# Length of a curve

1. Oct 25, 2006

### blumfeld0

hi. i am trying to find the length of the curve
Find the length of the curve
x = (3) *y^(4/3) - (3/32)*y^(2/3) from -64<_y<_27

so i do the usual

calculate( 1 + (dx/dy)^2 ) ^(1/2)

then do the integral from -64 to 27 with respect to y

the only thing is i get an imaginary answer!!!

someone could please shed some light on this

thank you

2. Oct 26, 2006

### HallsofIvy

Staff Emeritus
If you expect some one to tell you what you did wrong you will need to show exactly what you did! All I can say now is that you must have done some step of the calculation incorrectly. For this problem, [itex]1+ (\frac{dx}{dy})^2} is a perfect square: there is no square root in the integral so you couldn't possibly have got an imaginary answer!

3. Oct 26, 2006

### blumfeld0

hi. well i did everything on mathematica

i took the derivative, squared it, added one and took the square root of the whole thing. then i integrated with respect to y from -27 to 64
i.e (1+ dx/dy^2)^1/2 from -27 to 64 with respect to y

can you please tell me what the perfect square is?

thanks

4. Oct 26, 2006

### BlackEagle

Didn't you say you needed to integrate from -64 to 27?

5. Oct 26, 2006

### blumfeld0

yeah sorry thats what i meant

-64 to 27

6. Oct 27, 2006

### HallsofIvy

Staff Emeritus
Well, if it isn't important enough for you to write out exactly what you did, it certainly isn't important enough for me to do that- it's not my problem!

7. Oct 27, 2006

### ObsessiveMathsFreak

Two things.

Firstly this curve is a multivalued function, there are multiple answers unless you restrict yourself to a certain set. On my software at least, inputting (-1)^(1/3) evaluates to 1/2 + i*sqrt(3)/2, so it seem to be choosing one of the imaginary values here. If that's the case... something terrible will happen. In your case, the integral is imaginary.

Secondly, even if you fix the first problem, this curve has a discontinuity in its derivative at y=0. His will also result in horribleness if I remeber correctly.

Where did you get this curve? Perhaps the original parameterisation will be more forgiving.

8. Oct 27, 2006

### Office_Shredder

Staff Emeritus
Doing some algebraic manipulations by hand can usually radically alter the output of the computer. Even if it's just something simple like factoring f1/3+f-1/3 =f1/3(1 + 1/f2/3).

Although this is from using maple, not mathematica. It may not help you

Last edited: Oct 27, 2006
9. Oct 28, 2006

### HallsofIvy

Staff Emeritus
If x = (3) y(4/3) - (3/32)y(2/3)
then x'= 4y1/3- (1/16)y-1/3
so x'2= 16y2/3- (1/2)+ (1/256)y-2/3
Then 1+ x'2= 16y2/3+(1/2)+ (1/256)y-2/3= (4y1/3+ (1/16)y-1/3)2
and its square root is
4y1/3+ 1/16 y-1/3

As others pointed out, that's an improper integral because it is not defined at y= 0 so you have to be careful about it.

That's the "principal third root of -1" but its odd that your software would give it as "the" root. In the real numbers, of course, the third root of -1 is -1.