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Math Help - integrals and length of a curve

  1. #1
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    integrals and length of a curve

    HI

    So let curve C have equation y = 4x^(3/2)
    --------
    3
    Let a > 2. The portion of C btwn the points (2, (4(sqrt 2)^2)/3) and (a,(4a^(3/2))/3) has length 49/3

    Find the value of a

    So if y = (4x^(3/2))/3 then dy/dx = 2x^(1/2)

    Correct me anywhere you see a mistake!!

    Then L = integral sign (S) a= 2 and b= a sq rt( 1 + (dy/dx)^2) dx
    = S a=2 b=a sq rt (1 + 4x) dx

    SO then I let u = 1 + 4x so du = 4 dx

    so when x = 2 then u = 9 and when x = a then u = 1 + 4a

    L = 4 S a= 9 and b=1+4a sq rt u du which is 4(2/3)u^(3/2) a= 9 b = 1+4a

    but when I do the substitution I get 265/8 = sq rt ((1+4a)^3) and since this is not a easy equation to solve I don't know what a is

    Any idea where I went wrong???

    Thanks

    Struggling calculus beginner
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  2. #2
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    y = \frac{4}{3}x^{\frac{3}{2}}

    y' = 2x^{\frac{1}{2}}

    (y')^2 = 4x


    \int_2^a \sqrt{1+4x} \, dx = \frac{49}{3}

    \frac{1}{6}\left[(1+4x)^{\frac{3}{2}}\right]_2^a = \frac{49}{3}<br />

    \left[(1+4x)^{\frac{3}{2}}\right]_2^a = 98<br />

    (1+4a)^{\frac{3}{2}} - 27 = 98

    (1+4a)^{\frac{3}{2}} = 125

    1+4a = 125^{\frac{2}{3}}

    1+4a = 25

    a = 6
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