How to know the orientation (whether clockwise or anticlockwise) of the curve C: $(3/2)(x+y)^2+(x-y)^2/2=1$ whose parametrization is given by $$x=\frac{1}{\sqrt 6}\cos \theta -\frac{1}{\sqrt 2}\sin\theta, ~~y=\frac{1}{\sqrt 6}\cos \theta +\frac{1}{\sqrt 2}\sin\theta$$
NB: Taking $x+y=\sqrt{2/3} \cos \theta$ and $x-y=-\sqrt 2 \sin \theta$
Answer:
On $xy$-plane, at $\theta=0$, $(x,y)=\frac{1}{\sqrt 6}(1,1)$ which is up and $(dx/d\theta,dy/d\theta)=\frac{1}{\sqrt 2}(-1,1)$ which is to the left. So the motion is anticlockwise.
My doubt:
- I am unable to understand the argument. What is the logic behind it? Please help me to understand.
- Why $x-y$ is not considered as $x-y=+\sqrt 2 \sin \theta$?
Edit: I know that "A curve has positive orientation if a region R is on the left when travelling around the boundary of R"