# Radius function from length of curve

I have the following function definition for the length of a curve: $$l(\theta) = {K_0 \times \sin(\theta) \over \cos(\theta) + K_1} \\ 0 \le \theta \lt \frac \pi 2 \\ K_0, K_1 \ge 0$$ I would like a function which describes the radius of curve so I can graph it.

(First question / post, so moderators please feel free to modify as necessary.)

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## 1 Answer

If I understand correctly, $l(\theta)$ expresses the length of the curve from its initial point to the point with polar angle $\theta$. Since the initial point is at $\theta=0$, the standard formula for length in polar coordinates gives $$l(\theta)= \int_0^\theta \sqrt{r'(t)^2+r(t)^2}\,dt \tag1$$ Differentiating both sides and squaring, we obtain a differential equation for $r$: $$r'(\theta)^2+r(\theta)^2 = (l\,'(\theta))^2 \tag2$$ where the function $(l'(\theta))^2$ is known. This is a nonlinear ODE with no obvious (to me) simplifications. I think numerical or approximate solutions are necessary here.

For example, I took $K_0=K_1=1$, which simplifies a lot: $l\,'(\theta)=1/(1+\cos\theta)$. Then outsourced the job to Maple:

sol:=dsolve([diff(r(theta),theta)= sqrt(ls-r(theta)^2), r(0)=0],numeric, range=0..Pi/2);
odeplot(sol, [r(theta)*cos(theta),r(theta)*sin(theta)],thickness=3);


With these constants $r(\theta)$ is pretty close to $\theta/2$, so the curve looks like an Archimedean spiral.

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