Question

For the following exercises, find the arc length of the curve on the indicated interval of the parameter.\(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta\) on the interval \([0,2 \pi)\) (the hypocycloid)

Short answer

The arc length is \(6a\).

Step-by-step solution

Recall the Formula for Arc Length of a Parametric Curve

The arc length of a parametric curve given by \(x=f(t)\) and \(y=g(t)\) over interval \([a,b]\) is calculated using the formula:\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Differentiate x and y with Respect to θ

Given \(x=a \cos^3 \theta\) and \(y=a \sin^3 \theta\), first find the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).- \(\frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos^3 \theta) = a \cdot 3\cos^2\theta (-\sin\theta) = -3a \cos^2\theta \sin\theta\)- \(\frac{dy}{d\theta} = \frac{d}{d\theta}(a \sin^3 \theta) = a \cdot 3\sin^2\theta (\cos\theta) = 3a \sin^2\theta \cos\theta\)

Setup the Arc Length Integral

Substitute the derivatives into the arc length formula:\[ L = \int_{0}^{2\pi} \sqrt{(-3a \cos^2\theta \sin\theta)^2 + (3a \sin^2\theta \cos\theta)^2} \, d\theta \]

Simplify the Expression Under the Square Root

Simplify the expression under the square root:\[ L = \int_{0}^{2\pi} \sqrt{9a^2 \cos^4\theta \sin^2\theta + 9a^2 \sin^4\theta \cos^2\theta} \, d\theta \]Factor out \(9a^2\cos^2\theta \sin^2\theta\):\[ L = \int_{0}^{2\pi} \sqrt{9a^2 \cos^2\theta \sin^2\theta (\cos^2\theta + \sin^2\theta)} \, d\theta \]Since \(\cos^2\theta + \sin^2\theta = 1\), it becomes:\[ L = 3a \int_{0}^{2\pi} \left|\cos\theta \sin\theta\right| \, d\theta \]

Evaluate the Integral

Consider \(\left|\cos\theta \sin\theta\right| = \frac{1}{2}\left|\sin(2\theta)\right|\). Because \(\sin(2\theta)\) is positive for \([0,\pi/2] \cup [\pi, 3\pi/2]\) and negative for \([\pi/2,\pi] \cup [3\pi/2, 2\pi]\), we split the integral:\[ L = 3a \left(\int_{0}^{\pi/2} \frac{1}{2}\sin(2\theta) \, d\theta - \int_{\pi/2}^{\pi} \frac{1}{2}\sin(2\theta) \, d\theta + \int_{\pi}^{3\pi/2} \frac{1}{2}\sin(2\theta) \, d\theta - \int_{3\pi/2}^{2\pi} \frac{1}{2}\sin(2\theta) \, d\theta \right)\]Evaluating each, noting symmetries and periodicity, results in the total:\[ L = 6a \].

Conclusion

The arc length of the given parametric curve over the interval \([0, 2\pi)\) is \(6a\).

Key concepts

Parametric Curves

Parametric curves are a fascinating way to represent a curve in the plane. Unlike the traditional format of representing a curve using the relationship between x and y, parametric representation uses a separate parameter, usually denoted as \( t \) or \( \theta \), and defines both x and y using this parameter. This allows us to describe complex and intricate paths, where tracing direction and speed come into play.

In our example, the parametric equations are \( x = a \cos^3 \theta \) and \( y = a \sin^3 \theta \). Here, \( \theta \) serves as our parameter, and as \( \theta \) varies over the interval \([0, 2\pi)\), the shape known as a hypocycloid is traced out.

This representation provides a powerful way to express more complex curves that are not functions in the traditional sense. By understanding and utilizing parametric curves, we can model and analyze a wide range of real-world paths and movements that cannot be encapsulated by standard functions alone.

Calculus

Calculus is the tool we use to understand change and motion, making it particularly useful in analyzing parametric curves where change is inherent. The heart of calculus lies in the concepts of differentiation and integration, which we leverage to find values such as arc length.

For our problem, the task is to find the arc length of the hypocycloid described by our parametric equations. To do so, calculus gives us the arc length formula specific for parametric curves: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ dt \]This formula considers each tiny segment along the curve, measuring the actual distance traveled as \( t \), or in this case, \( \theta \), varies through its interval. Calculus allows us to rigorously determine this summation of infinitesimal distances as you "walk" along the curve.

Differentiation

Differentiation is the process that calculus uses to tackle change. It's how we understand slopes, rates of change, and the behavior of curves. In the context of parametric curves, differentiation helps us understand how each coordinate changes concerning the parameter \( \theta \).

For our specific problem, we first need to differentiate \( x = a \cos^3 \theta \) and \( y = a \sin^3 \theta \) with respect to \( \theta \). Here's how we approach it: - For \( x \), the chain rule helps in deriving \( \frac{dx}{d\theta} = -3a \cos^2\theta \sin\theta \).- Similarly, for \( y \), we find \( \frac{dy}{d\theta} = 3a \sin^2\theta \cos\theta \).Working through these steps, differentiation reveals the instantaneous behavior of each component of the parametric curve, necessary for correctly applying the arc length formula. With these derivatives, we integrate to capture the full length of the curve over interval \([0, 2\pi)\), leading to our result \( 6a \). Differentiation thus opens the door to deeper insights into complex curves.