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!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

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

Have something to add?