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Homework Help: Finding speed from position equation

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    We just had a test in my calc 3 class, and I'm pretty sure my teacher has the wrong solution to one of the answers. The question is about finding the speed of a particle given a space curve function r(t) = (cos2t)i + (3t - 1)j + (sin2t)k.

    2. Relevant equations
    v(t) = (x',y',z')
    |v(t)| = √(x' + y' + z')

    3. The attempt at a solution
    For velocity I get (-2sin2t)i + (3)j + (2cos2t)k. Now, for speed I put √(4sin22t + 4cos22t + 9) = 2(sin2t + cos2t) + 3. He marked it wrong saying that it can be reduced to √(4+9) = √(13). I was under the impression that the sin2x + cos2x = 1 identity only works if the variable stands alone in the function, as sin2t/cos2t calls for a double angle formula.
    Last edited: Oct 5, 2011
  2. jcsd
  3. Oct 5, 2011 #2


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    [itex]\sqrt{a^2+b^2+c^2}\ne a+b+c[/itex]

    Under the radical, you had [itex]4\sin^22t + 4\cos^22t + 9[/itex]

    This is equal to [itex]4(\sin^22t + \cos^22t) + 9\ =\ 4(1)+9[/itex]

    Look like your professor was right.
  4. Oct 5, 2011 #3
    Hmm I really never knew you could have any coefficient in front of x/t/whatever and still use the sin2x + cos2x = 1 identity. Oh well thanks a lot for the input
  5. Oct 5, 2011 #4


    Staff: Mentor

    I assume that you really meant sin2x + cos2x = 1.

    If you have something like 9sin2(4x) + 9cos2(4x), you can factor out the 9 to get
    9(sin2(4x) + cos2(4x)) = 9 * 1 = 9.

    You don't have to expand sin(2x) or sin(4x) or whatever. The identity is sin2<whatever> + cos2<whatever> = 1, as long as the <whatever> is the same in both places.
  6. Oct 5, 2011 #5
    Ahh okay thank you both. That's very helpful for the future, and I guess my ignorance here is due to the fact that I've never really worked with a function like that before, but upon reviewing the derivation for the identity it does make sense. I feel pretty stupid for asking something so trivial, but thanks again
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