Homework Help: Proof of independence of position and velocity

Tags:
1. Oct 3, 2016

weezy

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. Oct 3, 2016

Orodruin

Staff Emeritus
You said it yourself, $r_i$ is assumed to be a function of the coordinates $q_k$, not of their time derivatives.

3. Oct 3, 2016

Staff: Mentor

In future posts, please do not delete the homework template with its three parts.

4. Oct 4, 2016

weezy

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?

5. Oct 4, 2016

weezy

Sorry for any inconvenience. I'll be careful.