- #1
sgh1324
- 12
- 0
let
df=∂f/∂x dx+∂f/∂y dy and ∂f/∂x=p,∂f/∂y=q
So we get
df=p dx+q dy
d(f−qy)=p dx−y dqand now, define g.
g=f−q y
dg = p dx - y dq
and then I faced problem.
∂g/∂x=p←←←←←←←←←←←←←←← book said like this because we can see g is a function of x and p so that chain rule makes ∂g/∂x=p
but I wrote directly ∂g/∂x so i can get a result,
∂g/∂x=∂f/∂x−y ∂q/∂x−q ∂y∂x=p−y ∂q∂x - 0(y is independent of x)
I know that variation y is independent of x, but I'm not sure that q is also the independent function of x. what if the func f is xy? if I set like that,
g=f−qy=xy−xy=0
it's not same with p!mathematical methods in the physical sciences, Mary L. Boas 3rd edition page231
http://www.utdallas.edu/~pervin/ENGR3300/Boaz.pdf
df=∂f/∂x dx+∂f/∂y dy and ∂f/∂x=p,∂f/∂y=q
So we get
df=p dx+q dy
d(f−qy)=p dx−y dqand now, define g.
g=f−q y
dg = p dx - y dq
and then I faced problem.
∂g/∂x=p←←←←←←←←←←←←←←← book said like this because we can see g is a function of x and p so that chain rule makes ∂g/∂x=p
but I wrote directly ∂g/∂x so i can get a result,
∂g/∂x=∂f/∂x−y ∂q/∂x−q ∂y∂x=p−y ∂q∂x - 0(y is independent of x)
I know that variation y is independent of x, but I'm not sure that q is also the independent function of x. what if the func f is xy? if I set like that,
g=f−qy=xy−xy=0
it's not same with p!mathematical methods in the physical sciences, Mary L. Boas 3rd edition page231
http://www.utdallas.edu/~pervin/ENGR3300/Boaz.pdf