Question

Sketch and find the area under one arch of the cycloid \(x=r(\theta-\sin \theta), y=r(1-\cos \theta)\).

Short answer

The area under one arch of the cycloid is \(3\pi r^2\).

Step-by-step solution

Understanding the Cycloid Equation

The cycloid is defined parametrically by the equations \(x = r(\theta - \sin \theta)\) and \(y = r(1 - \cos \theta)\). This represents the path traced by a point on the circumference of a circle of radius \(r\) rolling along a straight line.

Determine the Range for One Arch

For one complete arch of the cycloid, \(\theta\) ranges from 0 to \(2\pi\). As \(\theta\) goes from 0 to \(2\pi\), the point on the circle returns to its initial position after tracing the full arch.

Set Up the Area Integral

The area under one arch of the cycloid can be found using the integral of \(y\) with respect to \(x\). The integral is set as \( \int_{0}^{2\pi} y \frac{dx}{d\theta} d\theta\). Here, \(\frac{dx}{d\theta} = r(1 - \cos \theta)\).

Integration of Determinants

Now substitute \(y = r(1-\cos \theta)\) and \( \frac{dx}{d\theta}\) into the integral: \[ \int_{0}^{2\pi} r(1 - \cos \theta) r (1 - \cos \theta) d\theta = r^2 \int_{0}^{2\pi} (1 - \cos \theta)^2 d\theta \].

Solving the Integral

Evaluate the integral: \[ \int_{0}^{2\pi} (1 - \cos \theta)^2 d\theta \]. Use the trigonometric identity \((1 - \cos \theta)^2 = 1 - 2\cos \theta + \cos^2 \theta\), where \(\cos^2 \theta = \frac{1+\cos(2\theta)}{2}\).Substitute and integrate the terms: \[ \begin{aligned} &\int_{0}^{2\pi} \left(1 - 2\cos \theta + \frac{1 + \cos 2\theta}{2}\right) d\theta \= &\int_{0}^{2\pi} \left(\frac{3}{2} - 2\cos \theta + \frac{1}{2}\cos 2\theta \right) d\theta \= &\left[ \frac{3}{2} \theta - 2\sin \theta + \frac{1}{4}\sin 2\theta \right]_{0}^{2\pi} \= &\left( 3\pi - 0 + 0 \right) - \left( 0 - 0 + 0 \right) \= &3\pi.\end{aligned} \]

Calculate the Area

Finally, multiply by \(r^2\) to account for the integral constants: \[\text{Area} = r^2 \cdot 3\pi = 3\pi r^2\].

Key concepts

Cycloid

A cycloid is a fascinating curve generated by the path of a point on the edge of a circular wheel as it rolls along a flat surface. Imagine placing a coin on a table and marking a point on its edge. As the coin rolls without slipping, the marked point traces a particular path known as the cycloid.

The cycloid is mathematically defined using parametric equations: \( x = r(\theta - \sin \theta) \) and \( y = r(1 - \cos \theta) \). This specific formulation helps in describing the motion of the point as the circle rolls.

Understanding the cycloid involves visualizing how a single point can create an arch, commonly called an arch of the cycloid, and the importance of this shape in various fields such as physics and engineering can be profound.

• Each cycle of the cycloid, called an arch, completes when the point returns to its initial contact position after the circle rolls through one complete revolution.
• Learning to sketch the cycloid helps to visualize this unique path and understand how its parameters function with each rotation.

Parametric Equations

Parametric equations are essential when describing curves that cannot easily be defined with a single equation. In the study of cycloids, these equations describe a set of related values using a parameter, typically \(\theta\), which represents the angle of rotation of the rolling circle.

In the context of the cycloid, the parametric equations are given by:

• \( x = r(\theta - \sin \theta) \)
• \( y = r(1 - \cos \theta) \)

The parameter \(\theta\) varies over an interval, usually from 0 to \(2\pi\), which denotes a full revolution of the circle.

These parametric equations allow us to handle more complicated shapes and motions, such as the cycloid, where there isn't a straightforward y-intercept equation like in linear functions.

Parametric equations thus enable the representation of complex dynamic systems and help in calculations involving those systems.

Integral Calculus

Integral calculus is a major mathematical field that focuses on accumulation and area under curves. In the case of finding the area under one arch of a cycloid, integral calculus is used to sum up tiny slices or differential elements of space under the curve.

The key to solving the area under the cycloid lies in setting up the correct integral equation, which combines elements from the parametric formulation with calculus. As shown in our exercise, the integral required is:
\[\int_{0}^{2\pi} y \frac{dx}{d\theta} d\theta\]

This is where integral calculus meets parametric equations, showing how each mathematical concept relies on others to solve real-world and theoretical problems.

Additionally, integral calculus provides tools such as substitution and trigonometric identities necessary to solve more complex integrals encountered, such as those involving trigonometric functions like \(1 - \cos(\theta)\).

Area Calculation

Calculating the area under the cycloid is a classic application of both parametric equations and integral calculus. The process involves breaking the area into manageable parts and finding their sum using integrals. For our cycloid, the required calculation can be formed as:

\[ \text{Area} = r^2 \cdot \int_{0}^{2\pi} (1 - \cos \theta)^2 d\theta \]

This integral takes into account both the parametric expression for \(y = r(1 - \cos \theta)\) and the derivative \( \frac{dx}{d\theta} = r(1 - \cos \theta) \). Through integration, we transform the geometrical problem of finding the area into an algebraic one that can be solved step-by-step.

Using trigonometric identities, such as \((1 - \cos \theta)^2 = 1 - 2\cos \theta + \cos^2 \theta\), simplifies solving the integral.

After solving and evaluating the integral, multiplying by \(r^2\) yields the total area under one arch as \(3\pi r^2\). This result reveals how powerful and efficient mathematical tools can be in solving geometrical problems.