Observe the graph and choose the correct option from following:

A = \(\frac{dN}{dt}\) = rN.

B = \(\frac{dN}{dt}\) = rN(\(\frac{K-N}{K}\))

The correct answer is Option (1) -

A = \(\frac{dN}{dt}\) = rN.

B = \(\frac{dN}{dt}\) = rN(\(\frac{K-N}{K}\))

A represents 'exponential growth' : 

If in a population of size N, the birth rates (not total number but per capita births) are represented as b and death rates (again, per capita death rates) as d, then the increase or decrease in N during a unit time period t (dN/dt) will be

\(\frac{dN}{dt}\)= (b – d) × N

Let (b–d) = r,

then \(\frac{dN}{dt}\) = rN.

The r in this equation is called the ‘intrinsic rate of natural increase’.

 B represents 'logistic growth' :

A population growing in a habitat with limited resources show initially a lag phase, followed by phases of acceleration and deceleration and finally an asymptote, when the population density reaches the carrying capacity. A plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth.

    = \(\frac{dN}{dt}\) = rN(\(\frac{K-N}{K}\))