partial differentiation of a variable w.r.t. its time derivative

Question

What will be the partial derivative :

$$\frac{\partial\theta}{\partial\dot\theta}$$ and $$\frac{\partial\dot\theta}{\partial\theta}$$

where, $\theta = \theta(t)$ and $\dot\theta = \frac{d\theta}{dt}$

would these be zero or something else??

Answer 1

The expressions you wrote are meaningless, at least without context.

If I have a function $f(x,y,z)$, I can only calculate the derivative of $f$ with respect to $x$, $y$ or $z$. Deriving always means with respect to one of the variables.

Answer 2

If you interpret them as functions, as in mathematical analysis, what you wrote makes no sense (it's not even zero).

If they are quantities of some kind (e.g. coordinates), like in physics or other applications, then you can write different functions, but which represent the same quantity. Take locally the inverse $\theta\mapsto t(\theta)$, and using the Leibniz notation: $$ \frac{dt}{d\theta} = \frac{1}{\dot\theta}. $$

This is a function of $\theta$, let's call it $\theta\mapsto f(\theta)$, which as a quantity is equal to $1/\dot\theta$.

Now, as a quantity: $$ \dot\theta(t)= \dot\theta(t(\theta)), $$

and: $$ \frac{d\dot\theta}{d\theta} = \frac{d\dot\theta}{dt}\frac{dt}{d\theta} = \ddot\theta\frac{1}{\dot\theta} = \frac{\ddot\theta}{\dot\theta} $$

Viceversa, $$ \frac{d\theta}{d\dot\theta} = \frac{\dot\theta}{\ddot\theta}. $$

Lastly, if the quantity $d\theta/d\dot\theta$ appears in calculation with Lagrangians, or in general functions or functionals in the form: $$ L = L(\theta, \dot\theta, t,\dots), $$

then $\theta$ and $\dot\theta$ are simply different variables, and: $$ \frac{\partial\dot\theta}{\partial\theta} = \frac{\partial\theta}{\partial\dot\theta} = 0, $$

just like, if $F(x,y)=x$, $\frac{\partial x}{\partial y}=\frac{\partial F}{\partial y}=0$.