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Help with a vector proof about perpedndicular unit vectors

  1. Sep 13, 2010 #1
    Prove that if a(t) is a unit vector whose components are functions of t, then the vector b(t) = da(t)/dt is perpendicular to a(t)



    here i have the parameters that make a(t) a unit vector:

    1 = f(t)^2 + g(t)^2




    i know of the example sinxi + cosxj = a(t)
    but that is the only one i can get to work. I need a general proof though so:

    a(t) = f(t)^2 i + g(t)^2 j

    b(t) = f '(t) i + g '(t) j

    (dot) product of a(t) and b(t) is:

    f(t)*f '(t) + g(t)*g '(t)

    i tried to take the derivative of a(t) and solve for both f'(t) and g'(t)

    1 = f(t)^2 + g(t)^2

    0 = 2*f(t)*f'(t) + 2*g(t)*g'(t)

    i treated this like one big equation and solved each for g'(t) and f'(t)

    f ' (t) = [-g(t)*g ' (t)]/ f(t)

    g ' (t) = [-f(t)*f ' (t)]/ g(t)

    when i carry out the dot product i still dont get zero, i need some advice, am i even on the right track.

    the only other way i could think of was to setting each component equal to zero to solve for the derivative of the functions, but i did not know if this was mathematically legal

    ex:

    2*g(t)*g'(t) = 0

    g'(t) = 0
     
    Last edited: Sep 13, 2010
  2. jcsd
  3. Sep 13, 2010 #2

    cepheid

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    Use this expression to express g(t) in terms of f(t). That will help greatly.


    I don't understand the part in red. Why have you squared the functions? They should not be squared if your definition of a is to be consistent with your definition of b. The rest of what you have doesn't make sense either. See if following my advice above helps.
     
  4. Sep 13, 2010 #3
    did you mean this one: 1 = f(t)^2 + g(t)^2

    so f(t) = (1 - g(t)^2)^1/2

    and g(t) = (1- f(t)^2)^1/2

    then the derivative of the above:

    f ' (t) = -[g(t)*g ' (t)] / [1-g(t)^2]
    g ' (t) = -[f(t)*f ' (t)] / [1-f(t)^2]

    so then if i take the dot product

    [f(t) * f ' (t)] + [ g(t) * g ' (t) ]

    (1 - g(t)^2)^1/2 * -[g(t)*g ' (t)] / [1-g(t)^2] + (1- f(t)^2)^1/2 * -[f(t)*f ' (t)] / [1-f(t)^2]

    simplified :

    -[g(t)*g ' (t)] -[f(t)*f ' (t)]

    still dont understand it, where is my error?? :confused:
     
  5. Sep 13, 2010 #4

    cepheid

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    Yes.

    You just need one of these. The whole point is to eliminate one by expressing it in terms of the other. For example, if you express g in terms of f, then henceforth everything is in terms of f and you don't need g anymore. Do you understand?

    Here is the problem. This differentiation is wrong. You need to use the chain rule since:

    dg/dt = dg/df * df/dt

    What is g(f), and what is its derivative with respect to f? That is the first step. The second step is to multiply that by df/dt.

    Also, again, you only have to compute ONE of the above derivatives.
     
  6. Sep 13, 2010 #5
    got it finished, thanks. i wasnt thinking of eliminating terms correctly, like you said
     
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