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Math Help - arc length

  1. #1
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    arc length

    The problem

    x= t^3 y=3/2 t^2 0 <= t <= sqrt(3)

    using the formula

    integral sqrt(f'(t)^2 + g'(t)^2) dt

    so

    int 0 to sqrt(3) sqrt((3t^2)^2 + (3t)^2) dt

    so sqrt(9t^4 + 9t^2) dt

    where would I go from here??
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  2. #2
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    Quote Originally Posted by ur5pointos2slo View Post
    The problem

    x= t^3 y=3/2 t^2 0 <= t <= sqrt(3)

    using the formula

    integral sqrt(f'(t)^2 + g'(t)^2) dt

    so

    int 0 to sqrt(3) sqrt((3t^2)^2 + (3t)^2) dt

    so sqrt(9t^4 + 9t^2) dt

    where would I go from here??
    So your integral is

     <br />
\int_0^{\sqrt{3}} 3t\sqrt{t^2+1}\,dt<br />

    try a substitution.
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    So your integral is

     <br />
\int_0^{\sqrt{3}} 3t\sqrt{t^2+1}\,dt<br />

    try a substitution.
    I am sorry but I do not understand where you came up with that integral from..
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  4. #4
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    Quote Originally Posted by danny arrigo View Post
    So your integral is

     <br />
\int_0^{\sqrt{3}} 3t\sqrt{t^2+1}\,dt<br />

    try a substitution.
    Your integral is \int_0^{\sqrt{3}}\sqrt{9t^4+9t^2}\,dt right? So \int_0^{\sqrt{3}}\sqrt{9t^2\left(t^2+1\right)}\,dt . Now bring the 9t^2 out from the square root.
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    Your integral is \int_0^{\sqrt{3}}\sqrt{9t^4+9t^2}\,dt right? So \int_0^{\sqrt{3}}\sqrt{9t^2\left(t^2+1\right)}\,dt . Now bring the 9t^2 out from the square root.

    oooo ok yes thank you.
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