Math Problem Statement

A population of bacteria follows the continuous exponential growth model  Upper P left parenthesis t right parenthesis equals Upper P 0 e Superscript kt​, where t is in days. The relative​ (daily) growth rate is 3​%. The current population is 888.

A) Find the growth model.​ (the function that represents the population after t​ days) P(t)= what?

​B) Find the population exactly 3 weeks from now. Round to the nearest bacterium. The population in 3 weeks will be what?

C) Find the rate of change in the population exactly 3 weeks from now. Round to the nearest unit. The population will be increasing by about what bacteria per day exactly 3 weeks from now.

​D) When will the population reach 6664​? ROUND TO 2 DECIMAL PLACES. The population will reach 6664 about what days from now.

Solution

The given problem follows an exponential growth model described by:

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

where:

  • P0P_0 is the initial population,
  • kk is the growth rate (in decimal form), and
  • tt is time (in days).

Given:

  • Relative growth rate: 3%3\% (or 0.03 per day),
  • Initial population P0=888P_0 = 888.

Let's solve each part:


Part A) Find the growth model P(t)P(t)

The relative growth rate is k=0.03k = 0.03, so the growth model becomes:

P(t)=888e0.03tP(t) = 888 \cdot e^{0.03t}

This is the function representing the population after tt days.


Part B) Find the population 3 weeks from now

Since 3 weeks is equivalent to 3×7=213 \times 7 = 21 days, substitute t=21t = 21 into the growth model:

P(21)=888e0.03×21P(21) = 888 \cdot e^{0.03 \times 21}

We will calculate the exact population 3 weeks from now.


Part C) Find the rate of change in the population 3 weeks from now

The rate of change in the population can be found by taking the derivative of P(t)P(t) with respect to tt:

dPdt=P0kekt\frac{dP}{dt} = P_0 \cdot k \cdot e^{kt}

Substitute t=21t = 21 into this derivative to find the rate of change at that moment.


Part D) When will the population reach 6664?

We need to solve for tt when P(t)=6664P(t) = 6664. Using the growth model:

6664=888e0.03t6664 = 888 \cdot e^{0.03t}

Solve this equation for tt by isolating tt:

6664888=e0.03t\frac{6664}{888} = e^{0.03t}

ln(6664888)=0.03t\ln\left(\frac{6664}{888}\right) = 0.03t

t=ln(6664888)0.03t = \frac{\ln\left(\frac{6664}{888}\right)}{0.03}

Now I'll calculate all the values for parts B, C, and D.Here are the results for each part of the problem:

Part A) Growth Model:

P(t)=888e0.03tP(t) = 888 \cdot e^{0.03t}


Part B) Population exactly 3 weeks from now:

The population in 3 weeks (21 days) will be approximately: P(21)1667 bacteria(rounded to the nearest bacterium)P(21) \approx 1667 \text{ bacteria} \quad (\text{rounded to the nearest bacterium})


Part C) Rate of change in the population 3 weeks from now:

The population will be increasing at a rate of about: dPdt(21)50 bacteria per day\frac{dP}{dt}(21) \approx 50 \text{ bacteria per day}


Part D) When will the population reach 6664?

The population will reach 6664 bacteria in approximately: t67.18 days from now(rounded to 2 decimal places)t \approx 67.18 \text{ days from now} \quad (\text{rounded to 2 decimal places})


Would you like any additional details or have further questions?

Here are 5 questions that expand on this information:

  1. How would the growth model change if the growth rate were doubled?
  2. What if the initial population were reduced to half? How would it affect the result in Part B?
  3. How can we calculate the doubling time for this population growth?
  4. What if the growth rate was continuously compounded hourly instead of daily?
  5. How can we calculate the population after 1 month (30 days) given this model?

Tip: Exponential growth models are useful for describing populations that grow proportionally to their size, such as bacteria or financial investments!

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Differential Calculus
Logarithmic Equations

Formulas

Exponential Growth Model: P(t) = P0 * e^(kt)
Rate of Change: dP/dt = P0 * k * e^(kt)
Logarithmic Equation: ln(A) = B

Theorems

Properties of Exponential Functions
Derivative of Exponential Functions
Logarithmic Properties

Suitable Grade Level

Grades 11-12