Question

The population of bacteria in a culture grows at a rateproportional to the number of bacteria present at time t. After 3hours it is observed that 400bacteria are present. After 10hours 2000bacteria are present. What was the initial number of bacteria?

Short answer

The initial number of bacteria is 201.

Step-by-step solution

Define growth and decay.

The initial-value problem, $\frac{\text{dx}}{\text{dt}}{\text{= kx, x(t}}_{\text{0}}{\text{) =x}}_{\text{0}}$where k is a constant of proportionality, serves as a model for diverse phenomena involving either growth or decay. This is in the form of a first-order reaction (i.e.) a reaction whose rate, or velocity, $\text{dx/dt}$is directly proportional to the amount x of a substance that is unconverted or remaining at time t.

Solve for first order growth and decay equation.

Let the linear equation with the population of a community as be,

$\frac{\text{dx}}{\text{dt}}\text{= kx}$… (1)

And with the conditions,$\text{x(t = 3 years) = 400}$ and $\mathrm{x}(\mathrm{t}=10\mathrm{years})=2000$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$\begin{array}{rcl}\frac{\mathrm{dx}}{\mathrm{x}}& =& \mathrm{kdt}\\ \int \frac{\mathrm{1}}{\mathrm{x}}\mathrm{dx}& =& \mathrm{k}\int \mathrm{dt}\\ \mathrm{lnx}& =& \mathrm{kt}+{\mathrm{c}}_{\mathrm{1}}\\ & & {\mathrm{e}}^{\mathrm{lnx}}={\mathrm{e}}^{\left(\mathrm{kt}+{\mathrm{c}}_{\mathrm{1}}\right)}\end{array}$

Then, the equation becomes,

$$\begin{gathered} {\text{x =e}}^{\text{kt}}{\text{e}}^{{\text{c}}_{\text{1}}} \\ {\text{= ce}}^{\text{kt}} \end{gathered}$$… (2)

Obtain the values of constants.

To find the values of constants, apply the point$\text{(x,t) = (400,3 hours)}$ in the equation (2), then

$$\begin{gathered} 400={\mathrm{ce}}^{3\mathrm{k}} \\ \mathrm{c}=\frac{\mathrm{400}}{{\mathrm{e}}^{3\mathrm{k}}} \end{gathered}$$

Substitute the value of in the equation (2).

$\mathrm{x}=\frac{400}{{\mathrm{e}}^{3\mathrm{k}}}{\mathrm{e}}^{\mathrm{kt}}$… (3)

$\begin{array}{rcl}\mathrm{dx}& =& 400{\mathrm{e}}^{-3\mathrm{k}}{\mathrm{e}}^{\mathrm{kt}}\\ & =& 400{\mathrm{e}}^{(\mathrm{kt}-3\mathrm{k})}\\ & =& 400{\mathrm{e}}^{\mathrm{k}(\mathrm{t}-3)}\end{array}$

Again, apply the other point$\text{(x,t) =}\left(\text{2000,10}\right.\text{years)}$ in the equation (3).

$\begin{array}{rcl}2000& =& 400{\mathrm{e}}^{\mathrm{k}(10-3)}\\ \frac{2000}{400}& =& {\mathrm{e}}^{7\mathrm{k}}\\ 5& =& {\mathrm{e}}^{7\mathrm{k}}\\ \ln \left(5\right)& =& 7\mathrm{k}\\ \mathrm{k}& =& \frac{\ln \left(5\right)}{7}\\ & =& \frac{1.61}{7}\\ & =& 0.23\end{array}$

Substitute the value of in the equation (3).

$\mathrm{x}=400{\mathrm{e}}^{0.23(\mathrm{t}-3)}$… (4)

Obtain the initial number of bacteria.

Substitute the valueinto the equation (4).

$\begin{array}{rcl}\mathrm{x}& =& 400{\mathrm{e}}^{0.23(\mathrm{t}-3)}\\ & =& 400{\mathrm{e}}^{0.23(0-3)}\\ & =& 400{\mathrm{e}}^{-0.69}\\ & =& 400\times 0.5017\\ & \approx & 201\end{array}$