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From http://www.mathpages.com/home/kmath523/kmath523.htm

Variations in $x,y,z$ and $X$ at constant $t$ are independent of $t$ (since each of these variables is strictly a function of $t$), so we have $${\frac{\partial x}{\partial X}=\frac{\partial{\dot x}}{\partial{\dot X}}}$$

(This is just after equation (5) on the page.)

I'm having trouble making sense of this. If each of these variables is strictly a function of $t$, and $t$ is held constant, how does a time derivative (${\dot x}$) make sense?

If it were to mean that $t$ is replaced with a constant after the differentiation, then I could take the example of

$$ x(y)=X^{3/4}=t^{3}$$ $$ X(t)=t^{4}$$

I could then calculate

$$ {\frac{\partial x}{\partial X}=\frac{\dot x}{\dot X} = \frac{3}{4t}}$$


$$ {\frac{\partial {\dot x}}{\partial {\dot X}}=\frac{\ddot x}{\ddot X}=\frac{1}{2t}}$$

...and those are not equal for any $t$.

What is the justification for this step?

Note: This is a follow up to my first question about this, but that error was resolved. Rates of change for functions dependent on same variable

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1 Answer 1

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  • Between eq. (4) and eq. (5)

By definition


and not $x=x(X(t),Y(t),t)$; so

$$\frac{dx}{dt}=\frac{\partial x}{\partial X}\frac{dX}{dt}+ \frac{\partial x}{\partial Y}\frac{dY}{dt},$$ from which follows $$\frac{\partial }{\partial X}\left(\frac{dx}{dt}\right)= \frac{\partial^2 x}{\partial X^2}\frac{dX}{dt}+ \frac{\partial^2 x}{\partial X\partial Y}\frac{dY}{dt},$$

as $\frac{\partial }{\partial X}\frac{dX}{dt}=\frac{\partial }{\partial X}\frac{dY}{dt}=0.$ On the other hand

$$\frac{d}{dt}\left(\frac{\partial x}{\partial X}\right):= \frac{d}{dt}g(X(t),Y(t))= \frac{\partial g}{\partial X}\frac{dX}{dt}+ \frac{\partial g}{\partial Y}\frac{dY}{dt} = \frac{\partial^2 x}{\partial X^2}\frac{dX}{dt}+ \frac{\partial^2 x}{\partial X\partial Y}\frac{dY}{dt}, $$


$$\frac{\partial }{\partial X}\left(\frac{dx}{dt}\right) =\frac{d}{dt}\left(\frac{\partial x}{\partial X}\right).$$

  • Just after eq. (5)

As $$x=x(X(t),Y(t)),$$ then $\frac{\partial x }{\partial \dot{X}}=0$ and $\frac{\partial x }{\partial \dot{X}}\left(\frac{\partial x}{\partial X} \right)=0,$ as well. Using the notation $\dot{x}=\frac{dx}{dt} $ it follows that

$$\frac{\partial\dot{x} }{\partial \dot{X}}= \frac{\partial }{\partial \dot{X}}\left(\frac{\partial x}{\partial X}\dot{X}+ \frac{\partial x}{\partial Y}\frac{dY}{dt},\right)=\frac{\partial x}{\partial X},$$

as claimed.

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