Fredrik said:
My point is that while dx is always just a real number, dy usually refers to something different: a first order estimate of how much f changes as a result of changing x (while Δy would be the actual change).
O.K. as I understand you
\frac{dx}{dy} = \frac{\Delta x}{dy}
is always true, but the following are not necessarily true:
\frac{dx}{dy} = \frac{ x}{\Delta y}
\frac{dx}{dy} = \frac{\Delta x}{\Delta y}
HallsofIvy has just gone and complicated things by introducing \delta x and saying it is not the same thing as dx. So does that mean:
\frac{dx}{dy} = \frac{\delta x}{\delta y}
is also not necessarily true?
I always thought one was just a informal shorthand for the other.
So let's take a practical example that might make things clearer to me.
We have a particle initially at rest at (x1,t1) = (0,0) that accelerates to a final instantaneous velocity of 50 m/s in 10 seconds over a distance of 200m so that (x2,t2) = (200,10).
The average velocity of the particle is:
\frac{\Delta x}{\Delta t} = (200/10) = 20 m/s
The initial instantaneous velocity of the particle is:
\frac{dx_1}{dt_1} = 0 m/s
The final instantaneous velocity of the particle is:
\frac{dx_2}{dt_2} = 50 m/sWhere does \frac{\delta x}{\delta t} fit into this picture?
HallsofIvy seems to be using \delta x as the small interval \Delta x that is sufficiently small that it is a good approximation to the conceptual dx. Is that correct?
I am primarily trying to get the notation right in my head, but the concepts are of interest too.