Does this function have a name?

[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?

 

[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.

 

[1MileCrash]

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.

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.

 

[pbuk]

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

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.

Does it have a name?

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

 

[guysensei1]

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.

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

 

[pbuk]

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

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).

 

[skiller]

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 $\sqrt{1 + y^2}$

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.

 

[Nugatory]

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.

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

 

[pbuk]

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.

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

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

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).