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).
Since y is a periodic function of θ with period 2π, it is sufficient to trace the curve for θ∈[0,2π]. For θ∈[0,2π],x and y are well defined.
Trace the curve x=a(θ+sinθ),y=a(1−cosθ),0≤θ≤2π,a>0.
1. Symmetry
f(−θ)=a(−θ+sin(−θ))=−f(θ),
g(−θ)=a(1−cos(−θ))=g(θ)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,
y=g(θ)=a(1−cosθ)=0
{sinθ=−θcosθ=1=>{sinθ=−θθ=0 or θ=2π=>θ=0A curve passes through the origin.
Derivatives:
dxdy=dθdxdθdy=a(1+cosθ)asinθ=1+cosθsinθ=tan(θ/2) At θ=0,dxdy=0. Tangent to the curve at θ=0 is x -axis.
At θ=π,dxdy=∞. Tangent to the curve at θ=π 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(θ)=a(1−cosθ)=0,0≤θ≤2π,a>0.
cosθ=1
θ=0 or θ=2π
x=0 or x=2πa
Then (x,y)=(0,0),(x,y)=(2πa,0).
Intersection with y-axis: The points of intersection of the curve with the y-axis are given by the roots of f(θ)=a(θ+sinθ)=0,0≤θ≤2π,a>0.
θ+sinθ=0
θ=0
y=0
Then (x,y)=(0,0).
4. Asymptotes
x=a(θ+sinθ),y=a(1−cosθ).
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(1−cosθ).
Note that y≥0. Entire curve lies above the y -axis (0≤y≤2a).
6. First derivative
x=a(θ+sinθ),y=a(1−cosθ).
dxdy=dθdxdθdy=a(1+cosθ)asinθ=1+cosθsinθ=tan(θ/2) At θ=0,dxdy=0. Tangent to the curve at θ=0 is x -axis.
At θ=π,dxdy=∞. Tangent to the curve at θ=π is perpendicular to x -axis.
For 0<θ<π,dxdy>0.
The function y(x) is increasing in this interval.
For π<θ<2π,dxdy<0.
The function y(x) is decreasing in this interval.
8. Second derivative
x=a(θ+sinθ),y=a(1−cosθ).
dx2d2y=dθdxdθd(dxdy)=a(1+cosθ)2cos2(θ/2)a=4cos4(θ/2)1
For 0<θ<2π,dx2d2y>0, => y is concave upward.
θxydy/dx0000π/2a(π/2+1)a1πaπ2a∞3π/2a(3π/2−1)a−12π2aπ00
Comments