Problem 24 - Thomas Calculus

Question

The length of an astroid 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 entire astroid is approximately 6 units.

Step 1: Understand the Equation

The astroid in question is given by the equation: \(x^{2/3} + y^{2/3} = 1\). In the first quadrant, we are particularly interested in the portion where \(y = \left(1-x^{2/3}\right)^{3/2}\) over the interval \(\frac{\sqrt{2}}{4} \leq x \leq 1\).

Step 2: Set up the Integral for Length

To find the arc length of a curve defined by \(y = f(x)\) from \(x = a\) to \(x = b\), we use the formula: \[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\, dx\]For this problem, the curve is \(y = \left(1-x^{2/3}\right)^{3/2}\).

Step 3: Differentiate y with respect to x

Differentiate \(y = \left(1-x^{2/3}\right)^{3/2}\) to find \(\frac{dy}{dx}\). Let \(u = 1-x^{2/3}\), then \(y = u^{3/2}\). The derivative \(\frac{dy}{dx} = \frac{3}{2}u^{1/2} \cdot \left(-\frac{2}{3}x^{-1/3}\right) = -x^{-1/3}(1-x^{2/3})^{1/2}\).

Step 4: Substituting into the Arc Length Formula

Substitute \(\frac{dy}{dx}\) into the arc length formula:\[L = \int_{\frac{\sqrt{2}}{4}}^{1} \sqrt{1 + \left(-x^{-1/3}(1-x^{2/3})^{1/2}\right)^2} \, dx.\]This simplifies to:\[L = \int_{\frac{\sqrt{2}}{4}}^{1} \sqrt{1 + x^{-2/3}(1-x^{2/3})} \, dx.\]

Step 5: Simplify the Expression under the Integral

Further simplify the expression under the square root:\[L = \int_{\frac{\sqrt{2}}{4}}^{1} \sqrt{1 + x^{-2/3} - x^{-2/3}x^{2/3}} \, dx = \int_{\frac{\sqrt{2}}{4}}^{1} x^{-1/3} \, dx\]because the inside simplifies to \(1/(1-x^{2/3})\).

Step 6: Evaluate the Integral

Now, integrate \( \int_{\frac{\sqrt{2}}{4}}^{1} \left(x^{-1/3}\right) \, dx \):\[L = \left[ \frac{3}{2} x^{2/3} \right]_{\frac{\sqrt{2}}{4}}^{1}.\]Upon evaluating, it simplifies to:\[L = \frac{3}{2} \left(1 - \left(\frac{\sqrt{2}}{4}\right)^{2/3}\right).\]

Step 7: Final Calculation for the Whole Astroid

Multiply the length of half the first-quadrant portion by 8 (since the astroid is symmetric and has 8 symmetrical parts total):\(L = 8 \times \left(\frac{3}{2} \left(1 - \left(\frac{\sqrt{2}}{4}\right)^{2/3}\right)\right)\).

Step 8: Solve the Expression

Calculate the value of the expression:For \(\left(\frac{\sqrt{2}}{4}\right)^{2/3}\) and simplify the expression to find the total arc length. It simplifies to approximately 6.

Key Concepts

Parametric Equations

Parametric equations provide a way to describe curves in the plane using parameters, typically in terms of another variable, such as time \( t \). For example, an astroid can be represented by parametric equations \( x = \cos^3(t) \) and \( y = \sin^3(t) \) for \( t \) ranging from 0 to \( 2\pi \). By varying the parameter \( t \), we get all the points on the curve.

Parametric equations are incredibly useful because they allow for the representation of more complex curves that might be difficult to describe using standard \( y = f(x) \) forms. This is particularly relevant for cycles, loops, and—like in this case—starlike structures such as an astroid.

When dealing with parametric curves, one often computes derivatives and integrals differently, taking into account the rate of change of each parameter.

Arc Length

The concept of arc length is central in determining the total 'distance' along a curve. In the context of the astroid, we're interested in finding the arc length of the path defined by the parametric equations. The formula for calculating the arc length \( L \) of a curve defined by \( y = f(x) \) over an interval \([a, b]\) is given by:

\[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\, dx \]

This formula emerges from the Pythagorean theorem, where we sum infinitesimal line segments over the length of the curve. For parametric equations, the arc length \( L \) from \( t = a \) to \( t = b \) is:

\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt \]

Such calculations are crucial for more than just academic exercises. They find applications in engineering, physics, and computer graphics, where understanding the length of curves is essential.

Integral Calculus

Integral calculus allows us to find the total accumulation of quantities. In the case of the astroid, we use integration to sum up infinitesimal lengths along the curve to find the total arc length.

The integral we set up is based on the arc length integral formula, which always involves a square root expression. The limits of integration, such as from \( \frac{\sqrt{2}}{4} \) to \( 1 \), specify the section of the curve we are interested in.

Evaluating integrals often requires simplification of the expressions within, as seen with the astroid problem. This involves algebraic manipulation and sometimes substitution to make the integral solvable using standard calculus techniques.

• Integral calculus is indispensable for areas like calculating the area under curves, solving differential equations, and even in physics for determining quantities like work and mass.

Understanding this concept gives a strong foundation for solving problems that involve summing continuous quantities.

Differentiation

Differentiation is the process of finding the rate at which a function changes at any point. In the case of an astroid, we differentiate the function \( y = \left(1-x^{2/3}\right)^{3/2} \) to facilitate the calculation of arc length.

This involves using the chain rule, a fundamental technique in differentiation. Let \( u = 1-x^{2/3} \); then \( y = u^{3/2} \). We find \( \frac{dy}{dx} \) by differentiating explicitly with respect to \( x \), yielding:

\[ \frac{dy}{dx} = -x^{-1/3}(1-x^{2/3})^{1/2} \]

Calculating \( \frac{dy}{dx} \) accurately allows us to use the appropriate integrand in the arc length formula. Differentiation is not just about calculating slopes; it's about understanding functions deeply, knowing how they behave, and using this knowledge to solve real-world problems.