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This is an excellent question. It is pedantic to whine about non differentiability ,however there is in fact a point to be made on that topic . We seem to be conflating the derivative dy/dx with the time derivative dx/dt(x is the position here). One may have a differentiable path but still the instantaneous velocity can remain undefined. Whether the converse ...


No, it's not possible, because one of the underlying assumptions of kinematics is that all paths are at least twice differentiable. Before you complain about this requirement, remember that physics is about building models that can be used to describe and predict measurements. Measurements always have some amount of uncertainty, and even if you suppose that ...


The length along any segment of the Koch snowflake is infinite. It has finite area but infinite perimeter. So, for a particle to move from one place on the snowflake to another it would have to travel an infinite distance. This is why differentiability is important.


A resistor is defined as the circuit element for which the voltage across is proportional to the current through and the constant of proportionality is the resistance $R$: $$V_R = R\cdot I_R $$ Clearly, for this linear relationship, it is also true that $$\frac{dV_R}{dI_R} = R$$ However, for general circuit elements, the derivative of $V(I)$ is not a ...


$R(V,I) = \frac{V}{I}$ by definition, it is not a gradient. $r = \frac{dV}{dI}$ is called the fractional, differential, dynamical or small-signal resistance. It just happens that for resistors $R(V,I) = R_0$ is a constant, thus the two quantities are the same: $r = R_0$.


The problem is that you are equating too many things to $\dot{q_k}$. Usually $\dot{q_k} = \frac{dq_k}{dt}$, a total derivative, as opposed to a partial derivative. If $q_k$ has no explicit time-dependence, i.e. it does not depend directly on $t$ itself, then $\frac{\partial q_k}{\partial t} = 0.$ In this case, the Poisson bracket reduces to: $ ...


$\frac{dM}{dt} = \frac{\partial{M}}{\partial{t}}+\frac{\partial{M}}{\partial{x}}\frac{d{x}}{d{t}} = \frac{\partial{M}}{\partial{t}}+v\cdot\nabla{M}$ (with no assumption on what is M) . So if $v\cdot\nabla{M} \neq0$ you can have one of $\frac{dM}{dt}$ and $\frac{\partial{M}}{\partial{t}}$ that is zero when the other is not. ...

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