# 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 $AC$, the speed is defined as the increase of the distance $\Delta d$ (or, the covered distance) divided by the increase of time $\Delta t$.$$\displaystyle v_m=\frac{\Delta d}{\Delta t}=\frac{20 \mbox{ km }}{10 \mbox{ min }}=120 \mbox{ km/h }$$

## 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. $$\displaystyle v(t)=\lim_{\Delta t \to 0 }\frac{\Delta d}{\Delta t}=\lim_{\Delta t \to 0}\frac{f(a+\Delta t)-f(a)}{\Delta t}$$ The speed is the function derivative of the position (or space).

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 $\Delta t= 5s$. We computed the covered distance:$$\Delta d= d(t=5)-d(t=0)=22-2 \mbox{ metros }$$ Therefore, $$\displaystyle v_m=\frac{20 \mbox{ m}}{5\mbox{ s}}= 4\mbox{ m/s}$$

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

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:$$d'(t)=2t-1 \Rightarrow d'(2)=2\cdot 2-1=3 \mbox{ m/s}$$