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Arclength parameter question

  1. Sep 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the arclength parameter s=s(t) for the path
    x(t)=e^(at)cos(bt)i + e^(at)sin(bt)j + e^(at)k

    2. Relevant equations
    [itex] s(t)=\int_{a}^{t}\left | \mathbf{x}(\tau ) \right |d\tau [/itex]

    3. The attempt at a solution
    I took the derivative and squared it, arriving at the equation [itex] (-be\sin bt)^{2}+(ae\cos bt))^{2}+(be\cos bt)^{2}+(ae\sin bt)^{2}+(ae^{at})^{2}[/itex]. How would I go about simplifying this in order to take the square root and integrate it? It seems extremely complex to be able to integrate at the moment. Thanks.
     
  2. jcsd
  3. Sep 30, 2015 #2

    SteamKing

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    Well, what have you tried? With a lot of sines and cosines being squared and all, are there any trig identities you think which might be useful in simplifying this integrand?
     
  4. Sep 30, 2015 #3
    The cos and sin for the ae terms cancel to 1, but then I am left with a -be and a be, which would let me factor out ##-\sin ^{2}(bt)+cos^{2}(bt)##, which I am unsure how to simplify. Also, not too sure what to do with the (ae^ae)^2 term.
     
  5. Sep 30, 2015 #4

    SteamKing

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    You're squaring everything. How can you have any negative quantities in the integrand?
     
  6. Sep 30, 2015 #5
    Ah, I see. So, I'd be left with ##2ae^{at^{2}}+be^{2}## under the square root?
     
  7. Sep 30, 2015 #6

    SteamKing

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    What is (eat)2 ? Remember the laws of exponents.
     
  8. Sep 30, 2015 #7
    It would be ##2ae^{2at}##. So, under the square root we have ##2ae^{2at}+be^{2}##?
     
  9. Sep 30, 2015 #8

    SteamKing

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    You seem to have lost some terms along the way. How did you start with eat and ebt and wind up with just plain ae and be ?

    Remember, many a calculus student's career was derailed not by the calculus, but by algebra or arithmetic.
     
  10. Sep 30, 2015 #9
    Going back, it seems like I should have ##be^{2at}+2ae^{2at}##. Correct?
     
  11. Sep 30, 2015 #10

    SteamKing

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    That's better. How about a little factoring, now, when you put this expression under the radical?
     
  12. Sep 30, 2015 #11
    Factoring? What do you mean, exactly?
     
  13. Sep 30, 2015 #12

    SteamKing

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    Don't you notice any common terms in your expression under the radical? You know, things which could be factored out from under the square root sign?
     
  14. Sep 30, 2015 #13
    e^(at) could be factored out, right?
     
  15. Sep 30, 2015 #14

    SteamKing

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    Yes, yes it could.
     
  16. Sep 30, 2015 #15
    So, in the end we're left with ##e^{at}\sqrt{2a^{2}+b^{2}}##. Am I correct in assuming that the things under the radical can no longer be simplified?
     
  17. Sep 30, 2015 #16

    SteamKing

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    You are correct. The stuff left under the radical is just another constant anyway.

    Now does this expression look like it can be integrated?
     
  18. Sep 30, 2015 #17
    Yes! I'm assuming ##\sqrt{2a^{2}+b^{2}} ## can be pulled out of the integral and we just have to integrate e^{at}. Doing this integral from 0 to t, I obtained ##\frac{1}{a}e^{at}\sqrt{2a^{2}+b^{2}}-\frac{1}{a}\sqrt{2a^{2}+b^{2}}##. Does this seem reasonable?
     
  19. Sep 30, 2015 #18

    SteamKing

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    Yes, but I believe this expression can be simplified. :wink:
     
  20. Sep 30, 2015 #19
    I've arrived at ##\sqrt{2a^{2}+b^{2}}(\frac{1}{a}e^{at}-\frac{1}{a})## by factoring out the 1/a
     
  21. Sep 30, 2015 #20
    Factoring out the radical i mean
     
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