Population growth of a species can be modelled as $$ \frac{dN(t)}{dt} = rN(t)\left(1 - \frac{N(t)}{K}\right) $$ where $N(t)$ is the population size at time $t$; $r$ is the growth rate; and $K$ is the carrying capacity of the environment. For $K = 9000$, $\frac{dN(t)}{dt}$ is maximized at $N =$ _____ (Answer in integer)
Maximizing Population Growth Rate in Logistic Model
The population growth is modelled by the logistic equation:
$ \frac{dN(t)}{dt} = rN(t)\left(1 - \frac{N(t)}{K}\right) $Let the growth rate function be $f(N) = rN\left(1 - \frac{N}{K}\right)$. To find when this rate is maximized, we need to find the value of $N$ where the derivative of $f(N)$ with respect to $N$ is zero.
Finding the Maximum Growth Rate
- Expand the function:
$f(N) = rN - \frac{rN^2}{K}$
- Differentiate $f(N)$ with respect to $N$:
$ \frac{df}{dN} = \frac{d}{dN}\left(rN - \frac{rN^2}{K}\right) = r - \frac{2rN}{K} $
- Set the derivative to zero to find critical points:
$ r - \frac{2rN}{K} = 0 $
$ r = \frac{2rN}{K} $
- Solve for $N$. Assuming $r \neq 0$:
$ 1 = \frac{2N}{K} $
$ N = \frac{K}{2} $
- Verify it's a maximum using the second derivative test:
$ \frac{d^2f}{dN^2} = -\frac{2r}{K} $
Since $r > 0$ and $K > 0$, $ \frac{d^2f}{dN^2} < 0 $, confirming a maximum.
- Substitute the given carrying capacity, $K = 9000$:
$ N = \frac{9000}{2} = 4500 $
Therefore, the population growth rate $\frac{dN(t)}{dt}$ is maximized when the population size $N$ is 4500.
Important Questions from Population growth curves
- A flask containing nutrient-rich media is seeded with 100 isogenic bacteria. Assuming that no bacteria die in the flask, after approximately how many generations will the population reach a size of $10^5$?
- The population size at which net recruitment is the highest is also when the greatest amount can be harvested, while ensuring the long-term survival of the population. The amount harvested at this population size is known as
The graphs shown represent the relationship between population size ($N$) and population growth rate ($\frac{dN}{dt}$). Which one of the following growth curves represents a density-dependent population that experiences a strong Allee effect?
Overfishing reduced food availability for sea lions in California, causing a decline in their population size. In 1972, under the US Endangered Species Act, fishing was banned from sea lion foraging areas. Subsequently, the population of sea lions increased in a logistic form as shown in the figure.
The per capita growth rate is highest in the interval __________ and the population growth rate is highest in the interval __________
- A population grows as per the equation $dn/dt = rn (1-n/1000)$ where $n$ is the population density, $r$ is the intrinsic growth rate and 1000 is the carrying capacity. The growth rate of the population is maximum at a population density of ________
