This is similar to this question, except I am finding the radius now using the Bisection method and then Newton's method for finding a zero.
This is a computer science for a Numerical Methods course. In the question the arc length is 4, and the chord length is 3.
I derived a function such that $f(r) = 0$ when $r$ is the radius like this:
$$L = r\theta$$
$$4 = r\theta$$
Using the Law of Cosines:
$$3^2 = 2r^2(1 – \cos\theta)$$
$$ \Rightarrow \theta = \cos^{-1}(1 – 9/2r^2)$$
$$ \Rightarrow 4 = r\cos^{-1}(1 – 9/2r^2)$$
So $f(r) = 4 – r\,\cos^{-1}(1 – 9/2r^2)$ will be $0$ when $r$ is the radius.
This all works well so far, but for the two methods I need an initial approximation. For bisection I need a left and a right $x$, and for newtons method I should use the average of my two $x$ values from the bisection. It's easy to show that a lower bound for the radius is $1.5$ since the chord length is 3. If the radius was less than $1.5$ the chord length would be greater than the diameter, which is a contradiction. I am having a lot of difficulty finding an upper bound though. If I just pick a value like $1.7$ or $\pi/2$ then I get the correct answer, but although I'm not sure that it's required for my assignment, I'd like to know how I can find an upper bound that I can show is actually provably greater than the radius.
The actually radius assuming my program and equation is correct is $1.567769039908$
Best Answer
The arc length can be larger than the diameter. Even if you don't allow angles greater than $\pi$ radians the arc length can be $\pi r$, so you should take the minimum $r$ as arclength$/\pi$ (or maybe even divide by $2\pi$-safer and it only costs one bisection). Having found the minimum, you can just keep doubling it until $f(r)$ changes sign.
Added: If $c$ is the chord, $\frac{c}{2r}=\sin \frac{\theta }{2}$. As we want $r$ large, we want $\sin \frac{\theta }{2}$ to be small, so use $\frac{\theta }{2}-\frac{\theta ^3}{48}$ to get an upper bound (alternating series theorem). This gives $c=L-\frac{L\theta ^2}{24}$ and with $L=r\theta$ you get $r$.