Problem 23 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The length of the curve is \(6\ln(1 + \sqrt{2})\).

Step 1: Understanding the Problem

We need to find the length of the curve given in polar coordinates by the equation \(r = \frac{6}{1 + \cos\theta}\) for \(0 \leq \theta \leq \frac{\pi}{2}\).

Step 2: Formula for Arc Length in Polar Coordinates

The formula for the length of a curve given by a polar equation \(r = f(\theta)\) from \(\theta = a\) to \(\theta = b\) is \(L = \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta\).

Step 3: Differentiate the Function

Find \(\frac{dr}{d\theta}\) for \(r = \frac{6}{1 + \cos\theta}\). Use the quotient rule: \( \frac{d}{d\theta}\left(\frac{u}{v}\right) = \frac{v\frac{du}{d\theta} - u\frac{dv}{d\theta}}{v^2}\) with \(u = 6\) and \(v = 1 + \cos\theta\).

Step 4: Apply the Quotient Rule

By applying the quotient rule, we compute \( \frac{dr}{d\theta} = \frac{-6\sin\theta}{(1+\cos\theta)^2} \).

Step 5: Calculate the Expression for Length

Substitute \(r = \frac{6}{1 + \cos\theta}\) and \(\frac{dr}{d\theta} = \frac{-6\sin\theta}{(1+\cos\theta)^2}\) into the arc length formula: \(L = \int_0^{\pi/2} \sqrt{\left(\frac{6}{1 + \cos\theta}\right)^2 + \left(\frac{-6\sin\theta}{(1+\cos\theta)^2}\right)^2} \, d\theta\).

Step 6: Simplify the Integrand

Simplify the expression inside the square root: \(\left(\frac{6}{1 + \cos \theta}\right)^2 + \left(\frac{-6\sin \theta}{(1 + \cos \theta)^2}\right)^2 = \frac{36 (2 + 2\cos \theta)}{(1 + \cos \theta)^4}\).

Step 7: Perform the Integration

Carefully integrate the simplified expression from \(0\) to \(\frac{\pi}{2}\) to find the total length: \(L = 6\ln(1 + \sqrt{2})\) by substituting variables and using integration techniques.

Key Concepts

Polar Coordinates

Polar coordinates are a way to represent points in a plane using a radius and angle instead of traditional x-y coordinates. This system is particularly useful in scenarios involving curves and circular patterns. In polar coordinates, each point is defined by:

• The radial distance from the origin, denoted as \(r\).
• The angle \(\theta\) measured from the positive x-axis.

For example, the equation \(r = \frac{6}{1 + \cos \theta}\) gives the shape of a curve, defining how far each point on the curve is from the origin at a given angle \(\theta\). This form is especially helpful for capturing non-linear shapes like spirals, circles, and arcs, making it a powerful tool in many fields like physics and engineering. When calculating lengths and areas in these systems, understanding the formulation in polar coordinates is key.

Arc Length Formula

The arc length formula in polar coordinates is crucial for calculating the length of a curve given in the equation form \(r = f(\theta)\). To find the arc length of a curve between two angles \(\theta = a\) and \(\theta = b\), we use:\[L = \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta\]This formula incorporates both the radial distance \(r\) and the rate of change of \(r\) with respect to \(\theta\) (denoted as \(\frac{dr}{d\theta}\)), combining these to accurately measure the actual path length of the curve. This is essential because curves in polar coordinates often don't follow straight lines; they can bend and twist, which is why a basic distance formula won't suffice. Knowing how to integrate this expression is fundamental to finding lengths of curves like spirals and circles.

Differentiation

Differentiation is a process in calculus that involves finding the rate at which a function changes. For a function expressed as \(r = f(\theta)\) in polar coordinates, this involves finding \(\frac{dr}{d\theta}\) to understand the curve's behavior as \(\theta\) changes. Differentiation helps determine how the shape of the curve changes as one moves around the path. In our exercise, we differentiate the curve \(r = \frac{6}{1 + \cos \theta}\) to understand its steepness and behavior by finding \(\frac{dr}{d\theta}\). This step is important for the overall calculation of the arc length since the degree of tilt (or steepness) at every point along the curve affects how far the curve stretches. Thus, careful differentiation aids in accurately applying the arc length formula.

Quotient Rule

The quotient rule is a technique used in differentiation when you have a function that is the ratio of two other functions. For a function \(\frac{u(\theta)}{v(\theta)}\), the derivative is calculated as follows:\[\frac{d}{d\theta} \left( \frac{u}{v} \right) = \frac{v\frac{du}{d\theta} - u\frac{dv}{d\theta}}{v^2}\]In our step-by-step approach, the quotient rule is applied to the given function \(r = \frac{6}{1 + \cos \theta}\), where we let \(u = 6\) and \(v = 1 + \cos \theta\). By using the quotient rule, we derive \(\frac{dr}{d\theta} = \frac{-6\sin\theta}{(1 + \cos \theta)^2}\). This manipulation is significant, as it gives us the expression needed to plug into the arc length formula to find the total length of the curve. Mastery of the quotient rule is vital for handling similar calculus problems involving division of functions.