos\theta} \). Substituting back into the integral gives:\[L = \int_{0}^{\frac{\pi}{2}} \frac{6\sqrt{2} \sqrt{1 + \cos\theta}}{(1+\cos\theta)^2} \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{6\sqrt{2}}{\sqrt{1+\cos\theta}} \, d\theta\]

Step 7: Calculate the Integral

To solve the integral:\\[L = 6\sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1+\cos\theta}} \, d\theta\]This integral involves using a substitution method, such as letting \( u = 1 + \cos\theta \), \( du = -\sin\theta \, d\theta \), but this integral generally requires special functions or advanced calculus techniques beyond basic substitutions in standard settings.After evaluation, the integral yields a numerical solution often involving elliptic integral forms.**Numerical or advanced techniques are recommended for further simplification and evaluation here in typical settings where arcsin or closed forms aren't readily applicable without computational or specific knowledge extensions.**

Step 8: Express the Result

The length of the parabolic segment can be expressed as a numerical approximation based on integral evaluations, which is a domain evaluation using more advanced calculus or computationally to derive a specific numeric approximation or value for this situation.

Key Concepts

Quotient Rule Differentiation

Differentiating functions that are given as a quotient is an essential skill, and the quotient rule is the main tool for this. Imagine you have a function expressed as the ratio of two other functions, for instance, \( r(\theta) = \frac{u(\theta)}{v(\theta)} \). Here, \( u(\theta) \) and \( v(\theta) \) are both functions of \( \theta \). The quotient rule helps us differentiate \( r(\theta) \) using the following formula:

• \[ \frac{d}{d\theta} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \]

This rule instructs us to take the derivative of the numerator \( u(\theta) \) and the derivative of the denominator \( v(\theta) \). Then, multiply the derivative of the numerator by the denominator and subtract the product of the numerator with the derivative of the denominator. All of this is divided by the square of the denominator.
To illustrate, consider the function \( r(\theta) = \frac{6}{1+\cos\theta} \). Here, \( u(\theta) = 6 \) and \( v(\theta) = 1 + \cos\theta \). Applying the quotient rule gives us:

• \[ \frac{dr}{d\theta} = \frac{(1+\cos\theta) \cdot 0 - 6 \cdot (-\sin\theta)}{(1+\cos\theta)^2} = \frac{-6\sin\theta}{(1+\cos\theta)^2} \]

Differentiating using the quotient rule can simplify solving polar curve problems and ensure accurate derivative results.

Polar Coordinates

Polar coordinates might seem complex at first, but they're a powerful tool for representing curves, especially those with circular and angular components. Unlike Cartesian coordinates, which use \( (x,y) \) to denote a point's position, polar coordinates use \( (r,\theta) \):

• \( r \) is the distance from the origin to the point.
• \( \theta \) is the angle between the positive x-axis and the line connecting the origin to the point.

These coordinates are especially useful when dealing with curves that have inherent symmetry or periodicity, such as circles, spirals, and, in our case, conic sections like the parabola segment.
The equation \( r = \frac{6}{1+\cos\theta} \) describes a polar curve known as a limacon, which is a kind of conic section. We can separate different parts of the curve by varying \( \theta \). Hence, understanding polar coordinates allows for elegant and often simplified expressions of otherwise complex curves.

Arc Length Formula for Polar Curves

Finding the arc length of a curve can be a bit challenging, but the formula for polar curves helps streamline this process. For a polar curve described by \( r = f(\theta) \), the length \( L \) from \( \theta = a \) to \( \theta = b \) is determined by:

• \[ L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta \]

What this formula does is extend the concept of a curve's length, utilizing the polar coordinates structure. Essentially, it accounts for both the changes in \( r \) and angular progressions along \( \theta \).
Applying it involves some calculus acumen: we first calculate \( \frac{dr}{d\theta} \), then substitute it together with \( r \) into the integral. This integral can get complex and sometimes requires numerical methods for a solution. Nonetheless, understanding the arc length formula provides a key insight into measuring the size and progression of intricate shapes described by polar curves, providing both depth and breadth to your calculus toolkit.