# Arc Length (Astroid)

1. Feb 3, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
the graph of the equation x^(2/3) + y^(2/3) = 7^(2/3) is one of the family of curves called astroids.

2. Relevant equations
find the length of first-quadrant and multiply by 8.
1. y=(7^(2/3) - x^(2/3))^(3/2)) ; 7sqrt(2)/4 <= x <= 7

3. The attempt at a solution
1. i found dy/dx, (dy/dx)^2, and now i'm at the length integral.
2. L = integral sqrt(1+7^(2/3) * x^(2/3) - x^(4/3))

i'm not sure how to evaluate this integral (#2) without using a calculator, but that doesn't give me the exact answer that i need.

2. Feb 3, 2013

### LCKurtz

I get a simpler expression for $1 + y'^2$. Please show your work.

3. Feb 3, 2013

### whatlifeforme

dy/dx = -x^1/3 (7^2/3 - x^2/3)^(1/2)
(dy/dx)^2 = x^(2/3) * (7^(2/3) - X^(2/3)V ----> 7^(2/3)* X^(2/3) - X^(4/3)

L = ∫ (x=a to x=b) SQRT(1 + 7^(2/3) * x^(2/3) - x^(4/3))

Last edited: Feb 3, 2013
4. Feb 3, 2013

### haruspex

Instead of converting the equation to the form y = f(x), just differentiate the equation as given. It's less messy, so less prone to error.

5. Feb 4, 2013

### whatlifeforme

it's already solved for y thats how the want you to solve it.

6. Feb 4, 2013

### LCKurtz

That is not the easy way to solve it and is probably why you are having trouble. Differentiate the equation $x^{2/3} + y^{2/3} = 7^{2/3}$ implicitly with respect to $x$ to get $y'$ and calculate $1 + y'^2$. Show us what you get.

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