Problem 47 - Thomas Calculus

Question

Cycloid a. Find the length of one arch of the cycloid \(x=a(t-\sin t), \quad y=a(1-\cos t)\) b. Find the area of the surface generated by revolving one arch of the cycloid in part (a) about the \(x\) -axis for \(a=1 .\)

Step-by-Step Solution

Answer

a. Length is \(2\pi a\). b. Surface area is \(8\pi^2\) for \(a=1\).

Step 1: Understand the Cycloid Parameters

A cycloid is the path traced by a point on the circumference of a circle as it rolls along a straight line. For this problem, we have the parametric equations: \(x=a(t-\sin t)\) and \(y=a(1-\cos t)\). An arch of the cycloid is completed as \(t\) goes from 0 to \(2\pi\).

Step 2: Length of One Arch of the Cycloid

The formula for the arc length of a parametric curve \((x(t), y(t))\) from \(t = t_1\) to \(t = t_2\) is given by \(L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). Calculate the derivatives: \(\frac{dx}{dt} = a(1-\cos t)\) and \(\frac{dy}{dt} = a\sin t\). Substitute these into the formula and evaluate \(L = \int_{0}^{2\pi} \sqrt{(a(1-\cos t))^2 + (a\sin t)^2} \, dt\). Simplifying, we find \(L = \int_{0}^{2\pi} a \, dt = 2\pi a\).

Step 3: Setup the Surface Area Formula

The formula for the surface area when revolving a parametric curve around the \(x\)-axis is\[SA = \int_{t_1}^{t_2} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt.\] Use \(y = a(1-\cos t)\) and the previously calculated expressions for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

Step 4: Integrate the Surface Area Expression

Substitute \(a = 1\) and integrate: \(SA = \int_{0}^{2\pi} 2\pi (1-\cos t) \times \sqrt{(1-\cos t)^2 + \sin^2 t} \, dt\). This simplifies to \(SA = \int_{0}^{2\pi} 2\pi (1-\cos t) \times 1 \, dt\) or \(SA = \int_{0}^{2\pi} 2\pi (1-\cos t) \, dt\). Evaluating this integral gives \(SA = 2\pi\left[ t - \sin t \right]_{0}^{2\pi} = 8\pi^2\).

Key Concepts

Parametric Equations

A parametric equation is a way of defining a curve in terms of a parameter, usually denoted as \(t\). It's like letting \(t\) run freely and as it does, it draws out the path of the curve. For the cycloid in our exercise, the parametric equations are \(x=a(t-\sin t)\) and \(y=a(1-\cos t)\). These equations describe how a point on the edge of a circle traces out a cycloid as the circle rolls along a straight line.

• The parameter \(t\) often represents time.
• \(x(t)\) and \(y(t)\) describe the position of the point at any time \(t\).

The advantage of parametric equations is they are powerful for modeling curves that don't fit typical \(y=f(x)\) forms. In the cycloid, the circle rolls without slipping, meaning the distance covered in one roll is exactly the circle's circumference, \(2 \pi a\), guiding \(t\) to range from \(0\) to \(2 \pi\) for one arch.

Arc Length

The arc length of a curve described by parametric equations is a measure of how long the curve is. Imagine you have a piece of string lying along the route of the cycloid, then straightening it would give you the arc length. We find it using the formula: \[L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]For our cycloid, the differentials \(\frac{dx}{dt} = a(1-\cos t)\) and \(\frac{dy}{dt} = a\sin t\) are substituted into the formula. This simplifies the integral to \[L = \int_{0}^{2\pi} a \, dt\]Resulting in the arc length of one complete arch being \(2\pi a\).

• Understandably, finding arc length requires dealing with derivatives and integrations.
• The result \(2\pi a\) confirms the neatness of parametric equations as they navigate through complex shapes.

Surface Area Revolution

The surface area due to the revolution of a curve in 3D space is about examining what happens when a curve is spun around an axis. In this problem, one arch of the cycloid revolves around the \(x\)-axis. The idea is similar to spinning a string of lights to see the lit-up surface item. The formula for surface area when the curve moves around the \(x\)-axis is:\[SA = \int_{t_1}^{t_2} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]Plug in \(y = a(1-\cos t)\), and the derivatives we calculated. Simplified further, when \(a=1\), the expression becomes:\[SA = \int_{0}^{2\pi} 2\pi (1-\cos t) \, dt\]Evaluation tells us the surface area from this revolution is \(8\pi^2\).

• This surface area concept transforms abstract curves into visual, rotational surfaces.
• Understanding it blends calculus and geometry, offering complete insight into the shape's outer area.