Firstly, we define a regular homotopy between regular closed curves as a continuous map $F: I \times I \to \mathbb{R}^n$ satisfying the following conditions:
(i) for each fixed $u \in I$, the map $\gamma_{u}(t)=F(u,t)$ is a regular closed curve.
(ii) All derivatives $\gamma_{u}^{(k)}(t)$ (with respect to $t$) depend continuously on $u$.
So here's the question:
Let $\gamma : [0, 2\pi] \to \mathbb{R}^2$ be given by $\gamma(t)=((1+2 \cos(t) ) \cos(t) , (1+2 \cos(t) ) \sin(t))$. By finding a suitable regular homotopy compute the rotation index of the curve. You also need to justify the regularity of your homotopy.
Now, I know that if 2 curves are regularly homotopic, then they have the same rotation index, and to check regularity I must check that the derivative never vanishes. But I am a little unsure of how to find a regular homotopy.
My first thoughts involved splitting up the function into $$\gamma(t)=(\cos(t) , \sin(t) ) +(2\cos^{2}(t), 2\cos(t) \sin(t) ).$$ And then perhaps trying $$F(u,t) =(\cos(t) , \sin(t) )+(1-u)(2\cos^{2}(t), 2\cos(t) \sin(t) ) \ ?$$ Or maybe $$F(u,t) =(1-u)(\cos(t) , \sin(t) )+(2\cos^{2}(t), 2\cos(t) \sin(t) ) \ ?$$
All help is greatly appreciated.
OK, so you've settled on $F(u,t) = (\cos t,\sin t) + 2(1-u)(\cos^2t,\cos t\sin t)$. To me it is evident that this is a smooth (infinitely differentiable) function of $t$ and $u$. What you really need to check is that $\dfrac{\partial F}{\partial t} \ne 0$ for every value of $t$ and $u$. This is the regularity condition for the curve $\gamma_u$. It might help to write $$\frac{\partial F}{\partial t} (u,t)= (-\sin t,\cos t) + 2(1-u)(-\sin 2t,\cos 2t)$$ and think a bit. Are these two vectors ever parallel?
Here's a picture of $\gamma$ for you, so you can verify that the rotation index is actually $2$.
For the homotopy you had, to turn a circle into this curve, one must introduce a cusp or kink when the second loop is born (thinking in reverse time here):