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Homework Help: Simple Arc Length Problem

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data

    x = 1+3t^2
    0<= x <=4

    2. Relevant equations

    L = integral from a to b of [tex]\sqrt{[dx/dt]^2 + [dy/dt]^2}[/tex] dx

    3. The attempt at a solution

    dx/dt = 6t
    dy/dt = 6t^2

    L = integral from 0 to 4 of [tex]\sqrt{(6t)^2 +(6t^2)^2}[/tex] dx
    = " [tex]\sqrt{36t^2 +36t^4}[/tex] dx
    = " [tex]\sqrt{(36)(t^2 +t^4)}[/tex] dx
    = " 6 [tex]\sqrt{(t^2)(1 +t^2)}[/tex] dx
    = " 6t [tex]\sqrt{1 +t^2}[/tex] dx

    let u = 1+t^2
    du = 2tdt
    1/2 du = tdt

    = " 3 (u)^(1/2)du
    = 3(2/3) [ u^(3/2)] from 0 to 4
    = 2[(1+t^2)^(3/2)] from 0 to 4
    = 2[(17)^(3/2)-1]

    Is that correct? Just want to make sure.
  2. jcsd
  3. Mar 10, 2009 #2


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    Staff: Mentor

  4. Mar 10, 2009 #3
    Haha. I apologize. The question I asked was actually very simple compared to the other topics being posted in this forum.

    The link you sent me looks like cases in which you would use trig substitution (though, that doesn't seem to be how the problems are worked out). But in the problem above, I'm integrating 6 sqrt(1+t^2) t dt, which is easy. I only ask for confirmation because I'm really bad at catching my own mistakes, and had already messed up on the last few steps of this problem three times before getting what I have now.
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