For animation purposes (see below) I need to use a parametrization which 'slows down' near a specific value;

More precisely, I am looking for a 'nice' monotonically increasing function $r:[0,1] \to [0,1]$ satisfying $ r(0)=0, r(1)=1, $ whose derivative is relatively small around the point $r^{-1}(\frac{1}{2})$.

After some playing around with polynomials, I came up with the following example: $$ r(t)=\frac{1}{2} \big((2 t - 1)^3 + 1\big). $$ This choice of $r$ satisfies $r'(t)=0 \iff t=\frac{1}{2}=r^{-1}(\frac{1}{2})$, so $r'(r^{-1}(\frac{1}{2}))=0.$

However, this is "too slow" for me - it turns out that having derivative zero makes the animation looks like it "stopped" (depending of course on the uniform discrete time-step size that I am using).

Question: How can I make an adaptation which makes the animation still slow but not "too slow" around $r^{-1}(\frac{1}{2})$?

Here is my animation: It is a shrinking disk - which starts at radius $1$, and shrinks to a point. After it reaches to $r=\frac{1}{2}$ it starts 'twisting' as well as shrinking (the twisting and shrinking occurs simultaneously beyond that stage). I want to focus attention on the part around where the twisting starts.

(If it's relevant I am working with Mathematica, but I guess there should be an analytic platform-independent solution to this problem.)

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    $\begingroup$ I don't think it is necessary for $r(t)$ to be smooth. Just make it piecewise linear. $\endgroup$ – Wolfgang Bangerth Feb 2 at 20:40
  • $\begingroup$ Thanks, you are completely right. I thought that it might be possible to just add a small perturbative term to my original polynomial parametrization. I guess I can keep it smooth except around $\frac{1}{2}$, where I could make it piecewise-linear and glue. $\endgroup$ – Asaf Shachar Feb 3 at 11:31

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