0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ What are your doubts? I haven't checked the details but that's pretty much what I would expect for the answer. $\endgroup$ – David Apr 18 '19 at 1:49
  • $\begingroup$ Do you have to use polar coordinates? There’s a somewhat simpler parameterization centered at $(-1,0)$. $\endgroup$ – amd Apr 18 '19 at 2:12
  • $\begingroup$ 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) $\endgroup$ – AaronTBM Apr 18 '19 at 2:14
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.