Q.What would be the per cent growth or birth rate per individual per hour for the same population mentioned in the previous question (Question 10)?

Since the growth rate is proportional to the number present, $x$ satisfies $\dfrac{dx}{dt} = kx$, giving the exponential relation $x = 100000\,e^{kt}$.

The rate of growth of bacteria is proportional to the number of bacteria present at time $t$:

$\frac{dx}{dt} = kx$

Separating variables and integrating,

$\int \frac{dx}{x} = \int k\,dt \implies \ln x = kt + C$

At $t=0$, $x = 100000$ (the initial count), so $C = \ln(100000)$. Substituting back,

$\ln x - \ln(100000) = kt \implies x = 100000\,e^{kt}$

Using $x = 100000e^{kt}$ with a 10% increase at $t=2$ gives $k = \tfrac12\ln(1.1) \approx 0.0477$ per hour.

From the previous part, $x = 100000\,e^{kt}$.

The bacteria increase by 10% in 2 hours, so at $t = 2$, $x = 100000 + 10\%\text{ of }100000 = 110000$.

$110000 = 100000\,e^{2k}$

$e^{2k} = 1.1$

$2k = \ln(1.1)$

$k = \frac{1}{2}\ln(1.1)$

Numerically, $k \approx \dfrac{0.0953}{2} \approx 0.0477$ per hour.

Doubling time follows from $x=100000e^{kt}$: setting $x=200000$ gives $t = \dfrac{\ln 2}{k} = \dfrac{2\ln 2}{\ln 1.1} \approx 14.55$ hours.

We want the time $t$ at which $x = 200000$, starting from $x = 100000\,e^{kt}$:

$200000 = 100000\,e^{kt}$

$e^{kt} = 2$

$kt = \ln 2$

$t = \frac{\ln 2}{k}$

Using $k = \dfrac{1}{2}\ln(1.1)$ from the previous part,

$t = \frac{\ln 2}{\tfrac12 \ln(1.1)} = \frac{2\ln 2}{\ln(1.1)}$

Numerically, $\ln 2 \approx 0.6931$ and $\ln 1.1 \approx 0.0953$, so

$t \approx \frac{2(0.6931)}{0.0953} \approx \frac{1.3863}{0.0953} \approx 14.55 \text{ hours}$

Geometric (exponential) growth with unlimited resources produces a J-shaped curve where population density accelerates upward without bound; the answer is (C).

The differential equation $f'(x)=f(x)$ is solved by separation of variables, giving the exponential function.

We have $\dfrac{df}{dx} = f$. Separating variables (assuming $f\ne 0$):

$\dfrac{df}{f} = dx$

Integrating both sides:

$\ln|f| = x + c_1$

$f(x) = Ce^{x}$, where $C = \pm e^{c_1}$ is an arbitrary constant.

The simplest such function is $f(x)=e^{x}$, which indeed satisfies $f'(x)=e^x=f(x)$.

Populations can grow exponentially (J-curve, unlimited resources) or logistically (S-curve, limited resources/carrying capacity).

When resources (food, space) in a habitat are unlimited, a population grows exponentially, and a plot of population size against time gives a J-shaped curve (dN/dt = rN). In nature, resources are finite, so growth eventually slows as the population approaches the habitat's carrying capacity (K); a plot of this pattern gives an S-shaped (sigmoid) curve, described by the logistic growth equation dN/dt = rN[(K − N)/K]. The logistic (S-shaped) curve is considered a more realistic model of population growth in nature.

The maximum population size that a given habitat can sustainably support is called the carrying capacity of that habitat for the species.

Every habitat has a finite amount of resources (food, space, nesting sites, etc.), so a population of a species cannot keep growing indefinitely within it. The largest population size that the habitat's resources can support — beyond which the population cannot be sustained (due to resource limitation, competition, increased mortality, etc.) — is termed the carrying capacity (K) of that habitat for that species. It is a central concept in the logistic model of population growth, where population growth rate slows as population size (N) approaches K, and growth effectively stops once N = K.