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Differentiation of vector function

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data
    suppose f is a differentiable mapping of R1 into R3 such that |f(t)|=1 for every t. Prove that [tex]f'(t)\cdot f(t)=0[/tex].
    I guess it is more proper to write [tex](\nabla f)(t) \cdot f(t)=0[/tex], where [tex](\nabla f)(t)[/tex] is the gradient of f ant t.


    2. Relevant equations



    3. The attempt at a solution
    it is then equivalent to prove [tex]\sum\;(Df_{i})(t)\cdot f_{i}(t)=0[/tex], but I've no idea of how to use the |f(t)|=1 to deduce the disired results, although the equation has an easy geometrically interpretation
    Any hint may help, thanks a lot.
     
    Last edited: Dec 15, 2008
  2. jcsd
  3. Dec 15, 2008 #2

    Dick

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    The f'(t) notation is fine. f(t) is a vector. You generally use the gradient notation for scalar functions, not vectors. And |f(t)|=1 is the same as saying f(t).f(t)=1, right?
     
  4. Dec 15, 2008 #3
    just differentiate both side and it is done....Thanks!
    and thanks for the explanation about gradient...I misunderstood it...
    Here is why I thought it improper..f'(t) is a linear transformation from R^1->R^3, whose basis are different things from the basis of R^3, so I thought it is weird to apply the dot product to f'(t) and f(t)....
     
  5. Dec 15, 2008 #4

    Dick

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    For functions of a single variable it's simpler just to think of f'(t) as an element of R^3 rather than as a linear transformation. If f(t)=(x(t),y(t),z(t)), f'(t)=(x'(t),y'(t),z'(t)).
     
  6. Dec 15, 2008 #5
    it took me years to open and refresh this forums....my connecting to some websites sucks...
    Thinking of it as a vector in R3 is okay and easy. okay, I found it more sensible in this way.
    take 1 as a basis of R^1. So f'(t)(1) = [tex]\sum^{3}_{1} (Df_{i})(t)e_{i}[/tex], where e_i is the basis of R^3, and dot product makes sense
     
  7. Dec 15, 2008 #6

    Dick

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    Sure, the linear transformation is f'(t)(h)=(x'(t),y'(t),z'(t))*h. That can be identified with a vector in R^3 in a pretty obvious way. I would say you are just throwing the 'h' away altogether.
     
  8. Dec 15, 2008 #7

    HallsofIvy

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    Saying |f(t)|= 1 is the same as saying [itex]f(t)\cdot f(t)= 1[/itex]. Use the product rule.
     
  9. Dec 15, 2008 #8
    it's been solved, Thanks anyway:)
     
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