Question

sections 4.5 - 4.6
score: 9/15 answered: 9/15
question 10
the doubling period of a bacterial population is 10 minutes. at time t = 80 minutes, the bacterial population was 70000.
what was the initial population at time t = 0?
find the size of the bacterial population after 5 hours.
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Explanation:

Step1: Recall the doubling - time formula

The formula for a population with a doubling period \(T\) is \(P(t)=P_0\cdot2^{\frac{t}{T}}\), where \(P(t)\) is the population at time \(t\), \(P_0\) is the initial population, and \(T\) is the doubling period. Here, \(T = 10\) minutes. We know that at \(t = 80\) minutes, \(P(80)=70000\).

Step2: Find the initial population \(P_0\)

Substitute \(t = 80\), \(T = 10\) and \(P(80)=70000\) into the formula \(P(t)=P_0\cdot2^{\frac{t}{T}}\):
\(70000=P_0\cdot2^{\frac{80}{10}}\)
\(70000 = P_0\cdot2^{8}\)
Since \(2^{8}=256\), we can solve for \(P_0\) by \(P_0=\frac{70000}{256}=\frac{70000\div8}{256\div8}=\frac{8750}{32}=\frac{8750\div2}{32\div2}=\frac{4375}{16}=273.4375\approx273\) (we can also keep it as a fraction \(\frac{4375}{16}\) or a decimal).

Step3: Find the population after 5 hours

First, convert 5 hours to minutes. Since 1 hour = 60 minutes, 5 hours = \(5\times60 = 300\) minutes.
Now use the formula \(P(t)=P_0\cdot2^{\frac{t}{T}}\) with \(P_0=\frac{4375}{16}\), \(t = 300\) and \(T = 10\).
\(P(300)=\frac{4375}{16}\cdot2^{\frac{300}{10}}\)
\(P(300)=\frac{4375}{16}\cdot2^{30}\)
We know that \(2^{30}=1073741824\)
\(P(300)=\frac{4375\times1073741824}{16}\)
\(4375\times1073741824 = 4375\times1073741824\)
\(4375\times1073741824=(4000 + 375)\times1073741824=4000\times1073741824+375\times1073741824\)
\(4000\times1073741824 = 4294967296000\)
\(375\times1073741824=(300 + 75)\times1073741824=300\times1073741824+75\times1073741824\)
\(300\times1073741824 = 322122547200\)
\(75\times1073741824 = 80530636800\)
\(322122547200+80530636800 = 402653184000\)
\(4294967296000+402653184000 = 4697620480000\)
Then \(P(300)=\frac{4697620480000}{16}=293601280000\)

Another way:
Since we know that \(P(80) = 70000\) and the doubling period is 10 minutes. The number of 10 - minute intervals from \(t = 80\) to \(t=300\) is \(\frac{300 - 80}{10}=\frac{220}{10}=22\). So the population at \(t = 300\) is \(P(300)=P(80)\cdot2^{22}\)
\(2^{22}=4194304\)
\(P(300)=70000\times4194304 = 70000\times4194304=293601280000\)

For the initial population:
We can also think in terms of going back in time. From \(t = 80\) to \(t = 0\), the number of 10 - minute intervals is \(\frac{80-0}{10}=8\). So we divide the population at \(t = 80\) by \(2^{8}\) (because we are going back 8 doubling periods). So \(P_0=\frac{70000}{2^{8}}=\frac{70000}{256}=273.4375\)

Part 1: Initial population

Step1: Use the doubling - time formula

The formula for population growth with doubling time \(T\) is \(P(t)=P_0\times2^{\frac{t}{T}}\), where \(P(t)\) is population at time \(t\), \(P_0\) is initial population, \(T = 10\) min, \(t = 80\) min, \(P(80)=70000\).

Step2: Substitute values and solve for \(P_0\)

Substitute into \(70000 = P_0\times2^{\frac{80}{10}}\), so \(70000=P_0\times2^{8}\). Since \(2^{8}=256\), then \(P_0=\frac{70000}{256}=\frac{4375}{16}=273.4375\) (or approximately 273).

Step1: Convert 5 hours to minutes

5 hours \(=5\times60 = 300\) minutes.

Step2: Use the population growth formula

We know that \(P(t)=P_0\times2^{\frac{t}{T}}\), and we can also use the fact that from \(t = 80\) to \(t = 300\), the number of 10 - minute intervals is \(\frac{300 - 80}{10}=22\). So \(P(300)=P(80)\times2^{22}\), where \(P(80)=70000\) and \(2^{22}=4194304\).

Step3: Calculate the population

\(P(300)=70000\times4194304 = 293601280000\)

Answer:

(Initial population):
\(\frac{4375}{16}\) (or \(273.4375\) or \(273\) if we round to the nearest whole number)

Part 2: Population after 5 hours