Math Problem Statement

A bacteria culture starts with 800 bacteria and grows at a rate proportional to its size. After 5 hours, there are 9000 bacteria. 1. Find an expression for the number of bacteria after t hours. 2. Find the number of bacteria after 6 hours. 3. Find the growth rate after 6 hours. 4. After how many hours will the population reach 30000?

Solution

Given the problem, the bacteria culture follows exponential growth, where the population grows at a rate proportional to its size. This can be modeled using the differential equation:

dPdt=kP\frac{dP}{dt} = kP where:

  • P(t)P(t) is the population at time tt,
  • kk is the growth constant,
  • P0P_0 is the initial population (800 bacteria).

Step 1: Expression for the number of bacteria after tt hours

The solution to the differential equation is:

P(t)=P0ektP(t) = P_0 e^{kt}

Using the initial condition P0=800P_0 = 800, the equation becomes:

P(t)=800ektP(t) = 800 e^{kt}

Now, we use the given information that after 5 hours, there are 9000 bacteria:

9000=800e5k9000 = 800 e^{5k}

Solving for kk:

e5k=9000800=11.25e^{5k} = \frac{9000}{800} = 11.25 5k=ln(11.25)5k = \ln(11.25) k=ln(11.25)50.5057k = \frac{\ln(11.25)}{5} \approx 0.5057

Thus, the expression for the number of bacteria after tt hours is:

P(t)=800e0.5057tP(t) = 800 e^{0.5057t}

Step 2: Number of bacteria after 6 hours

Substitute t=6t = 6 into the expression:

P(6)=800e0.5057×6P(6) = 800 e^{0.5057 \times 6} P(6)800e3.0342800×20.78516628P(6) \approx 800 e^{3.0342} \approx 800 \times 20.785 \approx 16628

So, the number of bacteria after 6 hours is approximately 16,628.

Step 3: Growth rate after 6 hours

The growth rate is given by dPdt=kP(t)\frac{dP}{dt} = kP(t). Using k=0.5057k = 0.5057 and P(6)16628P(6) \approx 16628:

dPdtt=6=0.5057×166288410bacteria/hour\frac{dP}{dt} \bigg|_{t=6} = 0.5057 \times 16628 \approx 8410 \, \text{bacteria/hour}

So, the growth rate after 6 hours is approximately 8410 bacteria per hour.

Step 4: Time to reach 30,000 bacteria

We want to find tt when P(t)=30000P(t) = 30000:

30000=800e0.5057t30000 = 800 e^{0.5057t} e0.5057t=30000800=37.5e^{0.5057t} = \frac{30000}{800} = 37.5 0.5057t=ln(37.5)0.5057t = \ln(37.5) t=ln(37.5)0.50577.504t = \frac{\ln(37.5)}{0.5057} \approx 7.504

Thus, the population will reach 30,000 after approximately 7.5 hours.


Would you like more details on any of these steps or have any questions? Here are five related questions:

  1. How does changing the initial population affect the growth model?
  2. What happens to the growth rate as time increases indefinitely?
  3. How would the results change if the growth rate kk were higher or lower?
  4. What is the doubling time of the bacterial culture?
  5. How would the equation change if there was a limiting population capacity?

Tip: In exponential growth, the rate of increase gets faster over time, which can lead to rapid growth after a certain point.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Differential Equations
Natural Logarithms

Formulas

Exponential Growth Model: P(t) = P_0 * e^(kt)
Population at time t: P(t) = 800 * e^(0.5057t)
Growth Rate: dP/dt = k * P(t)

Theorems

Exponential Growth Law
Natural Logarithm Properties

Suitable Grade Level

Grades 10-12