Question #291267

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


1
Expert's answer
2022-01-28T13:19:29-0500


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

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

Trace the curve x=a(θ+sinθ),y=a(1cosθ),0θ2π,a>0.x=a(\theta + \sin \theta), y = a(1-\cos \theta), 0\leq\theta\leq 2\pi, a>0.

1. Symmetry


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

g(θ)=a(1cos(θ))=g(θ)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(θ)=a(θ+sinθ)=0,(0, 0), x=f(\theta)=a(\theta + \sin \theta)=0,


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

{sinθ=θcosθ=1=>{sinθ=θθ=0  or  θ=2π=>θ=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:


dydx=dydθdxdθ=asinθa(1+cosθ)=sinθ1+cosθ=tan(θ/2)\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 θ=0,dydx=0.\theta=0, \dfrac{dy}{dx}=0. Tangent to the curve at θ=0\theta=0 is xx -axis.

At θ=π,dydx=.\theta=\pi, \dfrac{dy}{dx}=\infin. Tangent to the curve at θ=π\theta=\pi is perpendicular to xx -axis.


3. Intercepts

Intersection with xx-axis: The points of intersection of the curve with the xx-axis are given by the roots of g(θ)=a(1cosθ)=0,0θ2π,a>0.g(\theta)=a(1-\cos \theta)=0, 0\leq \theta \leq 2\pi, a>0.


cosθ=1\cos \theta=1

θ=0 or θ=2π\theta=0\ or\ \theta=2\pi

x=0 or x=2πax=0\ or\ x=2\pi a


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


Intersection with yy-axis: The points of intersection of the curve with the yy-axis are given by the roots of f(θ)=a(θ+sinθ)=0,0θ2π,a>0.f(\theta)=a(\theta+\sin \theta)=0, 0\leq \theta \leq 2\pi, a>0.

θ+sinθ=0\theta+\sin \theta=0

θ=0\theta=0

y=0y=0


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


4. Asymptotes

x=a(θ+sinθ),y=a(1cosθ).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(θ+sinθ),y=a(1cosθ).x=a(\theta+\sin \theta), y=a(1-\cos \theta).

Note that y0.y\geq 0. Entire curve lies above the yy -axis (0y2a).(0\leq y\leq 2a).


6. First derivative

x=a(θ+sinθ),y=a(1cosθ).x=a(\theta+\sin \theta), y=a(1-\cos \theta).


dydx=dydθdxdθ=asinθa(1+cosθ)=sinθ1+cosθ=tan(θ/2)\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 θ=0,dydx=0.\theta=0, \dfrac{dy}{dx}=0. Tangent to the curve at θ=0\theta=0 is xx -axis.

At θ=π,dydx=.\theta=\pi, \dfrac{dy}{dx}=\infin. Tangent to the curve at θ=π\theta=\pi is perpendicular to xx -axis.

For 0<θ<π,dydx>0.0<\theta<\pi,\dfrac{dy}{dx}>0.

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

For π<θ<2π,dydx<0.\pi<\theta<2\pi,\dfrac{dy}{dx}<0.

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


8. Second derivative

x=a(θ+sinθ),y=a(1cosθ).x=a(\theta+\sin \theta), y=a(1-\cos \theta).


d2ydx2=ddθ(dydx)dxdθ=a2cos2(θ/2)a(1+cosθ)=14cos4(θ/2)\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<θ<2π,d2ydx2>0,0<\theta<2\pi, \dfrac{d^2y}{dx^2}>0, => y is concave upward.


θ0π/2π3π/22πx0a(π/2+1)aπa(3π/21)2aπy0a2aa0dy/dx0110\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}



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