Exercise 9.3 · Q22: Bacteria up 10% in 2 hours: 2,00,000 after $\frac{2\log 2}{\log(11/10)}$ hours

Basics you need

Solution

1. Name the variables and write the model

   Let $t$ be the time in hours from the start and $N$ the number of bacteria at time $t$⁠. The rate of growth is proportional to the number present:

   $$\frac{dN}{dt} = kN,$$

   where $k > 0$ is a constant (per hour).

   "Proportional to" means "equal to a constant times". The count is increasing, so $k > 0$⁠; its value is unknown and will come from the 10% data. The equation has order 1 and degree 1. (A count is a whole number, but the model treats $N$ as a smooth quantity, as the book does.)
   A rate of change is a derivative. Basics: e^{kt} and its rate

2. Separate the variables and integrate

   $$\begin{aligned} \frac{dN}{N} &= k\,dt \\ \log N &= kt + C. \end{aligned}$$
   Dividing by $N$ is allowed since $N > 0$⁠. $\int\frac{dN}{N} = \log N$ and $\int k\,dt = kt$⁠; the two constants combine into one $C$⁠.
   Section 9.4.1; Section 7.2, formula (xii). Lesson: separate, then integrate

3. Use the starting count

   At $t = 0$⁠, $N = 100000$⁠: $\log 100000 = C$⁠. So

   $$\begin{aligned} \log N - \log 100000 &= kt, \\ \text{i.e.}\quad \log\frac{N}{100000} &= kt. \end{aligned}$$
   The condition fixes $C$⁠. In exponential form this is $N = 100000\,e^{kt}$⁠.
   Section 9.3: a particular solution from a condition. Basics: one constant, one condition

4. Use the 10% rise to find $k$

   After 2 hours the count is $100000 \times 1.1 = 110000$⁠. Put $t = 2$⁠, $N = 110000$⁠:

   $$\log\frac{110000}{100000} = 2k \;\Rightarrow\; k = \frac{1}{2}\log\frac{11}{10}.$$
   $\frac{110000}{100000} = \frac{11}{10}$⁠. This one equation has the one unknown $k$⁠. Numerically $k \approx 0.0476551$ per hour.
   Solving for the rate constant. Lesson: a culture that grows 10% every 2 hours

5. Find when the count reaches 2,00,000

   Let $N = 200000$ at $t = T$⁠. Then $\frac{N}{100000} = 2$⁠, and

   $$\begin{aligned} \log 2 = kT &= \frac{T}{2}\log\frac{11}{10} \\ \Rightarrow\; T &= \frac{2\log 2}{\log\frac{11}{10}}. \end{aligned}$$

   Numerically $T \approx 14.55$ hours, about 14 hours 33 minutes.

   The starting count cancels: only the ratio 2 matters, so this is the doubling time. Since $\log\frac{11}{10} > 0$⁠, $T > 0$⁠.
   $\log$ undoes $e$⁠. Basics: e^{kt} and its rate

6. Check, and see the growth

   With $k = \frac{1}{2}\log\frac{11}{10}$⁠, $e^{kt} = \left(\frac{11}{10}\right)^{t/2}$⁠, so $N = 100000\,(1.1)^{t/2}$⁠. Then $N(2) = 110000$ and $\frac{T}{2} = \frac{\log 2}{\log 1.1}$ gives $1.1^{T/2} = 2$⁠, so $N(T) = 200000$⁠. Every 2 hours the count is multiplied by 1.1; after 14 hours it is $100000 \times 1.1^7 \approx 194872$⁠, still below 2,00,000, so $T$ just over 14 hours is right.

   Substituting back: $\frac{dN}{dt} = kN$ holds for $N = 100000e^{kt}$⁠, and both data points are on the curve. The picture shows the growth speeding up as the population grows.
   $\frac{d}{dt}e^{kt} = ke^{kt}$⁠. Basics: e^{kt} is k times itself

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