# Does this function have a name?

1. Jul 21, 2014

### guysensei1

A function f(x) where f(x)=length of the graph curve/line from 0 to x

Can this function be expressed in algebraic form or some other form?

Does it have a name?

2. Jul 21, 2014

### HallsofIvy

You are talking about a particular way of getting a function, not a specific function so, no, it does not have a name. As one learns in Calculus, the length of the graph of y= f(x), from 0 to x is given by $\int_0^x \sqrt{1+ (f'(t))^2} dt$.

3. Jul 21, 2014

### 1MileCrash

It's without a doubt a specific function, one whose domain is the Cartesian product of x and the function space. I don't think that this is a good reason for it to not have a name.

4. Jul 21, 2014

### MrAnchovy

Not in general, but the solutions for a straight line and for a circle should be fairly obvious. Closed form solutions do exist for a few more complex curves e.g. parabola, catenery and cycloid but not for most others, even the humble ellipse.

The calculation of the length of a curve between two points is called rectification, or simply calculating arc length.

5. Jul 21, 2014

### guysensei1

What I was looking for is a function that gives itself when the length of curve function is applied.

6. Jul 22, 2014

### MrAnchovy

Try looking for this function. Clearly f(0) = 0. Let y = f(1). The length of the arc between (0, 0) and (1, y) is given by $\sqrt{1 + y^2}$ so we have $y = \sqrt{1 + y^2}$ or $y^2 = 1 + y^2$ which has no solution - the function you are looking for does not exist (over any non-zero domain).

7. Jul 22, 2014

### skiller

Why is that? That surely just gives the length of the straight line from (0, 0) to (1, y). The curve you are looking for is not going to be of that form.

8. Jul 22, 2014

### Staff: Mentor

No, but its length has to be strictly greater than the length of the straight line between the two points.

9. Jul 22, 2014

### MrAnchovy

Oh, I thought one thing and wrote something slightly different, let's try again.

Try looking for this function. Clearly f(0) = 0. Let y = f(1). The shortest arc between (0, 0) and (1, y) is simply the diagonal of length $\sqrt{1 + y^2}$, and so y must be at least as large as that. So we have $y \ge \sqrt{1 + y^2}$ or $y^2 \ge 1 + y^2$ which has no real solution. The function you are looking for does not exist (over any non-zero domain).