Populations of two species $A$ and $B$ at time $0$ are equal. If the instantaneous rates of growth of populations of species $A$ and $B$ are $u$ and $u + 1$ respectively, $u > 0$, then at time $1$ the population of species $B$ would be
(a) twice the population of species $A$
(b) $log 10$ times of the population of species $A$
(c) $e^{u}$ times the population of species $A$
(d) $e$ times the population of species $A$
I tries to understand instantaneous growth on the inernet but i found it very confusing, any help will be appreciated!
Hint:
Population at time t : $N_o e^{\lambda t}$ where $N_o$ is the initial population and $\lambda$ is the rate of growth.
Thus the answer should be $e$ times the population of $A$
This was derived by:
$$\frac{dP}{dt}=k P$$ $$\int\frac{dP}{P}=\int k dt$$ $$\ln P=k t$$
Note, that population growth is proportional to current population