1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    [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

    Mark44

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook