Gradient Representations

Question

Is $\frac{dy}{dx} \equiv \frac{\Delta y}{\Delta x}$

$\frac{\Delta y}{\Delta x}$ is often used in physics, for example: $I=\frac{\Delta Q}{\Delta t}$

I was just wondering, isn't this the same things as the derivative of $Q$ with respect to $t$?

Answer 1

In general ${\Delta y\over \Delta x}$ is a "macroscopic" variation. If you think to the current $I={\Delta Q\over\Delta t}$, this is the mean value of the current during a "macroscopic" time $\Delta t$.

If you take the limit:

$$\lim_{\Delta x \to 0}{\Delta y\over \Delta x}\equiv{dy\over dx}$$

you can identify the formulas and in the case of the current you get the instantaneous current, i.e. the derivative of the charge respect to the time:

$$I=\lim_{\Delta t \to 0}{\Delta Q\over \Delta t}={dQ\over dt}.$$