Math 61

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Find the arc length of the curve y^3=8x^2 fr0m x=1 to x=8.

  • Math 61 -

    since y^3 = 8x^2,
    y = 2x^(2/3)
    y' = 4/3 x^(-1/3)
    y'^2 = 16/(9x^(2/3))

    arc length is thus
    ∫[1,8] √(1+16/(9x^(2/3))) dx

    u=x^-2/3
    then let v = u^1/2
    then let 4w = 3tanv

    and you finally wind up with

    1/27 √(9+16/x^2/3) (9x + 16x^1/3) [1,8]
    = 1/27 ((√(9+4)(72+32))-(√(16+9)(9+16)))
    = 1/27 (104√13 - 125)

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