Arc Length

Arc Length

The length of a curve $y = f(x)$ from $a$ to $b$:

$L = int_a^b sqrt{1 + [f'(x)]^2} , dx$

Why It Works

At each $x$, the curve moves right by $dx$ and up by $f'(x)dx$. By the Pythagorean theorem, the infinitesimal piece has length $sqrt{dx^2 + (f'dx)^2} = sqrt{1+f'^2} , dx$.

Examples

Straight line $y = mx$ on $[a, b]$: $f' = m$, so $L = (b-a)sqrt{1+m^2}$

• $y = x$ on $[0,3]$: $L = 3sqrt{2} approx 4.2426$
• $y = 3x$ on $[0,4]$: $L = 4sqrt{10} approx 12.6491$

Numerical Approach

Use the midpoint rule with a central difference derivative:

double dx = (b - a) / n;
for each midpoint x:
    double deriv = (f(x+h) - f(x-h)) / (2*h);
    sum += my_sqrt(1.0 + deriv * deriv);
return sum * dx;

Your Task

Implement double arc_length(double (*f)(double), double a, double b, int n, double h).

Use the midpoint rule with step $dx = (b-a)/n$ and central difference with step $h$ for the derivative. A Newton's-method sqrt helper is provided.