# why logistic growth differential equation is a differential equation?

why is the diffraction equation for logistic growth is written as

$${dP \over dt} = kP({1 - {P \over L}})$$

why is it supposed to be a derivative in the first place?

say for example

the growth rate (k) = 10% per day

and the carrying capacity (L) = 1000

and population(P) = 10

if you plug in the numbers $${dP \over dt} = 0.99$$

which is the change of population in (ONE DAY) so we should say $${ΔP \over Δt} = 0.99$$

however a differential equation(derivative) should a give us rate of change at a point (not over a period of time)

i hope i was clear , sorry for my english, it's not my native language

The reason why you get a rate (it's $$\mathrm dP/\mathrm dt$$ that's the growth rate, not $$k$$!) expressed per day is because you're measuring time in days. If you were to measure time in seconds, then you'd get the rate per unit second. However, not even this captures what a moment is -- which is mathematically an infinitesimal. Such a thing cannot actually exist in a real-world measurement, which is always based on some finite unit.
• @AmirWG I don't know what it represents in this particular model, but it's just the constant of proportionality scaling the quadratic in $P.$ Whatever the case is, we generally interpret the derivative in contexts like this as a rate of change. Your equation simply says the rate with with the population changes with time is proportional to $P(1+MP),$ where $M$ is some constant. – Allawonder Apr 18 '20 at 19:18
• @AmirWG In any physical application of any equation, the units on both sides must balance each other -- we say that the equation must be dimensionally consistent. In that sense the units of all the parameters in the equation are related. Yes, it's because $k$ is measured per day that you get a growth rate per day. In this case, the unit of time in $\mathrm dP/ \mathrm dt$ will always have to be equal to that of time in $k.$ – Allawonder Apr 19 '20 at 4:42
To get the right result, one must accept that time is continuous, and no real $$tau>0$$ achieves $$P(t+\tau)-P(t)=\frac{dP}{dt}\tau$$ exactly. The rate of change from $$t$$ to $$t+\tau$$ is actually $$\int_t^{t+\tau}\frac{dP(u)}{du}du$$, which approximates $$\left.\frac{dP}{du}\right|_{u=t}\tau$$ if $$\tau$$ is very small. So start at time $$t$$. A split-second later, $$P$$ has changed, which means $$\frac{dP}{dt}$$ has changed. A split-second after that, $$P$$ has changed again, but not the same way, because of the change in $$\frac{dP}{dt}$$. Of course, since $$P$$ has changed again, $$\frac{dP}{dt}$$ has also changed again. And so on.