Given any arbitrary function $f(x)$, its arc length in the interval $[a,b]$ is defined as:
$$
s=\int_a^b\sqrt{1+(\nabla f(x))^2}\,dx,
$$
i.e., we integrate the above expression from $a$ to $b$, using the first derivative of the function $f(x)$.
Question: if $a,s$ are known, how to get $b$? In other words, I already have the arc length, and I do know from where it was measured (the beginning of the interval) – how to get $b$? $f(x)$ and its derivative $f'(x)$ are not analytical expressions, rather, I am doing numerical computations.
Best Answer
Suppose that we know the derivative but $$I=\int \sqrt{1+[f'(x)]^2}\,dx$$ is not analytical. Suppose that $I$ is analytical but cannot be inversed. What can be done beside numerical calculations ?
But the is not bad in the first case since Newton will perfect. We want to find the zero of function $$F(b)=s -\int_a^b \sqrt{1+[f'(x)]^2}\,dx$$ $$F'(b)=-\sqrt{1+[f'(b)]^2}$$ $$b_{k+1}=b_k+\frac{s -\int_a^{b_k} \sqrt{1+[f'(x)]^2}\,dx } {\sqrt{1+[f'(b_k)]^2} }$$ the worst initial estimate being $$b_0=a+\frac s{\sqrt{1+[f'(a)]^2} }$$
Just for illustration, using $f(x)=\text{erf}(x)$, $a=-1$ and $s=123.456$ So $$F(b)=123.456-\int_{-1}^b \sqrt{1+\frac{4 }{\pi }e^{-2 x^2}}\,dx$$ Being very lazy, I shall start the above $b_0$ and the solution is obtained in $2$ iterations for $50$ exact significant figures.