Given a differentiable real-valued function f, the arclength of its graph on [a,b] is given by
∫ba√1+(f′(x))2dx
For many choices of f this can be a tricky integral to evaluate, especially for calculus students first learning integration. I’ve found a few choices of f that make the computation pretty easy:
- Letting f be linear is super easy, but then you don’t even need the formula.
- Taking f of the form (stuff)32 might work out nicely if stuff is chosen carefully.
- Calculating it for f(x)=√1−x2 is alright if you remember that ∫1x2+1dx is arctan(x)+C.
- Letting f(x)=ln(sec(x)) results in ∫sec(x)dx, which classically sucks.
But it looks like most choices of f suggest at least a trig substitution f′(x)↦tan(θ), and will be computationally intensive, and unreasonable to ask a student to do. Are there other examples of a function f such that computing the arclength of the graph of f won’t be too arduous to ask a calculus student to do?
Answer
Ferdinands, in his short note “Finding Curves with Computable Arc Length”, also comments on the difficulty of coming up with suitable examples of curves with easily-computable arclengths. In particular, he gives a simple recipe for coming up with examples: let
f(x)=12∫(g(x)−1g(x))dx
for some suitably differentiable g(x) over the desired integration interval for the arclength. The arclength over [a,b] is then given by
12∫ba(g(x)+1g(x))dx
g(x)=x10 and g(x)=tanx are some of the example functions given in the article that are amenable to this recipe.
Attribution
Source : Link , Question Author : Mike Pierce , Answer Author : J. M. ain’t a mathematician