Problem 24 - Thomas Calculus in SI Units

Question

The graph of the equation \(x^{2 / 3}+\) \(y^{2 / 3}=1\) is one of a family of curves called astroids (not "asteroids") because of their starlike appearance (see the accompanying figure). Find the length of this particular astroid by finding the length of half the first-quadrant portion, \(y=\left(1-x^{2 / 3}\right)^{3 / 2}\) \(\sqrt{2} / 4 \leq x \leq 1,\) and multiplying by 8

Step-by-Step Solution

Answer

The length of the astroid is \(6\).

Step 1: Parametrize the Curve

To find the length of the astroid, first parametrize the equation. Notice that an astroid equation can be parameterized as follows: for some angle \(\theta\), set \( x = \cos^3(\theta) \) and \( y = \sin^3(\theta) \). This is because raising \(\cos\theta\) and \(\sin\theta\) to the power of 3 will maintain the sum of powers equal to 1 required by the equation. Hence, the parametric equations are \( x = \cos^3(\theta) \) and \( y = \sin^3(\theta) \), where \(0 \leq \theta \leq \pi/2\) for the first quadrant.

Step 2: Find Derivatives

Find the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). If \(x = \cos^3(\theta)\), then \(\frac{dx}{d\theta} = -3\cos^2(\theta)\sin(\theta)\). Similarly, if \(y = \sin^3(\theta)\), then \(\frac{dy}{d\theta} = 3\sin^2(\theta)\cos(\theta)\). These derivatives will be used to find the arc length.

Step 3: Set Up Arc Length Integral

The formula for the arc length of a parametric curve is \( L = \int_a^b \sqrt{\left( \frac{dx}{d\theta} \right)^2 + \left( \frac{dy}{d\theta} \right)^2} \, d\theta \). Substitute \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) from Step 2 into this formula. This results in the integral \[ L = \int_{\theta_1}^{\theta_2} \sqrt{(-3\cos^2(\theta)\sin(\theta))^2 + (3\sin^2(\theta)\cos(\theta))^2} \, d\theta \].

Step 4: Simplify the Integrand

Simplify the integrand: \[ \sqrt{(-3\cos^2(\theta)\sin(\theta))^2 + (3\sin^2(\theta)\cos(\theta))^2} = 3\sqrt{\cos^4(\theta)\sin^2(\theta) + \sin^4(\theta)\cos^2(\theta)} \]. Factor out common terms: \[ = 3\cos(\theta)\sin(\theta)\sqrt{\cos^2(\theta) + \sin^2(\theta)} = 3\cos(\theta)\sin(\theta) \]. Thus, \( \int\) becomes \[ L = 3 \int_{\theta_1}^{\theta_2} \cos(\theta)\sin(\theta) \, d\theta \].

Step 5: Evaluate the Integral

Make the substitution \(u = \sin(\theta)\), \(du = \cos(\theta) \, d\theta\), transforming the integral to \(3 \int_{u_1}^{u_2} u \, du\). The limits for \(u\) change accordingly from \(u(\theta_1) = \sin(\theta_1)\) to \(u(\theta_2) = \sin(\theta_2)\). The integral evaluates to \[ \frac{3}{2} [u^2]_{u_1}^{u_2} = \frac{3}{2} [\sin^2(\theta_2) - \sin^2(\theta_1) ]\].

Step 6: Substitute Limits

For the problem's domain, set \(x = \frac{\sqrt{2}}{4}\). Using \(x = \cos^3(\theta)\), solve \( \cos(\theta) = \left(\frac{\sqrt{2}}{4}\right)^{1/3} \). Approximate it to find \(\theta_1\) and use \(\theta_2 = \frac{\pi}{4}\) (as found from symmetry study that yields easy intervals). Substitute these values back into your integrated form to obtain the length of a half quadrant.

Step 7: Calculate Total Length

Multiply the obtained length of the quarter arc \( 3 \int_{\theta_1}^{\theta_2} \cos(\theta)\sin(\theta) \, d\theta \) by 4 (as it covers half the arc in the first quadrant) and then by 2 (to account for the entire arc). This provides the total length of the astroid.

Key Concepts

Parametric Equations

Parametric equations offer a sleek way to represent curves in the plane. Instead of using a single equation like \( y = f(x) \), parametric equations separate the coordinates as functions of an independent parameter, often \( t \) or \( \theta \). In the case of the astroid, the parameter \( \theta \) is a common choice. Here, the equations are \( x = \cos^3(\theta) \) and \( y = \sin^3(\theta) \), which are used for the first quadrant for \( 0 \leq \theta \leq \frac{\pi}{2} \).
- These equations transform the complexity of the astroid into more familiar trigonometric forms.
- Parametric forms rely heavily on trigonometric identities that ease the representation and manipulation of curves.
- By employing parameters, it's easier to calculate derived quantities such as arc lengths or areas related to these curves.

Arc Length

Finding the arc length of a parametric curve is crucial in calculus for understanding the geometry of the curve. Given the parametric equations \( x(\theta) \) and \( y(\theta) \), the arc length is computed using the formula \[L = \int_a^b \sqrt{\left( \frac{dx}{d\theta} \right)^2 + \left( \frac{dy}{d\theta} \right)^2} \, d\theta.\]
- Here, \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) are the derivatives with respect to the parameter \( \theta \).
- This formula provides a systematic way to "walk along the path" defined by the parametric curve and add up the small segments to find the total length.
- For our astroid, simplifying the integrand using trigonometric identities reveals the beauty of parametric form: the expression simplifies to \(3 \cos(\theta) \sin(\theta)\), and the integral becomes more straightforward to evaluate.
Evaluating \[ \int \cos(\theta)\sin(\theta) \, d\theta \]leads to calculating simple trigonometric integrals, culminating in the complete arc length of the astroid after multiplying by 8 for symmetry.

Derivatives

Derivatives are powerful tools that help understand the rate of change of functions. When we deal with parametric forms, derivatives provide the gradients needed for calculating quantities like arc length. For the parametric equations \( x = \cos^3(\theta) \) and \( y = \sin^3(\theta) \), their derivatives are given by:
- \( \frac{dx}{d\theta} = -3\cos^2(\theta)\sin(\theta) \)- \( \frac{dy}{d\theta} = 3\sin^2(\theta)\cos(\theta) \)
These derivatives measure how \( x \) and \( y \) change as \( \theta \) changes, essential for understanding the slope and behavior of the curve at every point:
- They allow us to setup the integrand for length calculation in the arc length formula.
- The derivatives often reveal symmetry and periodic properties in trigonometric parametric equations, making complex integrals easier to compute.
Understanding these derivatives is crucial for applications not only in geometry and calculus but also in physics and engineering where precise motion paths are modeled parametrically.