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?
Polar and parametric curves
calculusparametricparametrizationpolar coordinates
Related Solutions
Polar coordinates $(r_1,\theta_1)$ and $(r_2,\theta_2)$ represent the same point if $$\theta_2=\theta_1+2k\pi\,,\ r_2=r_1\quad\hbox{or}\quad \theta_2=\theta_1+(2k+1)\pi\,,\ r_2=-r_1\,.$$ For your curve the first case gives $$\frac{2}{\theta_1}=\frac{2}{\theta_1+2k\pi}$$ which has no solution except for the trivial $k=0$. The second case gives $$\frac{2}{\theta_1}=-\frac{2}{\theta_1+(2k+1)\pi}\ ,$$ which gives $$\theta_1=-\Bigl(k+\frac{1}{2}\Bigr)\pi\ .$$ That is, for any integer $k$ $$\Bigr(\frac{-4}{(2k+1)\pi},\,-\frac{(2k+1)\pi}{2}\Bigr)\quad\hbox{and}\quad \Bigr(\frac{4}{(2k+1)\pi},\,\frac{(2k+1)\pi}{2}\Bigr)$$ are the same point.
Funny how this site works sometimes.
I was doing a little research on Desmos for this completely unrelated question when I realized that the method of restricting domain and range I ended up using to answer that question would also work perfectly here. I had to switch to the rectangular versions of the petal equations, but that's a small price to pay.
Just as a quick example, I ended up graphing $$x^2+y^2=1.21$$ $$\begin{align} \frac{(x\cos\phi_1-y\sin\phi_1-h_1)^2}{a_1^2}+\frac{(y\cos\phi_1+x\sin\phi_1-k_1)^2}{b_1^2}&=c_1^2\{x\le-0.5\}\{y\ge0.96465\} \\ \frac{(x\cos\phi_1-y\sin\phi_1-h_1)^2}{a_1^2}+\frac{(y\cos\phi_1+x\sin\phi_1-k_1)^2}{b_1^2}&=c_1^2\{x\ge-0.49999\}\{y\ge1.09571\} \\ \frac{(x\cos\phi_2-y\sin\phi_2-h_2)^2}{a_2^2}+\frac{(y\cos\phi_2+x\sin\phi_2-k_2)^2}{b_2^2}&=c_2^2\{x\le-1.29445\} \\ \frac{(x\cos\phi_2-y\sin\phi_2-h_2)^2}{a_2^2}+\frac{(y\cos\phi_2+x\sin\phi_2-k_2)^2}{b_2^2}&=c_2^2\{x\le-0.99377\}\{y\le0.47522\} \\ \frac{(x\cos\phi_2-y\sin\phi_2-h_2)^2}{a_2^2}+\frac{(y\cos\phi_2+x\sin\phi_2-k_2)^2}{b_2^2}&=c_2^2\{x\ge-0.72153\}\{y\ge0.84065\} \end{align}$$
to obtain the center and the first two petals, and the result looked exactly like what I had hoped for:
I wouldn't even dream of typing out all $40$ equations I ended up graphing, but here's the final result, which looks considerably nicer than what I started with:
And here's the link to the final version, in case anyone's curious.
Best 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.