I"m looking at "Principles of Dynamics: Second Edition" by Donald T Greenwood. I'm trying to figure out how he obtains Eq. (6-64)

$$\frac{\partial\dot x_j}{\partial\dot q_i} = \frac{\partial x_j}{\partial q_i}\tag{6-64}$$

from Eq (6-54)

$$\dot x_j = \sum_{i=1}^n \frac{\partial x_j}{\partial q_i}\dot q_i + \frac{\partial x_j}{\partial t}.\tag{6-54}$$

where the transformation equations from a set of $3N$ Cartesian coordinates to a set of $n$ generalized coordinates are of the form given by Eq (6-1)

$$x_1=f_1(q_1,q_2,...,q_n,t)$$ $$x_2=f_2(q_1,q_2,...,q_n,t)$$ $$\vdots$$ $$x_{3N}=f_{3N}(q_1,q_2,...,q_n,t).\tag{6-1}$$

When I differentiate Eq (6-54) with respect to $q_i$ I get second derivatives and I have no idea how the term $\frac{\partial^2 x_j}{\partial t\partial \dot q_i}$ is dealt with. Any insight appreciated.


1 Answer 1


Greenwood is asking how does $\dot x_j$ change if we vary $\dot q_i$ at some specified time $t$ and point $q_i$ (or equivalantly a specified point $x_i$ as there is a - perhaps time dependent - 1-1 relation between the $q$'s and $x$'s). At that specified point and time all the quantities on the RHS of your second equation, with the exception of the $\dot q$'s are to be treated as constants. The answer then, is exactly your first equation.

  • $\begingroup$ Ahhhhh, I think I see now. Just to clarify, $\frac{\partial\dot q_i}{\partial \dot q_j}=0 $, because $q_i$ is independent of $q_j$ for $i \ne j$? $\endgroup$
    – eball
    Commented Jun 14, 2019 at 17:14
  • $\begingroup$ Yes. You can vary the $\dot \q_i$'s independently of one another. $\endgroup$
    – mike stone
    Commented Jun 14, 2019 at 21:40
  • $\begingroup$ Makes sense. Thanks again!! $\endgroup$
    – eball
    Commented Jun 15, 2019 at 19:51

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