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I have a parametric function for an ellipse:

$$f\left(t\right)=\left(a\cos\left(t\right),b\sin\left(t\right)\right)$$

As the function goes linearly through t from 0 to 2pi, the point speeds up near the minor axis and slows down near the major axis. I don't want this, I want it to have a constant speed.

I have tried taking the derivative, but then I don't know what to do with the derivative to make the parametric function output a constant speed. I have tried plugging the linearly increasing value into functions that I put inside the parametric function, trying to find a function that increases at a rate inverse to the speed. I have been using Desmos for my work.

In other words, how can I derive a function that creates the same shape over time but has a constant speed or derivative of magnitude of velocity? As well as this, can the same method be applied to any parametric function?

Kitty Craft0
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  • I think you'd have to use elliptic integrals, which are non-elementary functions. – mr_e_man Nov 21 '23 at 19:49
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    The arclength $s$ is given by $$\left(\frac{ds}{dt}\right)^2=\lVert f'(t)\rVert^2=a^2\sin^2t+b^2\cos^2t$$ $$s=\int\sqrt{a^2\sin^2t+b^2\cos^2t},dt,$$ and then you'd have to invert this function, solving for $t$ in terms of $s$. – mr_e_man Nov 21 '23 at 19:56
  • Unit speed parametrization is equivalent to parametrization by arclength. Please refer to my post here. – Ng Chung Tak Nov 23 '23 at 00:09

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I'm trying to solve this particular problem for something else I'm working on. After some thought I came up with a not so satisfying conclusion (for which I have no proof): there exists a parametric function $u(t)$ such that the path traced by it is the extact same as $f(t)$ and the speed at which it traces the curve is constant, however, almost every curve you can think of results in a not so pretty $u(t)$.

My reasoning: Basically we want the derivative of $u(t)$ to have constant magnitude along the curve. This means

$$||u'(t)|| = k$$

for some $k$ constant. For simplicity, we can try for $k=1$. Then it follows that $u'(t)$ has to be on the unit circle, which means

$$u'(t) = (cos(\theta(t)),sin(\theta(t)))$$

for some function $\theta(t)$. After integrating, we arrive to the parametric function

$$u(t) = \left(\int_0^tcos(\theta(w))dw,\int_0^tsin(\theta(w))dw\right)$$

There are 2 problems with this: we don't who $\theta(t)$ is, and even if we did, it might not be possible to calculate the integral. For instance, take $\theta(t) = sin(t)$. You can picture the curve traced by $u(t)$ considering the fact that $\theta(t)$ oscilates between -1 and 1 (an image of said curve). The parametric function is written as

$$u(t) = \left(\int_0^tcos(\sin(w))dw,\int_0^tsin(\sin(w))dw\right)$$

With the help of numerical methods, we can graph the solution for this curve, however there exists no elemental solution for neither $\int_0^tcos(\sin(w))dw$ or $\int_0^tsin(\sin(w))dw$. There might be some obscure book talking about normalizing the velocity vector for parametric functions, but I haven't found any.

MaxProBer
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