Question

Suppose the function \(y=h(x)\) is nonnegative and continuous on \([\alpha, \beta],\) which implies that the area bounded by the graph of h and the x-axis on \([\alpha, \beta]\) equals \(\int_{\alpha}^{\beta} h(x) d x\) or \(\int_{\alpha}^{\beta} y d x .\) If the graph of \(y=h(x)\) on \([\alpha, \beta]\) is traced exactly once by the parametric equations \(x=f(t), y=g(t),\) for \(a \leq t \leq b,\) then it follows by substitution that the area bounded by h is $$\begin{array}{l}\int_{\alpha}^{\beta} h(x) d x=\int_{\alpha}^{\beta} y d x=\int_{a}^{b} g(t) f^{\prime}(t) d t \text { if } \alpha=f(a) \text { and } \beta=f(b) \\\\\left(\text { or } \int_{\alpha}^{\beta} h(x) d x=\int_{b}^{a} g(t) f^{\prime}(t) d t \text { if } \alpha=f(b) \text { and } \beta=f(a)\right)\end{array}$$. Find the area under one arch of the cycloid \(x=3(t-\sin t)\) \(y=3(1-\cos t)(\text { see Example } 5)\)

Short answer

#Answer# The area under one arch of the given cycloid with parametric equations \(x = 3 (t - \sin t)\) and \(y = 3 (1 - \cos t)\) is \(27\pi\).

Step-by-step solution

Find limits of integration (a, b)

To find the limits of integration, we need to find the values of \(t\) for which the cycloid completes one arch. An arch of the cycloid is completed when the rolling circle makes one full rotation. In this case, the circle has a radius of 3, so it takes \(2\pi\) radians for it to complete one rotation. Therefore, we have the limits of integration as \(a = 0\) and \(b = 2\pi\).

Find the derivative of f(t) with respect to t

The given parametric equation for \(x\) is: \(x = 3(t - \sin t)\). We need to find \(f'(t)\), which is the derivative of the function \(f(t)\). Using the chain rule, we get: $$f'(t) = \frac{dx}{dt} = 3(1 - \cos t)$$

Substitute the values into the integral formula

Now we have all the required information to substitute into the integral formula. So, we have: $$\text{Area} = \int_{0}^{2\pi} g(t) f^\prime(t) dt = \int_{0}^{2\pi} 3(1 - \cos t) \cdot 3 (1 - \cos t) dt$$

Simplify and integrate

Next, we simplify the integrand and perform the integration: $$\text{Area} = \int_{0}^{2\pi} 9 (1 - \cos t)^2 dt = 9 \int_{0}^{2\pi} (1 - 2\cos t + \cos^2 t) dt$$ Now, we know that \(\cos^2 t = \frac{1 + \cos 2t}{2}\). So, the integral becomes: $$\text{Area} = 9 \int_{0}^{2\pi} \left(1 - 2\cos t + \frac{1 + \cos 2t}{2}\right) dt$$ Now, we integrate each term separately: $$\text{Area} = 9 \left[\int_{0}^{2\pi} dt - 2\int_{0}^{2\pi} \cos t dt + \frac{1}{2}\int_{0}^{2\pi} (1 + \cos 2t) dt\right]$$ After integrating and simplifying, we get: $$\text{Area} = 9 \left[2\pi - 0 + \frac{1}{2}(2\pi)\right] = 9(3\pi)$$

Final answer

So, the area under one arch of the given cycloid is: $$\text{Area} = 27\pi$$ Thus, the area under one arch of the cycloid with parametric equations \(x = 3 (t - \sin t)\) and \(y = 3 (1 - \cos t)\) is \(27\pi\).

Key concepts

Parametric Equations

Parametric equations are an excellent mathematical tool used to express a set of quantities as explicit functions of a parameter. Unlike the standard representation, which directly relates two variables like \(x\) and \(y\), parametric equations introduce a third variable, usually \(t\), as a parameter. This technique helps describe more complex curves and motions.

For example, in the cycloid exercise, the parametric equations \(x=3(t-\sin t)\) and \(y=3(1-\cos t)\) describe the coordinates of a point moving along a path. Here, \(t\) represents the angle through which a circle rotates to generate the cycloid curve. This approach simplifies the description of curves that are hard to define with a single equation.

Parametric equations are particularly useful when dealing with

• Circular or elliptical paths.
• Curves traced over time, like cycloids and Lissajous patterns.
• Physical phenomena where motion is involved.

Understanding these equations allows for a deeper insight into the behavior of dynamic systems and curves.

Cycloid

A cycloid is a fascinating curve traced by a point on the circumference of a circle as it rolls along a straight line. Imagine a pebble stuck to the rim of a bicycle wheel that rolls along a flat road. The path that pebble traces in the air is a cycloid. The cycloid is known for several interesting properties that make it a subject of study in both calculus and physics.

In our exercise, the parametric equations \(x=3(t-\sin t)\) and \(y=3(1-\cos t)\) represent the path of the cycloid generated by a circle with a radius of 3 units. Here, the equation for \(x\) describes the horizontal displacement, and \(y\) describes the vertical displacement.

Cycloids are more than just math curiosities; they have practical applications:

• In the design of gear teeth for conjugate gear pairs.
• In architecture, for designing arches like the famous cycloidal arches.
• As tautochrones, where a bead sliding down a cycloid reaches the bottom in the same time, regardless of where it starts.

Delving into cycloids offers a unique glimpse into geometry and its intersections with real-world applications.

Integration

Integration is a central concept in calculus that involves finding the accumulation of a quantity, often interpreted as the area under a curve. When dealing with functions defined by parametric equations, integration helps calculate the area bounded by these curves within specific limits.

In the exercise, we employ integration to compute the area under one arch of a cycloid. By using the formula \[\text{Area} = \int_{a}^{b} g(t) f'(t) \, dt\]we can determine this area. This approach involves taking the product of \(g(t)\) and the derivative \(f'(t)\), ensuring we calculate the area bounded accurately.

The process consists of a few key steps:

• Identifying the limits of integration that correspond to one cycle of the parametric curve.
• Calculating the derivatives of the x and y components.
• Substituting and simplifying within the integral.
• Finally, evaluating the definite integral to find the solution.

The use of integration here adds depth to understanding the dimensions and size of regions enclosed by curves in a parametric setting.

Area under a curve

Finding the area under a curve is a common problem in calculus, allowing for the quantification of the space enclosed by a curve against a baseline, typically the x-axis. With parametric equations like those in the cycloid example, this process becomes an exploration of calculus where each curve's unique path requires an equally unique approach.

In our specific problem, we calculate the area under one arch of a cycloid using a parametric integration formula. This involves evaluating the integral of the curve from one limit to another. By doing this, we can ascertain the entire space enclosed within a single cycle of the cycloid.

This methodology is crucial for:

• Ship designers to calculate hull areas, crucial for buoyancy and balance.
• Architects to understand building floor plans and lot usage better.
• Economists to model and analyze area measures in various graphs, depicting data visually.

Calculating the area under a curve, especially in a parametric context, extends beyond theoretical mathematics and into everyday applications in numerous fields. It provides the analytical backbone for assessing shape, curves, and their implications effectively.