Problem 27 - Thomas Calculus

Question

Find the lengths of the curves in Exercises \(21-28 .\) The curve \(r=\cos ^{3}(\theta / 3), \quad 0 \leq \theta \leq \pi / 4\)

Step-by-Step Solution

Answer

The length of the curve is \(\frac{\pi + 3}{8}\).

Step 1: Understanding the Formula for Arc Length in Polar Coordinates

To find the length of a curve in polar coordinates defined by \(r = f(\theta)\), the formula for arc length \(L\) is given by: \[ L = \int_{\alpha}^{\beta} \sqrt{ r^2 + \left( \frac{dr}{d\theta} \right)^2 } \, d\theta \] where \( [\alpha, \beta] \) is the interval of \(\theta\) along the curve. In this problem, \(\alpha = 0\) and \(\beta = \frac{\pi}{4}\).

Step 2: Calculating \(\frac{dr}{d\theta}\)

Given \(r = \cos^3(\theta/3)\), first find \(\frac{dr}{d\theta}\) using the chain rule. The derivative of \( \cos^3(u) \) with respect to \(u\) is \(3\cos^2(u)(-\sin(u))\) and \(u = \theta/3\), so \(\frac{du}{d\theta} = 1/3\). Thus, \(\frac{dr}{d\theta} = 3 \cos^2(\theta/3) \times (-\sin(\theta/3)) \times (1/3) = -\cos^2(\theta/3) \sin(\theta/3)\).

Step 3: Substituting into Arc Length Formula

Substitute \(r = \cos^3(\theta/3)\) and \(\frac{dr}{d\theta} = -\cos^2(\theta/3) \sin(\theta/3)\) into the arc length formula:\[L = \int_{0}^{\pi/4} \sqrt{ \cos^6(\theta/3) + [-\cos^2(\theta/3) \sin(\theta/3)]^2 } \, d\theta \] Simplify the expression under the square root as:\[ \cos^6(\theta/3) + \cos^4(\theta/3) \sin^2(\theta/3) \]\[= \cos^4(\theta/3)(\cos^2(\theta/3) + \sin^2(\theta/3)) \]\[= \cos^4(\theta/3) \] (since \(\cos^2(x) + \sin^2(x) = 1\)).This simplifies our integral to:\[ L = \int_{0}^{\pi/4} |\cos^2(\theta/3)| \, d\theta \].

Step 4: Solving the Integral

Since \(\cos^2(\theta/3)\) is always positive in the given interval, the absolute value can be removed:\[ L = \int_{0}^{\pi/4} \cos^2(\theta/3) \, d\theta \] Using the substitution \( u = \theta/3 \), \( du = (1/3) d\theta \) or \( d\theta = 3du \), change the limits of integration: when \( \theta = 0 \), \( u = 0 \), and when \( \theta = \pi/4 \), \( u = \pi/12 \). Therefore, the integral becomes:\[ L = 3 \int_{0}^{\pi/12} \cos^2(u) \, du \]Use the identity \( \cos^2(u) = \frac{1 + \cos(2u)}{2} \):\[ L = 3 \int_{0}^{\pi/12} \frac{1 + \cos(2u)}{2} \, du \]\[= \frac{3}{2} \int_{0}^{\pi/12} (1 + \cos(2u)) \, du \]\[= \frac{3}{2} \left[ u + \frac{1}{2}\sin(2u) \right]_{0}^{\pi/12} \]\[= \frac{3}{2} \left[ \frac{\pi}{12} + \frac{1}{2} \sin(\frac{\pi}{6}) - (0 + 0) \right] \]\[= \frac{3}{2} \left[ \frac{\pi}{12} + \frac{1}{2} \times \frac{1}{2} \right] \]\[= \frac{3}{2} \left[ \frac{\pi}{12} + \frac{1}{4} \right] \]\[= \frac{3}{2} \times \frac{\pi + 3}{12} \]\[= \frac{3(\pi + 3)}{24} \]\[= \frac{\pi + 3}{8} \].

Step 5: Final Result

The length of the curve \(r = \cos^3(\theta/3)\) over the interval \(0 \leq \theta \leq \pi/4\) is \(\frac{\pi + 3}{8}\).

Key Concepts

Polar Coordinates

Polar coordinates are a way of representing points in a plane using a radius and an angle. Unlike the traditional Cartesian coordinates which use x and y values, polar coordinates specify a point with radius \(r\) and angle \(\theta\). This system is particularly useful for problems involving circular and rotating symmetries or patterns.
In the given exercise, the curve is defined by \(r = \cos^3(\theta / 3)\). Here, \(r\) is a function of \(\theta\), which means that as the angle \(\theta\) varies, the radius \(r\) also changes, tracing out a specific path or shape. Polar coordinates allow us to easily describe this type of curve, where the relationship between the radius and angle is essential for describing the geometry of the path.
To find the length of a curve in polar coordinates, we rely on a special formula that accounts for changes in both \(r\) and \(\theta\). This formula involves integration, which we'll talk about next.

Integration Techniques

Integration is a fundamental concept in calculus used to find areas, volumes, central points, and many other important mathematical quantities. In the context of finding arc lengths, integration helps us to accumulate infinitesimally small lengths along the curve to find the total length.
The specific formula for arc length in polar coordinates is:
\[L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta\]
This formula might look complex but it essentially adds up small pieces of straight lines that approximate the curve. We compute \( \frac{dr}{d\theta} \), which is the derivative of \( r \) with respect to \( \theta \), to understand how \( r \) changes as \( \theta \) changes.
In our problem, once \( \frac{dr}{d\theta} \) is calculated, it plugs back into the formula to compute the integral and thus the length of the curve. One important technique used here involves simplifying expressions under the square root to make the integral manageable. Then, using substitution like \( u = \theta/3 \), we adjusted the limits of integration and used trigonometric identities to solve the integral effectively.

Chain Rule in Calculus

The chain rule is a powerful tool in calculus used to compute the derivative of composite functions. It's essential when dealing with nested functions, where a function is applied within another function.
In the solution to the exercise, the chain rule helps us to compute \( \frac{dr}{d\theta} \) when \( r = \cos^3(\theta/3) \). Due to the structure of the function \( r \), which is composed of the cosine function raised to a power and the argument divided by a number, the chain rule guides us to find its derivative efficiently.
Specifically, if we have \( u = \theta/3 \) and \( r = \cos^3(u) \), the chain rule allows us to differentiate this by first finding the derivative of \( \cos^3(u) \) with respect to \( u \) and then multiplying by the derivative of \( u = \theta/3 \) with respect to \( \theta \), which is \( 1/3 \). This results in:
\[\frac{dr}{d\theta} = 3 \cos^2(\theta/3)(-\sin(\theta/3)) \times \frac{1}{3} = -\cos^2(\theta/3) \sin(\theta/3)\]
This step is crucial as it directly feeds into the arc length formula and is indicative of how calculus handles complex functions smoothly with the help of rules like the chain rule.