Math Problem Statement

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.

Solution

This problem involves exponential growth, where the rate of change in the population of bacteria is proportional to the current population. The general formula for exponential growth is:

N(t)=N0ektN(t) = N_0 e^{kt}

Where:

  • N(t)N(t) is the number of bacteria at time tt,
  • N0N_0 is the initial number of bacteria,
  • kk is the growth constant,
  • tt is the time in hours.

Step 1: Finding the growth constant kk

We are told that the number of bacteria triples in 5 hours. This means at t=5t = 5, N(5)=3N0N(5) = 3N_0. Substituting this into the exponential growth equation:

3N0=N0e5k3N_0 = N_0 e^{5k}

Dividing both sides by N0N_0:

3=e5k3 = e^{5k}

Taking the natural logarithm (ln) of both sides:

ln(3)=5k\ln(3) = 5k

k=ln(3)5k = \frac{\ln(3)}{5}

Now, let's calculate kk:

k1.09865=0.2197k \approx \frac{1.0986}{5} = 0.2197

Step 2: Finding the number of bacteria after 10 hours

Now that we know kk, we can find the number of bacteria after 10 hours. The equation is:

N(10)=N0e0.2197×10N(10) = N_0 e^{0.2197 \times 10}

Simplifying:

N(10)=N0e2.197N(10) = N_0 e^{2.197}

Using the value of e2.1979.0e^{2.197} \approx 9.0:

N(10)=9N0N(10) = 9N_0

Thus, after 10 hours, the number of bacteria will be 9 times the initial population.

Step 3: Finding the time when the population is 10 times the initial population

We need to find the time tt such that N(t)=10N0N(t) = 10N_0. Substituting into the growth equation:

10N0=N0e0.2197t10N_0 = N_0 e^{0.2197t}

Dividing by N0N_0:

10=e0.2197t10 = e^{0.2197t}

Taking the natural logarithm of both sides:

ln(10)=0.2197t\ln(10) = 0.2197t

t=ln(10)0.2197t = \frac{\ln(10)}{0.2197}

Since ln(10)2.3026\ln(10) \approx 2.3026, we can calculate tt:

t2.30260.219710.48 hourst \approx \frac{2.3026}{0.2197} \approx 10.48 \text{ hours}

Thus, it will take approximately 10.48 hours for the population to become 10 times the initial number.

Summary:

  1. After 10 hours, the population will be 9 times the initial population.
  2. It will take approximately 10.48 hours for the population to become 10 times the initial number.

Would you like more details or have any questions?

Here are 5 related questions to explore this topic further:

  1. What happens to the population if the bacteria's growth rate doubles?
  2. How does the initial number of bacteria affect the time for tripling?
  3. What if the bacteria's growth slows down, how would the model change?
  4. How would you calculate the population after half a day (12 hours)?
  5. Can this model be applied to real-life bacterial cultures with resource limitations?

Tip: In problems involving exponential growth, always check if the rate or time units need to be adjusted for the most accurate calculation.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Logarithms
Natural Logarithms

Formulas

N(t) = N_0 e^{kt}
k = ln(3) / 5
N(10) = N_0 e^{0.2197 * 10}
t = ln(10) / 0.2197

Theorems

Exponential Growth Theorem
Logarithmic Properties

Suitable Grade Level

Grades 10-12