Answer to Question #291267 in Calculus for NEO

Question

Trace the curve x=a(theta + sin theta), y = a(1-cos theta). State the properties you use for tracing it also.

Answer 1

We use the following properties for tracing the curve. We have the Cartesian curve defined by the parametric equations $x=f(\theta), y = g(\theta).$

Since $𝑦$ is a periodic function of $\theta$ with period $2\pi,$ it is sufficient to trace the curve for $\theta\in[0, 2\pi].$ For $\theta\in[0, 2\pi], x$ and $y$ are well defined.

Trace the curve $x=a(\theta + \sin \theta), y = a(1-\cos \theta), 0\leq\theta\leq 2\pi, a>0.$

1. Symmetry

$$f(-\theta)=a(-\theta+\sin(-\theta))=-f(\theta),$$

$$g(-\theta)=a(1-\cos(-\theta))=g(\theta)$$

Therefore, the curve is symmetrical about the 𝑦 −axis. Curve is not symmetrical about 𝑦- axis.

2. Origin

$$(0, 0), x=f(\theta)=a(\theta + \sin \theta)=0,$$

$$y=g(\theta) = a(1-\cos \theta)=0$$

$$\begin{cases} \sin\theta=-\theta \\ \cos\theta=1 \end{cases}=> \begin{cases} \sin\theta=-\theta \\ \theta=0\ \ or\ \ \theta=2\pi \end{cases}=>\theta=0$$

A curve passes through the origin.

Derivatives:

$$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{a\sin \theta}{a(1+\cos \theta)}=\dfrac{\sin \theta}{1+\cos \theta}=\tan(\theta/2)$$

At $\theta=0, \dfrac{dy}{dx}=0.$ Tangent to the curve at $\theta=0$ is $x$ -axis.

At $\theta=\pi, \dfrac{dy}{dx}=\infin.$ Tangent to the curve at $\theta=\pi$ is perpendicular to $x$ -axis.

3. Intercepts

Intersection with $x$-axis: The points of intersection of the curve with the $x$-axis are given by the roots of $g(\theta)=a(1-\cos \theta)=0, 0\leq \theta \leq 2\pi, a>0.$

$$\cos \theta=1$$

$$\theta=0\ or\ \theta=2\pi$$

$$x=0\ or\ x=2\pi a$$

Then $(x, y)=(0, 0), (x, y)=(2\pi a, 0).$

Intersection with $y$-axis: The points of intersection of the curve with the $y$-axis are given by the roots of $f(\theta)=a(\theta+\sin \theta)=0, 0\leq \theta \leq 2\pi, a>0.$

$$\theta+\sin \theta=0$$

$$\theta=0$$

$$y=0$$

Then $(x, y)=(0, 0).$

4. Asymptotes

$x=a(\theta+\sin \theta), y=a(1-\cos \theta).$

There is no vertical asymptote.

There is no horizontal asymptote.

There is no oblique asymptote.

5. Regions where no Part of the curve lies

$x=a(\theta+\sin \theta), y=a(1-\cos \theta).$

Note that $y\geq 0.$ Entire curve lies above the $y$ -axis $(0\leq y\leq 2a).$

6. First derivative

$x=a(\theta+\sin \theta), y=a(1-\cos \theta).$

$$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{a\sin \theta}{a(1+\cos \theta)}=\dfrac{\sin \theta}{1+\cos \theta}=\tan(\theta/2)$$

At $\theta=0, \dfrac{dy}{dx}=0.$ Tangent to the curve at $\theta=0$ is $x$ -axis.

At $\theta=\pi, \dfrac{dy}{dx}=\infin.$ Tangent to the curve at $\theta=\pi$ is perpendicular to $x$ -axis.

For $0<\theta<\pi,\dfrac{dy}{dx}>0.$

The function $y(x)$ is increasing in this interval.

For $\pi<\theta<2\pi,\dfrac{dy}{dx}<0.$

The function $y(x)$ is decreasing in this interval.

8. Second derivative

$x=a(\theta+\sin \theta), y=a(1-\cos \theta).$

$$\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{d\theta}(\dfrac{dy}{dx})}{\dfrac{dx}{d\theta}}=\dfrac{\dfrac{a}{2\cos^2(\theta/2)}}{a(1+\cos \theta)}=\dfrac{1}{4\cos^4(\theta/2)}$$

For $0<\theta<2\pi, \dfrac{d^2y}{dx^2}>0,$ => y is concave upward.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} \theta & 0 & \pi/2 & \pi & 3\pi/2 & 2\pi \\ \hline x & 0 & a(\pi/2+1) & a\pi & a(3\pi/2-1) & 2a\pi\\ \hdashline y & 0 & a & 2a & a & 0\\ \hdashline dy/dx & 0 & 1 & \infin & -1 & 0 \\ \hdashline \end{array}$$