Hard Math question for HW!?

x² = 8y³

differentiating wrt x

2x = 24 y² y'

square both sides

4x² = 576 y⁴ (y')²

(y')² = x² / 144y⁴

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x² = 8y³

y = ½ ∛x²

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∴

y⁴ = (½ ∛x²)⁴

y⁴ = ( ∛x⁸)/16

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(y')² = x² / 144y⁴

(y')² = 16x² / [ 144(∛x⁸) ]

(y')² = 1 / (9∛x²)

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s = ∫√[ 1 + (y')² ] dx

s = ∫√[ 1 + 1 / (9∛x²) ] dx

s = ∫√[ (9∛x² + 1 ) / (9∛x²) ] dx

`````````just some nasty algebra here. square root of the denominator (9∛x²) is equal to 3∛x

s = ∫(1/ 3∛x) [ √(9∛x² + 1 ) ] dx

use substitution u = ∛x²

s = ∫(1/ 3∛x) [ √(9u + 1 ) ] dx

u = ∛x²

du/dx = 2 /(3∛x)

dx = ½ (3∛x) du

s = ∫(1/ 3∛x) [ √(9u + 1 ) ] dx

s = ∫(1/ 3∛x) [ √(9u + 1 ) ] [½ (3∛x) du]

s = ∫½√(9u + 1 ) du

s = (1/27) √(9u + 1 )³

A(0,0) and B(64,8)

x = 0, u = ∛x² ; u = ∛0² = 0

x = 64, u = ∛64² = 16

s(0) = (1/27) √(9u + 1 )³ = (1/27) √(9*0 + 1 )³ = 0.037037037037037

s(16) = (1/27) √(9u + 1 )³ = (1/27) √(9*16 + 1 )³ = 64.66782273795863

arc length

s = s(16) - s(0) = 64.63

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int over y in [ 0 , 8 ] of { √ ( 1 + [dx / dy ]² ) dy } where dx / dy = 12 y² / x