# Quick Arc Length Equation

1. Feb 15, 2008

### chaotixmonjuish

I'm working on this problem

x^5/6+1/(10x^3) [1,2]

and I got the equation:

sqrt(1+(5x^4/6-3/10^4)^2) or
sqrt(1+25x^8/36+9/100x^8-1/2)

I'm not sure how to integrate the last part, is there some sort of obvious substitution I'm missing?

2. Feb 15, 2008

### CompuChip

Is the equation you get
$$\sqrt{1+ \frac{25 x^8}{36} + \frac{9}{100}x^8-\frac{1}{2}}$$
If so, it's easy to integrate (just collect polynomials and integrate term by term.
If not, please be a little more clear (use TeX if possible, or else at least put brackets when writing things like a/bc, which can be read as (a/b)*c or a/(b*c)).

3. Feb 15, 2008

### HallsofIvy

CompuChip, that's clearly not what is intended because he did use parentheses to indicat that the x3 in the original function was in the denominator.

Let's check: y= x5/6+ 1/(10x3)= x5/6+ (1/10)x-3 has derivative y= (5/6)x4- (3/10)x-4. Squaring that, (y')2= (25/36)x8- 2(5/6)(3/10)+ (9/100)x-8= (25/36)x8- (1/2)= (9/100)x-8. Adding 1 to that just changes the "-1/2" to "+1/2" and since the first was a square, this is also a square with the sign changed: (a- b)2= a2- 2ab+ b2 while (a+b)= a2+ 2ab+ b2: 1+ (y')2= (25/36)x8+ (1/2)+ (9/100)x-8= [(5/6)x4+ (3/10)x-4]2 and so $\sqrt{1+ (y')^2}= \sqrt{[(5/6)x^4+ (3/10)x^{-4}]^2}= (5/6)x^4+ (3/10)x^{-4}$.
That should be easy to integrate.

(Tip: in general length integrals are very difficult. That's why calculus professors (who are not actually monsters) like tricks like the above that ensure the integral will be easy. It's a good idea to always check to see if you have a "perfect square"!)

4. Apr 8, 2008

### dexteronline

Last edited by a moderator: Apr 23, 2017
5. Apr 8, 2008

### HallsofIvy

But that doesn't answer his question, does it?

6. Apr 8, 2008

### dexteronline

Yes it doesn't answer for integral to be found. Yet I thought he may want to know the length of the curve