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I am trying to divide some simple curves into equal segments. I managed to use arc length formula but then I got stuck. What to do next to divide any of these curves into let's say 10 equal segments?

I provided a link containing 4 examples that I would like to figure out: https://www.desmos.com/calculator/ks9o0p59tg

Having this solved in Desmos would be of great help. Desmos is a free online platform if you want to give a try and solve any of the provided examples. Thank you!

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Construct a parameterization of the curve $f(x)$ with respect to arc length. Since you did not do this, I assume you are unfamiliar with parameterization. This is a mathematical procedure in which we express the coordinates of each point on the curve as some function of a common parameter, say, $t$. As an example, the unit circle has a parameterization $x=cos(t)$, $y=sin(t)$ for $0\leq t\leq 2\pi$. Notice that in general parameterizations are not unique, i.e. we could let $2\pi\leq t\leq 4\pi$. The goal, then is to construct a parameterization where a 'step' in $t$ corresponds to a 'step' in arc length. You can use the arc length formula to find the arc length of the curve in terms of your parameterization variable $t$, and the formula I assume you're using is a special case of this more general form: $s(t)=\int_0^t\sqrt{x'(u)^2+y'(u)^2}du$ where $u$ is some dummy variable substitution $u=t$. This will spit out some function of $t$, and solving $s=f(t)$ for $t$ in terms of $s$ will give a function you plug back in to your parameterization to make the parameter (now $s$) exactly equal to arc length over some new interval. Divide this interval up into however many pieces, and then plug the value of the n-th boundary into your paraneterizations to get values of $x$ and $y$. Once I get off mobile I'll see if I can cook up some Desmos thing.

P S. Look into Mathematica; it's like 10 billion times better than Desmos.

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    $\begingroup$ Thank you for your detailed response. As you well guessed i indeed encounter a lot of unfamiliar aspects regarding calculus. This is something that i just begun to try and understand. Having your answer is of great help for me because now i have clues on what i should research and practice in order to achieve this division thing on a variety of curves. Thank you so much! $\endgroup$ Mar 8, 2020 at 10:41

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