# Arc length of a curve

1. Sep 27, 2014

### Cassi

1. The problem statement, all variables and given/known data
A curve has the equation y2 = x3. Find the length of the arc joining (1, - 1) to (1, 1).

2. Relevant equations

3. The attempt at a solution
I took the integral of the distance and tried to evaluate from -1 to 1.
L = [intergral (-1 to 1) sqrt (1+(dy/dx x^3/23/2)2 dx]
Evaluated I got [integral (-1 to 1) sqrt (1+9x/4) dx]
1/27 [4+9x]3/2 ] exaluated from -1 to 1

I know the answer is supposed to be (26sqrt(13) - 16)/27 and mine is not coming out to that. Is there a mistake somewhere?

2. Sep 27, 2014

### Staff: Mentor

The formatting in your first fomula looks broken.
This is undefined for some x and not symmetric in x, so the error has to be before that. It would help to see more steps of your work.

3. Sep 27, 2014

### HallsofIvy

Staff Emeritus
I presume you mean that the equation of the curve is $y^2= x^3$ (if you don't want to use "Latex", at least write it as y^2= x^3 to indicate the powers). You should know that the arclength is given by the integral $\int \sqrt{1+ (dy/dx)^2} dx$. Since $y^2= x^3$, $2y (dy/dx)= 3x^2$ so that $dy/dx= (3/2)(x^2/y)= (3/2)(x^2/(x^{3/2})= (3/2)x^{1/2}$. So $1+ (dy/dx)= 1+ (9/4)x$

mfb's complaint that "this I undefined for some x and not symmetric in x" does NOT apply because x is never negative. What is important (and may be what mfb was saying) is that x does NOT go from -1 to 1- it is y that goes from -1 to 1. You can do either of two things:
1- take the x- integral from 0 to 1 and double. (You can do that because the graph goes from x= 0 to 1 and is symmetric for y> 0 and y< 0.)

2- use $\int \sqrt{1+ (dx/dy)}dy$ for the arclength instead. $2y= 3x^2 (dx/dy)$ so $dx/dy= (2/3)(y/x^2)= (2/3)(y/y^{4/3})= (2/3)y^{-1/3}$. (Since the integral from -1 to 1 contains y= 0, that is an "improper" integral, but converges.)

4. Sep 27, 2014

### Staff: Mentor

Ah right, y changes its sign. Well, you still have to pay attention to it in some way.

5. Sep 28, 2014

### haruspex

$\int \sqrt{1+ (dx/dy)^2}dy$ ?