- #1
weezy
- 92
- 5
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.
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.