Calculating Arc Length of a Curve: y^2 = x^3, (1,-1) to (1,1)

[Cassi]

Homework Statement

A curve has the equation y2 = x3. Find the length of the arc joining (1, - 1) to (1, 1).

Homework Equations

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?

 

[mfb]

The formatting in your first fomula looks broken.

[integral (-1 to 1) sqrt (1+9x/4) dx]

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.

 

[HallsofIvy]

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.)

 

[mfb]

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

 

[haruspex]

2- use \int \sqrt{1+ (dx/dy)}dy for the arclength instead.

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