Question

Finding the Area under a Parametric Curve Find the area under the curve of the cycloid defined by the equations $$ x(t)=t-\sin t, \quad y(t)=1-\cos t, \quad 0 \leq t \leq 2 \pi $$

Short answer

The area under the cycloid curve is \(3\pi\).

Step-by-step solution

Understand the Problem

We need to find the area under the parametric curve given by the equations \( x(t) = t - \sin t \) and \( y(t) = 1 - \cos t \) over the interval \( 0 \leq t \leq 2\pi \). This means we integrate to find the area described as the parametric curve traces out as \( t \) evolves over this interval.

Identify the Formula for Area Under a Parametric Curve

The area \( A \) under a parametric curve is calculated using the integral formula: \[ A = \int_{t=a}^{t=b} y(t) \cdot \frac{dx}{dt} \, dt \] where \( x(t) \) and \( y(t) \) are the parametric equations of the curve.

Compute \( \frac{dx}{dt} \)

Differentiate \( x(t) = t - \sin t \) with respect to \( t \): \[ \frac{dx}{dt} = 1 - \cos t \]

Substitute in the Integral Formula

Substitute \( y(t) = 1 - \cos t \) and \( \frac{dx}{dt} = 1 - \cos t \) into the area formula. This gives:\[ A = \int_{0}^{2\pi} (1 - \cos t)(1 - \cos t) \, dt \]

Simplify the Integral

Simplify the expression:\[ A = \int_{0}^{2\pi} (1 - 2\cos t + \cos^2 t) \, dt \]

Evaluate the Expressions

We break down the integral into simpler parts:1. \( \int_{0}^{2\pi} 1 \, dt = [t]_{0}^{2\pi} = 2\pi .\) 2. \( \int_{0}^{2\pi} -2\cos t \, dt = [-2\sin t]_{0}^{2\pi} = 0 .\)3. For \( \int_{0}^{2\pi} \cos^2 t \, dt \), use the identity \( \cos^2 t = \frac{1+\cos 2t}{2} \), which simplifies to: \[ \int_{0}^{2\pi} \frac{1+\cos 2t}{2} \, dt = \frac{1}{2} \int_{0}^{2\pi}(1+\cos 2t) \, dt = \frac{1}{2} (\int_{0}^{2\pi} 1 \, dt + \int_{0}^{2\pi} \cos 2t \, dt) \]\( = \frac{1}{2} (2\pi + 0) = \pi\)

Add up the Evaluated Parts

Combine all these evaluated integrals to find the area:\[ A = 2\pi + 0 + \pi = 3\pi \]

Key concepts

Cycloid

A cycloid is a fascinating curve generated by a point on the circumference of a circle as the circle rolls along a straight line without slipping. Imagine a wheel rolling along the ground. A pebble attached to the rim will trace a cycloid curve.
The parametric equations of a cycloid are given as

• \( x(t) = t - \sin t \)
• \( y(t) = 1 - \cos t \)

These equations result from the horizontal and vertical displacements of the pebble, where \( t \) represents the angle through which the circle has rotated.
The stunning characteristic of a cycloid is its periodic nature and cusp points where the curve touches the line momentarily. These interesting points are at regular intervals on the horizontal axis whenever \( t \) is a multiple of \( 2\pi \).
In this exercise, we're exploring the area under this type of curve over one complete cycle, from \( t = 0 \) to \( t = 2\pi \). Understanding cycloids is crucial for a range of applications, from physics to engineering, where motion, cycles, or periodic behavior plays a critical role.

Integral Calculus

Integral calculus is a fundamental branch of mathematics focusing on integrals, which represent the accumulation of quantities. Conceptually, if we think of differentiation as breaking down a function to understand how it changes, integration is the piecing together to find the total or net value of a function over an interval.
When dealing with parametric equations like in this exercise, integrating the curve means finding the area it covers on a plane. We rely on the integral formula for such a task:

• \[ A = \int_{t=a}^{t=b} y(t) \cdot \frac{dx}{dt} \, dt \]

This formula computes the total area under the parametric curve by considering the rate of change of \( x \) with respect to \( t \) and the value of \( y(t) \). Calculating the integral involves:

• Finding \( \frac{dx}{dt} \) by differentiating the \( x \)-equation.
• Substituting \( y(t) \) and \( \frac{dx}{dt} \) into the integral.
• Evaluating the integral over the specified interval.

This method allows us to handle more complex curves that cannot be described by simple algebraic equations, essential for real-world problem-solving in dynamic systems.

Area Under a Curve

Finding the area under a curve is a practical problem solved with integral calculus, especially when the curve is defined with parametric equations. The area under the curve corresponds to the integral of the function over a given interval.
In this specific exercise with the cycloid:

• The area \( A \) is computed using \[ A = \int_{0}^{2\pi} (1 - \cos t)(1 - \cos t) \, dt \].

To grasp this process:

• We first expanded \((1 - \cos t)(1 - \cos t)\) to simplify the expression inside the integral.
• Then, used trigonometric identities, such as \( \cos^2 t = \frac{1 + \cos 2t}{2} \), to simplify further.
• The integral is split into simpler parts and solved separately.

Finally, the sum of these parts gives the total area under one arch of the cycloid \(3\pi\). Understanding this method is crucial in various applications, such as physics for projectile motion or planning in architectural designs where space and load distribution are analyzed.