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Homework Help: Is there something wrong?

  1. Dec 17, 2006 #1
    1. The problem statement, all variables and given/known data

    d( [tex]\vec {F}[/tex].[tex]\vec {F}[/tex])/dt=d(F*F)/dt=2*F*dF/dt (1)
    d( [tex]\vec {F}[/tex].[tex]\vec {F}[/tex])/dt=2* [tex]\vec{F}[/tex].d [tex]\vec{F}[/tex]/dt (2)
    so F*dF/dt=[tex]\vec{F}[/tex].d [tex]\vec{F}[/tex]/dt (3)
    ???????????
    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Dec 17, 2006
  2. jcsd
  3. Dec 17, 2006 #2

    dextercioby

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    Why would something be wrong in what you've written ?

    Daniel.
     
  4. Dec 17, 2006 #3

    do you mean it is right?
     
  5. Dec 17, 2006 #4
    [tex]\vec{F}[/tex] and d[tex]\vec{F}[/tex]/dt are in the same direction?
     
  6. Dec 17, 2006 #5

    dextercioby

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    What relevance does that have for your equation ?

    Daniel.
     
  7. Dec 17, 2006 #6
    but it really seems relevant
     
  8. Dec 17, 2006 #7

    dextercioby

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    Not to me. Why is it to you ?

    Daniel.
     
  9. Dec 17, 2006 #8
    so which one do you think the relerance is weak?
    (1) (2) (3)
     
  10. Dec 17, 2006 #9

    dextercioby

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    I dunno what you mean, I told you that for your derivation, it doesn't matter that the 2 vectors are in the same direction or not. It's simply "irrelevant".

    Daniel.
     
  11. Dec 17, 2006 #10

    arildno

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    Why do you believe this?? :confused:
     
  12. Dec 17, 2006 #11
    Because of the the equation of (3)
    it seems the angle between the two vector =0
     
  13. Dec 17, 2006 #12

    arildno

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    Why do you believe the rate of change of the magnitude of the vector F equals the magnitude of the vector that is the rate of change of the vector F??

    That's totally wrong!
    Here's why:
    [tex]\vec{F}=|F|\vec{i}_{r}, \vec{i}_{r}\cdot\vec{i}_{r}=1[/tex]
    Therefore,
    [tex]\frac{d\vec{F}}{dt}=\frac{d|F|}{dt}\vec{i}_{r}+|F|\frac{d\vec{i}_{r}}{dt}[/tex]
    Thus, if the unit vector [itex]\vec{i}_{r}[/itex] changes with time, then your result doesn't hold.
     
  14. Dec 17, 2006 #13
    yeah,it must be wrong.
    but i don'nt know in which equation,while they all obey the rules of vector.
    and in (2):

    d([tex]\vec{F}[/tex].[tex]\vec{F}[/tex])/dt=[tex]\vec{F}[/tex].d[tex]\vec{F}[/tex]/dt+d[tex]\vec{F}[/tex]/dt.[tex]\vec{F}[/tex]
    =2*[tex]\vec{F}[/tex].d[tex]\vec{F}[/tex]/dt

    tell me,please!
     
    Last edited: Dec 17, 2006
  15. Dec 18, 2006 #14

    dextercioby

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    I already told you that all 3 equations are correct.

    Daniel.
     
  16. Dec 18, 2006 #15
    [tex]\left |\frac{d\vec{F}}{dt} \right | \neq \frac{d|\vec{F}|}{dt}[/tex]

    [tex]F*\frac{dF}{dt}=|\vec{F}| \frac{d|\vec{F}|}{dt}=\vec{F}\cdot\frac{d\vec{F}}{dt}=|\vec{F}|\left |\frac{d\vec{F}}{dt} \right | \cos{\theta}[/tex]

    where [tex]\theta[/tex] is the angle between [tex]\dot{\vec{F}}[/tex] and [tex]\vec{F}[/tex]
     
    Last edited: Dec 18, 2006
  17. Dec 18, 2006 #16
    Hello,
    i Think also that the vectors F and dF/dt are perpendicular if the square of the modulus do not change in time, i.e. something that in differential geometry can be interpretated as a mobile 2-D basis among a given curve a=a(t). And with the introduction of a third vector, let's call him n, perpendicular to both of them we have the so famous "Triedro di Frenet".
    sorry for my bad english.

    since the scalar product (,):VxV--->R is a bilinear form defined on a vector space and has value on the Real field numbers.

    if we develop the calculus we obtain from a side:

    [tex] \frac{d\vec{F}\vec{F}}{dt}=2 \vec{F}\frac{d\vec{F}}{dt} [/tex]

    but from the other side:

    [tex] \vec{F}\vec{F}=|F|^{2}[/tex]

    and d/dt of this quantity is zero by hypothesis.

    we can recognize de def. of perpendicularity of the two vec. F and dF/dt.

    N.B.
    i did'n use the dot for the scalar product
    bye bye
    Marco
     
    Last edited: Dec 18, 2006
  18. Dec 18, 2006 #17
    i dont know whats going on with the tex compiler but i think everybody understood the meaning of my opinion.

    bye

    maRCO
     
  19. Dec 18, 2006 #18

    D H

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    The inner product of any constant-magnitude vector and its time derivative is identically zero. That's not an opinion, its an identity. (Note that this alone disproves the OP's misconception.) That [itex]\vect F \cdot \frac {d\vect F}{dt} = F \frac {dF}{dt}[/itex] is also an identity. That [itex]\vect F[/itex] is parallel to [itex]\frac{d\vect F}{dt}[/itex] is just plain wrong.
     
  20. Dec 18, 2006 #19
    Thank you!
    I can understand this problem thoroughly now.
    best wishes for the coming of Christmas
    o:) o:) o:) o:) o:) o:) o:) o:) o:) o:) o:)
     
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