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Easy multivariable derivative

  1. Aug 8, 2007 #1

    malawi_glenn

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    I am repeting some multivariable calculus.

    I want to know if have done right now:

    [tex]\mathbf{r} = \mathbf{r}(q_1, q_2, q_3)[/tex]

    [tex] \dfrac{\partial \mathbf{r}}{\partial q_1} = \left(\dfrac{\partial r_1}{\partial q_1} , \dfrac{\partial r_2}{\partial q_1} , \dfrac{\partial r_3}{\partial q_1} \right) [/tex]

    let

    [tex] \mathbf{r} = (q_1 + 2q_3, q_2 + 3q_1 - q_3, q_1 - q_3) [/tex]

    [tex] \dfrac{\partial \mathbf{r}}{\partial q_1} = (1,3,1) [/tex]
     
  2. jcsd
  3. Aug 8, 2007 #2

    HallsofIvy

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    If I understand what your problem is correctly- that r is a three dimensional vector function of 3 variables, then, yes, your answer is correct.
     
  4. Aug 8, 2007 #3

    malawi_glenn

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    thanx dude!

    How about this one?

    [tex] \dfrac{d}{d\theta}\left( \dfrac{d\theta}{dt}\right) = 0 \text{ ?} [/tex]
     
  5. Aug 8, 2007 #4

    CompuChip

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    Yes. What about it? What is [itex]\theta[/itex]? And what is the question?
     
  6. Aug 8, 2007 #5

    malawi_glenn

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    theta is a function, and I am wondering if

    [tex] \dfrac{d}{d\theta}\left( \dfrac{d\theta}{dt}\right) = 0 \text{ ?} [/tex]

    Is correct in this case: i forgot to post the link.

    http://en.wikipedia.org/wiki/Lagrangian_mechanics

    I am trying to figure out what is happening in "Pendulum on a movable support"
     
  7. Aug 8, 2007 #6
    According to information presented in the link
    you are right
    [tex] \dfrac{d\theta}{dt} [/tex] is only function of t so derivativee w.r.t. theta is 0
     
    Last edited: Aug 8, 2007
  8. Aug 8, 2007 #7

    malawi_glenn

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    great thanx! Have not done calculus for a while, so I am repeting a bit before next semester=)
     
  9. Aug 8, 2007 #8

    HallsofIvy

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    No, that is not true.

    In general, if [itex]\theta[/itex] is a function of t and f is any function of [itex]\theta[/itex] then it is also a function of t and
    [tex]\frac{df}{d\theta}= \frac{df}{dt}\frac{dt}{d\theta}[/tex]
    In particular, if [itex]f= d\theta /dt[/itex] then
    [tex]\frac{d}{dt}\frac{d\theta}{dt}= \frac{d^2\theta}{dt^2}\frac{dt}{d\theta}[/tex]
    which is not necessarily 0.

    To take an easy example, if [itex]\theta= e^t[/itex] then
    [tex]\frac{d\theta}{dt}= e^t= \theta[/itex]
    so that
    [tex]\frac{d}{d\theta}\frac{d\theta}{dt}= \frac{d\theta}{d\theta}= 1[/tex]
     
  10. Aug 8, 2007 #9
    Right but

    In the link L is function of theta, theta- dot and t

    So for L at least
    [tex] \frac{\partial}{\partial\theta}\left( \dfrac{d\theta}{dt}\right) = 0 [/tex]
    is true
    I have considered the lagrangian and was not carefull:blushing:
    thanks
     
  11. Aug 8, 2007 #10

    HallsofIvy

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    Try being careful!
     
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