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Finding a curve's unit tangent vector

  1. Feb 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Screen_Shot.jpg

    2. Relevant equations

    I know the equations. See question below.

    3. The attempt at a solution

    I am just wondering with this problem, how is it that they go from that derivative to the magnitude at the bottom of that image? I know the formula, but what I mean is they have no steps for something that to me seems extremely tedious, so I'm assuming they are applying some type of identity to save time and trouble, but I'm not seeing it? If someone can show me an easier way than squaring each of those, splitting it all up, factoring, then trying to make it so you can take the square root of it all, I'd really appreciate it.
     
  2. jcsd
  3. Feb 8, 2014 #2

    LCKurtz

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    Just how tedious is it? Have you tried it? I haven't worked it but it is surely workable and I would guess it uses the identity ##\sin^2t + \cos^2 t = 1##. Dive right in.
     
  4. Feb 8, 2014 #3
    Well, there are 9 problems on the assigned homework just like this one (on top of 12 others on other topics), so cumulatively, if I don't have a way to simplify things like this, it will be very tedious. Besides, I didn't want to waste time if there was a more efficient way I was missing out on.

    Anyway, I just took the time to factor it out and it's telling me my answer was wrong. I really, truly hate MyMathLab with a passion.

    **Update** - I did figure out where I went wrong and got the right answer, but I still feel it was extremely tedious and that if they are skipping all the work, they are most likely applying some type of identity, but maybe not. They don't typically skip that much work in an example though.
     
  5. Feb 8, 2014 #4

    LCKurtz

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    Did you get ##4\sqrt 2(1+t)##? That's what I get. But in any case, it was quicker to just work it than type it into this forum.
     
  6. Feb 8, 2014 #5
    Oh, my specific problem was a bit different. The one I posted was just an example. Same exact setup, different coefficients and a couple different signs. I do appreciate you working it out though.
     
  7. Feb 8, 2014 #6

    haruspex

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    As LCKurtz wrote, it's a matter of using cos2+sin2=1, but also of spotting that you can use it even before you write anything down.
    By inspection, the leading terms of the i.i and j.j contributions sum to 16. Similarly the trailing terms sum to 16t2. Next, the middle terms (+/- 32t sin(t)cos(t)) cancel. It remains to add the k.k term and factorise.
     
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