In my opinion, by the definition of ##\frac{\partial f} {\partial x}##, we should calculate the changes in ##f## over changes in ##x## under the condition that both ##a## and ##b## are fixed. But as you said, we can't fix ##a(x)## unchanged while changing ##x## in the case of ##a(x)=1+2x##. So I...
Dear Daz,
While I hold the same opinion with Shyan on his example that ##\frac{\partial f} {\partial x}## should be zero. Because according to our discussion, when calculating ##\frac{\partial f} {\partial x}##, we should keep ##a## and ##b## unchanged, which makes the change in ##f##...
Thanks a lot to both of you. Actuall, I got another understanding of this
suppose ##L=L(q,\dot{q})##, q may depend explicitly on ##x##
when we calculate ##\frac{\partial L}{\partial x}##, we could first bear in mind that ##L=L(q,\dot{q},x)##.
And in order to calculate ##\frac{\partial...
Thanks Shyan!
In my opinion,
if ##\vec v=\vec v(x,y,z,t)##, then according to ##T=\frac{1}{2} mv^2##,we have
$$\frac{\partial T}{\partial x}=mv\frac{\partial v}{\partial t}$$.
And in this case, ##\frac{\partial T}{\partial x}## is not zero.
I mean the T might depend explicitly on ##t##.
Hi guys,
I got one confusion when reading Goldstein's Classical Mechanics (page 20, third edition). After getting the equation,
then it says that
Note that in a system of Cartesian coordinates the partial derivative of T with respect to qj vanishes. Thus, speaking in the language of...