Physical interpretation of the derivative

The world of physics gives us a good tool for understanding derivatives.

Average Change = Average Speed

A driver covers $20$ km that separate her house from her office in $10$ minutes. What is the average speed?

Just as the $A C$ , the speed is defined as the increase of the distance $Δ d$ (or, the covered distance) divided by the increase of time $Δ t$ . $v_{m} = \frac{Δ d}{Δ t} = \frac{20 km}{10 min} = 120 km/h$

Derivative at a point = instantaneous Speed

The speed is not $120$ km/h during the whole trajectory. Most probably her speed will be changing (it does not leave the parking next to its house at $120$ km/h !).

The instantaneous speed is the speed at a precise moment in time. In other words, we are trying to find the distance covered by the driver as the time interval goes to zero. $v (t) = lim_{Δ t \to 0} \frac{Δ d}{Δ t} = lim_{Δ t \to 0} \frac{f (a + Δ t) - f (a)}{Δ t}$ The speed is the function derivative of the position (or space).

Example

The distance that a person covers given at a certain time is: $d (t) = t^{2} - t + 2$

Find the average speed for the first 5 seconds of the movement.

We have that $Δ t = 5 s$ . We computed the covered distance: $Δ d = d (t = 5) - d (t = 0) = 22 - 2 metros$ Therefore, $v_{m} = \frac{20 m}{5 s} = 4 m/s$

Now find the instantaneous speed after that at $t = 2 s$ .

The instantaneous speed is the derivative of the distance at the point $t = 2$ .

We compute the derivative (we can either use the definition of derivative or use more sophisticated techniques) and obtain: $d^{'} (t) = 2 t - 1 \Rightarrow d^{'} (2) = 2 \cdot 2 - 1 = 3 m/s$