Is it possible to vary the "speed" at which a function is drawn in a parametric equation? It is to my understanding that a parametric function produces the same amount of points per time. So, is it possible to accelerate/decelerate the rate at which the functions is "drawn"?
[Math] Varying “speed” in parametric equation
calculusparametric
Related Solutions
You're really looking for a cubic equation in one dimension (time).
$$ y = u_0(1-x)^3 + 3u_1(1-x)^2x + 3u_2(1-x)x^2 + u_3x^3 $$
Is all you need.
Walking $t$ at even intervals (say in steps of 0.1) takes evenly spaced points along the parametric curve.
So, the answer to your question is really quite simple. The parametric bezier curve provides 2 variables as the output, with only 1 variable as the input. To control an animation in time like these, that's only a 1 dimensional situation. So consider $t$ as time, and drop one variable (say drop $x$). Your animation ease curve is controlled by the $y$ value:
Now as $t=0,0.1..1$, you have an animation parameter that starts slowly, moves at medium speed in the middle, and slows down at the end.
Examples
Setting $u_0=0$, $u_1=0.05$, $u_2=0.25$, $u_3=1$ gives an ease-in curve (slow start, fast end)
Setting $u_0=0$, $u_1=0.75$, $u_2=0.95$, $u_3=1$ gives an ease-out curve (fast start, slow end)
First write $x-2t=-2\sin t,\,y-2=-2\cos t.$ Then square and add to get $$(x-2t)^2+(y-2)^2=4.$$ Finally, solve for $t$ in the second equation to get $\cos t=\frac{2-y}{2}.$ Then for values of $t$ such that $0\le t\le π,$ we have $$t=\arccos\left(\frac{2-y}{2}\right),$$ so that the equation becomes $$\left(x-2\arccos\left(\frac{2-y}{2}\right)\right)^2+(y-2)^2=4.$$
Apart from the fact that this probably doesn't trace out the whole curve, since there is a restriction on $y$ so that $4\ge y\ge 0,$ it is not very pretty to look at.
We are lucky in this case to be able to get such a relationship at all. It is not always possible to do this in general.
Best Answer
Yes, functions may be parameterized many different ways, for example: let$ \lambda \in \mathbb{R}, \overline{f}(\lambda) = \lambda\begin{pmatrix} 1 \\ 2 \\ \end{pmatrix} + \begin{pmatrix} 3 \\ 4 \end{pmatrix} $ and $\overline{g}(\lambda) = \lambda \begin{pmatrix}2 \\ 4 \end{pmatrix} + \begin{pmatrix} 3 \\ 4\end{pmatrix}$, where $-\infty < \lambda < \infty$ are both paramatarizations of the line $h(x) = 2x - 2$. Notice, of course that the point $(4,6)$ is of coursed reached by both $\overline{f}$ and $\overline{g}$, yet with $\overline{f}$ it is reached when $\lambda = 1$, and with $\overline{g}$ it is reached when $\lambda = \frac{1}{2}$