I want to prove that given a circle with a radius $R>0$ then its circumference length is a constant independent of $R$ .
For that purpose I have defined the function $$f(x)=\sqrt{R^2-x^2}, x\in[-R,R]$$ and tried to calculate the length of its curve. Since $f$ is differentiable, to do that, I need to solve $$\ell (f)=\int_{-R}^{R}\sqrt{1+(f'(t))^2}\mathrm{d}t=\int_{-R}^{R}\sqrt{1+R^2-t^2}\mathrm{d}t$$ I thought that the substitution $s=R\cos (s)$ would do the trick but I can't proceed with it. I thank you for any help/hint!
EDIT:
I have changed the subject since I'm not interested in the value of the constant but just to prove that it is a constant.
I have 2 questions in mind:
- Can we define the trigonometric functions with all their properties before knowing that the ratio is a constant? If so, then the substitution $s=R\cos (s)$ is justified.
Otherwise,
- Can we bound $\int_0^{R} \frac{\mathrm{d}t}{\sqrt{R^2-t^2}} $ without using the trig functions? The integrand is clearly unbounded and I wasn't able to find a different approach.
Thanks again.
I think you find f' incorrect now look at the picture that i send I hope it will useful