How to i find the arc length of X^8 + Y^8 = 1 ?

[Ibliam]

I have to find the arc length of the square like circle X^8 + Y^8 = 1

I know that the arc length formula is L = integral ( square root (1+[f'(x)]^2) dx from a to b

The length of the arc in each quadrant is the same so I should be able to find one and multiply it by 4.

X^8 + Y^8 = 1 => y = (1-X^8)^(1/8)
d/dx ((1-x^8)^(1/8)) = (-x^7)/(1-x^8)^(7/8)

so the arc length should be
4 * Limit as t=>1 of integral (square root(1 + ((-x^7)/(1-x^8)^(7/8))^2) dx from x=0 to t

but whatever I try to type into my calculator takes 30 minutes and tells me it's a non-real answer. I'm stuck.

 

[Deleted member 4993]

I have to find the arc length of the square like circle X^8 + Y^8 = 1

I know that the arc length formula is L = integral ( square root (1+[f'(x)]^2) dx from a to b

The length of the arc in each quadrant is the same so I should be able to find one and multiply it by 4.

X^8 + Y^8 = 1 => y = (1-X^8)^(1/8)
d/dx ((1-x^8)^(1/8)) = (-x^7)/(1-x^8)^(7/8)

so the arc length should be
4 * Limit as t=>1 of integral (square root(1 + ((-x^7)/(1-x^8)^(7/8))^2) dx from x=0 to t

but whatever I try to type into my calculator takes 30 minutes and tells me it's a non-real answer. I'm stuck.

See if your calculator can answer the question for a regular circle - where the equation is x² + y² = 1

 

[wjm11]

X^8 + Y^8 = 1 => y = (1-X^8)^(1/8)
d/dx ((1-x^8)^(1/8)) = (-x^7)/(1-x^8)^(7/8)

Your approach is correct.

 

[renegade05]

Your approach is correct. However, you've made a couple of errors in your derivative. You lost a 1/8, and 1/8 - 1 = -7/8.

I don't think he did. He wrote it in simplified form.

And I don't think any calc will be able to handle this, my TI-89 froze up as well. I tried to convert to polar, but that did no good as well.

You might want to try wolfram alpha.

 

[Ibliam]

You might want to try wolfram alpha.

I tried wolfram alpha but I don't think I entered it right. I'd never used it before.

I just tried entering "Limit ( integral (square root(1 + ((-x^7)/(1-x^8)^(7/8))^2) dx from x=0 to t ) x => 1" and it doesn't finish computing. Is that even remotely right to enter?