# Polar and parametric curves

I was solving a calculus problem on polar coordinates and I came across with some doubts, I don't know how to solve it. It says: "Given the curve $$C: (x+1)^2+y^2=1$$ parametrize the arc of a curve that intersects the points $$A=(-2;0)$$ and $$B=(-1,1)$$ and that doesn't intersect the point $$(0;0)$$. Do this using polar coordinates.". What I have done so far is writting the curve as a parametric equation using polar coordinates, and it looks like this: $$r(x;y)=2cos^2 (t); 2cos(t)sin(t) t\epsilon [\pi /2;3/2 \pi]$$. How should I go on?

• What are your doubts? I haven't checked the details but that's pretty much what I would expect for the answer. – David Apr 18 '19 at 1:49
• Do you have to use polar coordinates? There’s a somewhat simpler parameterization centered at $(-1,0)$. – amd Apr 18 '19 at 2:12
• I don't know how to go on. Because I don't know how to parametrize it in way that it intersects the points that it has to intersect. And in a way that it doesn't intersect (0;0) – AaronTBM Apr 18 '19 at 2:14

## 1 Answer

First, after expressing $$(x,y)$$ in polar coordinates, you get the equation $$(r\cos t+1)^2+(r\sin t)^2=1$$, which after a few computations turns out to be $$r(r+2\cos t)=0$$. So, $$r=0$$(which is the case when $$(x,y)=(0,0)$$, we are not interested in this) or $$(r+2\cos t)=0$$, i.e. $$r=-2\cos(t)$$. So, the parametrization that we get is $$(x,y)=(-2\cos^2(t),-2\cos t\sin t)$$ (so the parametrization that you got was missing a "-" sign.) Now, we are just left with finding the bounds for $$t$$ (i.e. find the interval). We want $$A=(-2,0)$$ and $$B=(-1,1)$$ to be in our parametrization. Note that $$A$$ occurs when $$t=0,\pi,2\pi,\cdots$$ and $$B$$ occurs when $$t=\pi/4,3\pi/4,\cdots$$. Now, you want your parametrization not containing $$(0,0)$$ (which occurs when $$t=\pi/2,3\pi/2,\cdots$$; so you just need to choose a good interval not containing these $$t$$-values.

For instance $$t\in[0,\pi/4]$$ works.