Why logistic growth differential equation is a differential equation

calculusderivatives

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
$${ΔP \over Δt} = 0.99$$

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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

thanks in advance

Best Answer

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.

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