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Arc length for these parametric equations

  1. Aug 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the arc length of the curve described by the parametric equations: x=2e^t & y=3e^3t/2 ln3≤t≤2ln3

    2. Relevant equations

    S = ∫(a->b) √[(dy/dt)^2 + (dx/dt)^2]dt

    3. The attempt at a solution

    Differentiated the two parametrics:
    dy/dt = 2e^t
    dx/dt = (3/2)*3e^3t/2 = (9/2)e^3t/2

    Plugged it all in and got:
    ∫(ln3->2ln3) √[(2e^t)^2 + ((9/2)e^3t/2)^2]dt
    = ∫(ln3->2ln3)√[(4e^2t + (81/4)e^3t]dt

    I'm stuck at this integral, I don't see any viable choice for u, and I don't think I'm allowed to approximate it.

    *sorry about the sloppy bound notation.
  2. jcsd
  3. Aug 4, 2009 #2


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    Try factoring out a [tex]e^{2t}[/tex] from inside the square root and then doing a u substitution. Keep in mind that [tex]\sqrt{e^{2t}}=e^{t}[/tex]
  4. Aug 4, 2009 #3

    Wow! Didn't even think of that.

    So the integrand simplifies to sqrt([e^2t](4+(81/4)e^t]) = (e^t)*sqrt(4 + 81/4)*e^t)). And u = e^t, du = e^t*dt, right? :D

    If so, I can totally handle it from there, thanks!

    Edit: meant e^t*dt, not du
    Last edited: Aug 4, 2009
  5. Aug 4, 2009 #4
    I did a u-sub, then a v-sub (v=[81/4]u+4), and got (2/3e^t + 32/243)|(ln2->3ln2)

    And then I got 9, is this answer correct?
  6. Aug 4, 2009 #5


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    That's not what I'm getting.
    [tex]\int_{\ln(2)}^{2\ln(3)} \sqrt{4e^{2t}+\frac{81}{4}e^{3t}}\,dt=\frac{9}{2}\int_{\ln(2)}^{2\ln(3)} e^{t}\sqrt{\frac{16}{81}+e^{t}}\,dt[/tex]

    Simplifies to
    [tex]\left[ 3(\frac{16}{81}+e^{t})^{\frac{3}{2}}\right]_{\ln(3)}^{2\ln{3}}[/tex]

    definitely not 9
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