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Vector Functions, Differentiation

  1. Apr 11, 2004 #1
    This doesn't exactly have to do with the title, but generally comes up in this sort of thing. I'm feeling pretty dense right now to not be able to understand what's going on.. but my text book isnt exactly helping.

    lets say there is a function which relates a particles position, r, with respect to time, t: (w is representative of the period)

    r=3cos(wt)i + 4cos(wt)j + 5sin(wt)k

    its derivative should give you its velocity

    v=-3wsin(wt)i - 4wsin(wt)j + 5wcos(wt)k

    the magnitude of the velocity vector should give the speed of the particle

    s= ?? (the book suggests the answer is s=5w)


    How does one arrive at the magnitude of vector v?? I can do this in a case where we are looking at simple coefficients of i,j, and k... but with the trig in there, I'm getting sorta confused. If someone would post up a step-by-step, it would be much appreciated!

    Thanks!! Sorry for asking such a simple question.. probably goes back to grade 10 math that I've just forgotten! lol
     
  2. jcsd
  3. Apr 11, 2004 #2
    The vector v can also be represented as (-3wsin(wt), -4wsin(wt), 5wcos(wt)). Use the expanded pythagoras theorem:
    |v|2 = [-3wsin(wt)]2 + [-4wsin(wt)]2 + [5wcos(wt)]2
    |v|2 = 9w2sin2(wt) + 16w2sin2(wt) + 25w2cos2(wt)
    |v|2 = w2[9sin2(wt) + 9cos2(wt) + 16sin2(wt) + 16cos2(wt)]
    |v|2 = 25w2
    |v| = 5w

    If you are not familiar with this method you can use the dot-product:
    |v|2 = v(dot)v
    You basically reach the same expression after multiplying the vector by itself. :smile:
     
    Last edited: Apr 11, 2004
  4. Apr 11, 2004 #3
    Thanks Chen,

    I probably should have shown the steps that I had come up with.

    I can reach this:
    |v|^2 = w^2[9sin^2(wt) + 9cos^2(wt) + 16sin^2(wt) + 16cos^2(wt)]

    and i can see how we get the w^2 in the next line
    |v|2 = 25w^2

    but how do we get the 25 from this mess: [9sin^2(wt) + 9cos^2(wt) + 16sin^2(wt) + 16cos^2(wt)] ??


    (btw, how did you manage superscript without tex?)
     
  5. Apr 11, 2004 #4
    ahh... OK.. I just got an idea.. good old trig identities.. lemme see if I can work it out

    EDIT: ahhh... makes perfect sense now :D

    9sin^2(wt) + 9cos^2(wt) + 16sin^2(wt) + 16cos^2(wt)

    =25sin^2(wt) + 25cos^2(wt)

    =25(sin^2(wt) + cos^2(wt))

    =25(1)

    =25
     
    Last edited: Apr 11, 2004
  6. Apr 11, 2004 #5
    Hint:
    sin2(x) + cos2(x) = 1
    See, you figured it our yourself. :smile:

    (Use [ sup ] and [ sub ] tags, without the spaces.)
     
  7. Apr 11, 2004 #6
    Heh, I'm not really sure why I split it to two pairs of 9 and 16, I could've used just one pair of 25. Old habit I guess. :wink:
     
  8. Apr 11, 2004 #7
    Yeah, in fact looking at it now, I think I would have just added them all to

    |v|2 = 25w2sin2(wt) + 25w2cos2(wt)

    then factored 25w2 all at once
     
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