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