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Line integral problem

  1. Apr 16, 2014 #1

    BiGyElLoWhAt

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    ok, my turn to ask a question.

    Problem: evaluate [itex]∫_{C}xyds[/itex] for [itex] x=t^2 and y = 2t from 0\leq t \leq 5[/itex]

    not sure what I did wrong, but here it goes:

    solve for ds:
    [itex] ds =\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} = \sqrt{4t^2+4}=2\sqrt{t^2+1}[/itex]

    substitute:
    [itex]∫_{0}^5 4t^3\sqrt{t^2+1}dt[/itex]

    substitute again:
    [itex] t=tan(\theta) → dt = sec(\theta)d\theta → ∫_{L}4tan^3(\theta)sec^2(\theta)d\theta[/itex]

    [itex]4∫_{L}tan^3(\theta)(tan^2(\theta)+1)d\theta =4 ∫_{0}^5t^3(t^2+1) =4∫_{0}^5t^5 +t^3 =4\frac{5^6}{6} + 5^4[/itex]
    ...
    But it's wrong
     
  2. jcsd
  3. Apr 16, 2014 #2

    jbunniii

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    Surely the fact that you transformed ##\int_{0}^{5}4t^3 \sqrt{t^2 + 1} dt## into ##\int_{0}^{5}4t^3 (t^2 + 1) dt## is a red flag. Where did the square root go?

    Double check your derivative of ##\tan(\theta)##.
     
  4. Apr 16, 2014 #3

    SammyS

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    What is the derivative of tan(Θ) ?
     
  5. Apr 16, 2014 #4

    BiGyElLoWhAt

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    ok, its sec*tan, stupid mistake, but i'm still not getting my root back. substituting my tan in under the root gimves me sqrt(sec^2)=sec, so i now have 1 exrtra power of tan, but dt is sec*tan, and that gives me 2 secants, one from the root, 1 from my substituted differential.... I do agree with you, I always remember the root coming back in calc 2... I'll check it again when i get off work
     
  6. Apr 16, 2014 #5

    BiGyElLoWhAt

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    Alright I got it, I was making the wrong substitution. I tried u=t^2 +1 and that got me the right answer. I just saw the sqrt(t^2 +1) and assumed a trig substitution, but once I actually wrote it down and looked at it, it was a useless substitution, as I either had all secants or all tangents, and no good differential to sub in.

    Thanks for catching my d/dtheta (tan(theta)) error guys =]
     
  7. Apr 16, 2014 #6

    jbunniii

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    The derivative of ##\tan(\theta)## is not ##\sec(\theta) \tan(\theta)##.

    If you are not sure, you can always calculate it by applying the quotient rule to ##\tan(\theta) = \sin(\theta)/\cos(\theta)##.
     
  8. Apr 16, 2014 #7

    BiGyElLoWhAt

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    Huh, your right... lol
    oops that's what I get for not actually doing it.
     
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