I'm trying to determine the relationship between the major and minor radii ($a$ and $b$, respectively) of an ellipse of constant perimeter and variable eccentricity, and I've been thinking that numerically solving a differential equation $\frac{db}{da}=f(a,b)$ is the way to go. I'd like to know if my derivation for $f(a,b)$ is correct.
I begin by considering the equation for the perimeter of an ellipse: $$C=4aE(e)$$ where E(e) is the complete elliptic integral of the second kind evaluated at eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}$. Differentiating $C$ with respect to $a$ yields: $$\frac{dC}{da}=0=4\left(E(e)+a\frac{dE(e)}{de}\frac{de}{da}\right)=E(e)+a\left(\frac{E(e)-K(e)}e\right)\frac d{da}\left(\sqrt{1-\frac{b^2}{a^2}}\right)$$ Through implicit differentiation, one may evaluate $\frac d{da}\left(\sqrt{1-\frac{b^2}{a^2}}\right)$, if $b=f(a)$: $$\frac d{da}\left(\sqrt{1-\frac{b^2}{a^2}}\right)=\frac12\left(1-\frac{b^2}{a^2}\right)^{-\frac12}\frac d{da}\left(1-\frac{b^2}{a^2}\right)=\frac12\left(1-\frac{b^2}{a^2}\right)^{-\frac12}\left(-a^{-2}\frac d{da}(b^2)+b^2\frac d{da}(-a^2)\right)=\frac{b^2-ab\frac{db}{da}}{a^3\sqrt{1-\frac{b^2}{a^2}}}$$ Making this substitution for $\frac d{da}\sqrt{1-\frac{b^2}{a^2}}$ in $\frac{dC}{da}$ yields: $$\frac{dC}{da}=E(e)+a\left(\frac{E(e)-K(e)}{\sqrt{1-\frac{b^2}{a^2}}}\right)\frac{b^2-ab\frac{db}{da}}{a^3\sqrt{1-\frac{b^2}{a^2}}}=E(e)+\left(\frac{E(e)-K(e)}{a^2-b^2}\right)\left(b^2-ab\frac{db}{da}\right)=0$$ Isolating $\frac{db}{da}$ yields: $$\frac{db}{da}=\frac ba+\left(\frac{E(e)}{E(e)-K(e)}\right)\left(\frac ab - \frac ba\right)$$
Is my derivation correct? Thanks!
There is something wrong. I agree on the implicit differentiation with $b=f(a)$.However when I repeat your computation I get \begin{equation} E(e)+(\frac{E(e)-K(e)}{2(a^2-b^2)})(b^2-a b \frac{db}{da})=0 \end{equation} The $2$ in the denominator comes when differentiating $E(e)$.