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Finding arc length of vector valued function

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Find arc length of the graph of
    r(t) = ti + ( (t6/6) - (6/t4) )j + t√3 k
    1≤t≤2

    2. Relevant equations

    Arc length = ∫ ||dr/dt|| dt
    (Integral from t0 to t1 of norm of derivative of r)

    3. The attempt at a solution

    dr/dt = i + (t5 + 24/(t3) )j + √3 k

    12 = 1
    (√3)2 = 3
    (t5 + 24/(t3) )2 = t10 + 48t2 + 576/t6

    ||dr/dt|| = √(1 + 3 + t10 + 48t2 + 576/t6)


    I get stuck here because I do not know how to integrate that. It doesn't seem anywhere close to a perfect square.
    Did I make a mistake somewhere, or is there some kind of property to integrate this?
     
  2. jcsd
  3. Jul 9, 2012 #2

    jbunniii

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    You did not differentiate the [itex]6/t^4[/itex] term correctly.
     
  4. Jul 9, 2012 #3
    Ok, so now it should be

    (t5 +24t-5)2
    (t10 + 48 + 576t-10)

    ||dr/dt|| = √( 1 + 3 + t10 + 48 + 576t-10 )
    ||dr/dt|| = √( t10 + 52 + 576t-10 )

    Which I am still not sure how to integrate, but it at least looks nicer.
     
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