Question

The number of bacteria doubles after every hour. If there were 30 bacteria originally find the number of bacteria after 3 days.

Answer 1

Answer on Question #53689 – Math – Differential Equations

The number of bacteria doubles after every hour. If there were 30 bacteria originally found the number of bacteria after 3 days.

Solution

At first, we must construct the differential equation:

$$\frac{dy}{dt} = ky$$

where $y = y(t)$ is the number of bacteria at time $t$, $k$ is the proportionality factor.

We will rewrite equation

$$\frac{dy}{y} = kdt$$

and integrate

$$\int \frac{dy}{y} = k \int dt.$$

After integrating we will have:

$$\ln y = kt + \ln C. \text{ So we have}$$ $$y(t) = y_0 e^{kt},$$

where $y_0 = \ln C = 30$ is the number of bacteria at time $t = 0$.

After 1 hour the number of bacteria doubles:

$$y(1) = 60 = 30 e^k. \text{ From this, we have: } e^k = 2. \text{ This gives that}$$ $$y(t) = 30 \cdot 2^t.$$

From the last formula we can find the number of bacteria after 3 days (3 days equals 72 hours):

$$y(72) = 30 \cdot 2^{72} \approx 30 \cdot 4.7 \cdot 10^{21} = 141 \cdot 10^{21}.$$

Answer: after 3 days the number of bacteria will be $141 \cdot 10^{21}$.

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