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.

Explanation:

Step1: Recall the exponential growth formula for doubling time

The formula for exponential growth with doubling time \( T \) is \( P(t) = P_0 \cdot 2^{\frac{t}{T}} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, and \( T \) is the doubling time. Here, \( T = 10 \) minutes.

Step2: Find the initial population \( P_0 \)

We know that at \( t = 80 \) minutes, \( P(80) = 70000 \). Substitute into the formula:
\( 70000 = P_0 \cdot 2^{\frac{80}{10}} \)
Simplify the exponent: \( \frac{80}{10} = 8 \), so \( 70000 = P_0 \cdot 2^8 \)
Since \( 2^8 = 256 \), we solve for \( P_0 \): \( P_0 = \frac{70000}{256} \approx 273.4375 \). We can keep it as a fraction or a decimal, but likely round to a reasonable number (or keep exact: \( \frac{70000}{256} = \frac{4375}{16} \approx 273.44 \)).

Step3: Find the population after 5 hours

First, convert 5 hours to minutes: \( 5 \times 60 = 300 \) minutes.
Use the formula \( P(t) = P_0 \cdot 2^{\frac{t}{T}} \), with \( P_0 = \frac{4375}{16} \), \( t = 300 \), \( T = 10 \).
Calculate the exponent: \( \frac{300}{10} = 30 \), so \( P(300) = \frac{4375}{16} \cdot 2^{30} \)
Simplify \( 2^{30} = (2^{10})^3 = 1024^3 = 1073741824 \), but easier to use \( 2^{30} = 2^{8 + 22} = 2^8 \cdot 2^{22} \), but actually, \( \frac{4375}{16} \cdot 2^{30} = 4375 \cdot 2^{30 - 4} = 4375 \cdot 2^{26} \) (wait, no: \( \frac{1}{16} = 2^{-4} \), so \( 2^{-4} \cdot 2^{30} = 2^{26} \)). So \( P(300) = 4375 \cdot 2^{26} \)
Calculate \( 2^{26} = 67108864 \), then \( 4375 \times 67108864 \). Wait, but maybe better to use the initial \( P_0 \approx 273.4375 \):
\( P(300) = 273.4375 \cdot 2^{30} \)
\( 2^{30} = 1073741824 \), so \( 273.4375 \times 1073741824 \). But let's use the exact \( P_0 = \frac{70000}{256} \):
\( P(300) = \frac{70000}{256} \cdot 2^{30} = 70000 \cdot 2^{30 - 8} = 70000 \cdot 2^{22} \) (since \( \frac{2^{30}}{2^8} = 2^{22} \))
\( 2^{22} = 4194304 \), so \( 70000 \times 4194304 = 70000 \times 4,194,304 = 293,601,280,000 \). Wait, that seems large, but let's check the exponent again. Wait, \( t = 300 \) minutes, doubling time 10 minutes, so number of doublings is \( 300 / 10 = 30 \) doublings. Starting from \( P_0 \approx 273.44 \), after 30 doublings: \( 273.44 \times 2^{30} \). \( 2^{10} = 1024 \), \( 2^{20} = 1024^2 = 1,048,576 \), \( 2^{30} = 1024^3 = 1,073,741,824 \). So \( 273.44 \times 1,073,741,824 \approx 273.44 \times 1.073741824 \times 10^9 \approx 293.6 \times 10^9 = 2.936 \times 10^{11} \), which matches the previous calculation.

But let's re-examine Step 2: when \( t = 80 \), number of doublings is 8, so \( P(80) = P_0 \times 2^8 \), so \( P_0 = 70000 / 256 = 273.4375 \), that's correct. Then for 5 hours (300 minutes), number of doublings is 30, so \( P(300) = 273.4375 \times 2^{30} \). Let's compute \( 2^{30} = 1073741824 \), so \( 273.4375 \times 1073741824 \):

First, \( 273.4375 = 273 + 0.4375 = 273 + \frac{7}{16} \)

\( 273 \times 1073741824 = 273 \times 1.073741824 \times 10^9 \approx 293.1315 \times 10^9 \)

\( 0.4375 \times 1073741824 = \frac{7}{16} \times 1073741824 = 7 \times 67108864 = 469,762,048 \)

Adding them together: \( 293,131,500,000 + 469,762,048 = 293,601,262,048 \), which is approximately \( 2.936 \times 10^{11} \) (or 293,601,262,048).

Wait, but maybe we made a mistake in the initial population. Let's check: \( 2^8 = 256 \), \( 70000 / 256 = 273.4375 \), that's correct. Then after 30 doublings, \( 273.4375 \times 2^{30} \). Alternatively, using the formula \( P(t) = P(t_0) \cdot 2^{\frac{t - t_0}{T}} \), where \( t_0 = 80 \), \( P(t_0) = 70000 \), and we want \( t = 0 \) (first part) and \( t = 300 \) (second part).

For \( t = 0 \): \( P(0) = P(80) \cdot 2^{\frac{0 - 80}{10}} = 70000 \cdot 2^{-8} = 70000 / 256 = 273.4375 \), which matches.

For \( t = 300 \): \( P(300) = P(80) \cdot 2^{\frac{300 - 80}{10}} = 70000 \cdot 2^{\frac{220}{10}} = 70000 \cdot 2^{22} \)
\( 2^{22} = 4194304 \), so \( 70000 \times 4194304 = 70000 \times 4,194,304 = 293,601,280,000 \) (which is \( 2.9360128 \times 10^{11} \)).

So the initial population is approximately 273 (or 273.44), and after 5 hours, it's approximately \( 2.936 \times 10^{11} \) (or 293601280000).

Answer (Initial Population):

\( \frac{4375}{16} \approx 273.44 \) (or 273 when rounded to nearest whole number, or exact fraction \( \frac{4375}{16} \))

Answer (Population after 5 hours):

\( 70000 \times 2^{22} = 293601280000 \) (or \( 2.9360128 \times 10^{11} \))

(Note: Depending on rounding, the initial population could be 273 or 273.44, and the 5-hour population is a large number as calculated.)

Answer:

(Initial Population):
\( \frac{4375}{16} \approx 273.44 \) (or 273 when rounded to nearest whole number, or exact fraction \( \frac{4375}{16} \))