elliptic functions

taking small sections that are diagonal lines from time $s$ to time $s+ds$

horizontally: $dx=x(s+ds)-x(s)=x^{\prime }(s)ds$, i.e. change in $x$ for change in $t$ which is derivative
vertically: same thing, so $dy=y^{\prime }(s)ds$

by pyth theorem, $dL$, infinitesimal portion of arc, is $\sqrt{(dx{)}^2+(dy{)}^2}$, or $\sqrt{[x^{\prime }(s){]}^2+[y^{\prime }(s){]}^2}ds$
trying to get full arc length; $L(t)={\int }_0^{t}dL={\int }_0^t\sqrt{[x^{\prime }(s){]}^2+[y^{\prime }(s){]}^2}ds$

two particles:

1. $(x_1(s),y_1(s))$ using $x_1(s)=A\cos (s)$, $y_1(s)=A\sin (s)$ for some $A>0$
2. $(x_2(s),y_2(s))$ using $x_2(s)=\cos (s)$, $y_2(s)=B\sin (s)$ for some $B>1$
   $L_1$ and $L_2$ are arc lengths for $x_1$, $y_1$ and $x_2$, $y_2$

for $t=1$, for every $A$, is there a value $B$ such that $L_1(1)=L_2(1)$

move to earlier time, e.g. $t=\frac{1}{2}$; is it still possible to find $B$ given $A$ so $L_1(\frac{1}{2})=L_2(\frac{1}{2})$

seems to be the same as before? since both $L_1$ and $L_2$ share bounds for the integral and the shared values are because the expression within the integral was the same in the previous answer.

same thing, but with $t\to 0^{+}$

for $t>1$…

would still work with an arbitrary $t$? since all cases of this worked with different $t$s also, including the last one that also had a different solution using the limit