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Proof of independence of position and velocity

  1. Oct 3, 2016 #1
    A particle's position is given by $$r_i=r_i(q_1,q_2,...,q_n,t)$$ So velocity: $$v_i=\frac{dr_i}{dt} = \sum_k \frac{\partial r_i}{\partial q_k}\dot q_k + \frac{\partial r_i}{\partial t} $$

    In my book it's given $$\frac{\partial v_i}{\partial \dot q_k} = \frac{\partial r_i}{\partial q_k}$$ without any proof. So I tried to take derivative of ##v_i## w.r.t ##\dot q_k##. I can only arrive at the proof if $$\frac{\partial r_i}{\partial \dot q_k} = 0?$$ Why is that?
    Is it because of explicit dependence of ##r_i##on ##\dot q_k##? Sorry if this question is too basic but I'm confused because I believe it could be written as $$\frac{\partial r_i}{\partial \dot q_k} = \frac{\partial r_i}{\partial q_k}.\frac{\partial q_k}{\partial \dot q_k}$$ and that $$\frac{\partial q_k}{\partial \dot q_k}$$ is not zero.
     
  2. jcsd
  3. Oct 3, 2016 #2

    Orodruin

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    You said it yourself, ##r_i## is assumed to be a function of the coordinates ##q_k##, not of their time derivatives.
     
  4. Oct 3, 2016 #3

    Mark44

    Staff: Mentor

    In future posts, please do not delete the homework template with its three parts.
     
  5. Oct 4, 2016 #4
    yes but can't that be written like ##
    \frac{\partial r_i}{\partial \dot q_k} = \frac{\partial r_i}{\partial q_k}.\frac{\partial q_k}{\partial \dot q_k}##?
    ##
    \frac{\partial r_i}{\partial \dot q_k} ## is not zero that's obvious so does it imply that ##
    \frac{\partial q_k}{\partial \dot q_k}
    ## is zero?
     
  6. Oct 4, 2016 #5
    Sorry for any inconvenience. I'll be careful.
     
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