Question: A bacteria culture grows with constant relative growth rate. The bacteria count was $2,880$ after $4$ hours and $2,949,120$ after $12$ hours. (a) What is the relative growth rate? Express your answer as a percentage. (Round your answer to two decimal places.) (b) What was the initial size of the culture? (c) Find an expression for the exact number of bacteria after $t$ hours, $y(t)$. (d) Find the number of bacteria after $7.5$ hours. (Round your answer to the nearest whole number.) (e) Find the rate of growth (in bacteria per hour) after $7.5$ hours. (Round your answer to the nearest whole number.) (f) How many hours did it take for the population to reach $117,000$? (Round your answer to two decimal places.) The table gives estimates of the world population, in millions, from $1750$ to $2000$. (Round your answers to the nearest million.) Year Population (millions) $1750$ $790$ $1800$ $980$ $1850$ $1260$ $1900$ $1650$ $1950$ $2560$ $2000$ $6080$ (a) Use the exponential model and the population figures for $1750$ and $1800$ to predict the world population (in millions of people) in $1900$ and $1950$. (Compare with the actual figures.) (b) Use the exponential model and the population figures for $1800$ and $1850$ to predict the world population (in millions of people) in $1950$. (Compare with the actual population.) (c) Use the exponential model and the population figures for $1900$ and $1950$ to predict the world population (in millions of people) in $2000$. (Compare with the actual population.)

1. **Problem Statement:** We have a bacteria culture growing with a constant relative growth rate. Given bacteria counts at two times, find the growth rate, initial size, expression for population at time $t$, population at $7.5$ hours, growth rate at $7.5$ hours, and time to reach $117,000$ bacteria. Also, use exponential models to predict world population at given years based on historical data. --- ### Part 1: Bacteria Growth 2. **Formula and Rules:** The bacteria growth follows the exponential model: $$y(t) = y_0 e^{rt}$$ where: - $y(t)$ is the population at time $t$ - $y_0$ is the initial population - $r$ is the relative growth rate (per hour) - $t$ is time in hours Given two points $(t_1, y_1)$ and $(t_2, y_2)$, we can find $r$ by: $$y_2 = y_1 e^{r(t_2 - t_1)} \implies r = \frac{1}{t_2 - t_1} \ln\left(\frac{y_2}{y_1}\right)$$ 3. **Given Data:** - $y(4) = 2880$ - $y(12) = 2,949,120$ 4. **Find $r$:** $$r = \frac{1}{12 - 4} \ln\left(\frac{2,949,120}{2880}\right) = \frac{1}{8} \ln(1024)$$ Since $1024 = 2^{10}$, $$r = \frac{1}{8} \times 10 \ln(2) = \frac{10}{8} \ln(2) = 1.25 \ln(2)$$ Using $\ln(2) \approx 0.693147$, $$r \approx 1.25 \times 0.693147 = 0.866434$$ Expressed as a percentage: $$0.866434 \times 100 = 86.64\% \text{ per hour}$$ 5. **Find initial size $y_0$:** Using $y(4) = y_0 e^{4r}$, $$y_0 = \frac{y(4)}{e^{4r}} = 2880 \times e^{-4r}$$ Calculate $e^{-4r}$: $$e^{-4 \times 0.866434} = e^{-3.465736} \approx 0.03125$$ So, $$y_0 = 2880 \times 0.03125 = 90$$ 6. **Expression for $y(t)$:** $$y(t) = 90 e^{0.866434 t}$$ 7. **Number of bacteria after $7.5$ hours:** $$y(7.5) = 90 e^{0.866434 \times 7.5} = 90 e^{6.498255}$$ Calculate $e^{6.498255} \approx 664.5$, $$y(7.5) \approx 90 \times 664.5 = 59,805$$ Rounded to nearest whole number: $59,805$ 8. **Rate of growth at $7.5$ hours:** The rate of growth is the derivative: $$\frac{dy}{dt} = r y(t) = 0.866434 \times 59,805 \approx 51,800$$ Rounded: $51,800$ bacteria per hour 9. **Time to reach $117,000$ bacteria:** Solve for $t$: $$117,000 = 90 e^{0.866434 t} \implies e^{0.866434 t} = \frac{117,000}{90} = 1300$$ Take natural log: $$0.866434 t = \ln(1300) \approx 7.17012$$ $$t = \frac{7.17012}{0.866434} \approx 8.27 \text{ hours}$$ --- ### Part 2: World Population Predictions 10. **Exponential model:** $$P(t) = P_0 e^{kt}$$ where $P(t)$ is population at time $t$, $P_0$ initial population, $k$ growth rate, and $t$ years since a base year. --- 11. **(a) Using $1750$ and $1800$ data:** - $P(0) = 790$ (year 1750) - $P(50) = 980$ (year 1800) Find $k$: $$980 = 790 e^{50k} \implies e^{50k} = \frac{980}{790} = 1.240506$$ $$50k = \ln(1.240506) = 0.215111 \implies k = 0.004302$$ Predict for 1900 ($t=150$): $$P(150) = 790 e^{0.004302 \times 150} = 790 e^{0.6453}$$ $$e^{0.6453} \approx 1.906$$ $$P(150) \approx 790 \times 1.906 = 1505 \, \text{million}$$ Actual: $1650$ million Predict for 1950 ($t=200$): $$P(200) = 790 e^{0.004302 \times 200} = 790 e^{0.8604}$$ $$e^{0.8604} \approx 2.364$$ $$P(200) \approx 790 \times 2.364 = 1867 \, \text{million}$$ Actual: $2560$ million 12. **(b) Using $1800$ and $1850$ data:** - $P(0) = 980$ (year 1800) - $P(50) = 1260$ (year 1850) Find $k$: $$1260 = 980 e^{50k} \implies e^{50k} = \frac{1260}{980} = 1.2857$$ $$50k = \ln(1.2857) = 0.2513 \implies k = 0.005026$$ Predict for 1950 ($t=150$): $$P(150) = 980 e^{0.005026 \times 150} = 980 e^{0.7539}$$ $$e^{0.7539} \approx 2.125$$ $$P(150) \approx 980 \times 2.125 = 2082 \, \text{million}$$ Actual: $2560$ million 13. **(c) Using $1900$ and $1950$ data:** - $P(0) = 1650$ (year 1900) - $P(50) = 2560$ (year 1950) Find $k$: $$2560 = 1650 e^{50k} \implies e^{50k} = \frac{2560}{1650} = 1.5515$$ $$50k = \ln(1.5515) = 0.4383 \implies k = 0.008766$$ Predict for 2000 ($t=100$): $$P(100) = 1650 e^{0.008766 \times 100} = 1650 e^{0.8766}$$ $$e^{0.8766} \approx 2.403$$ $$P(100) \approx 1650 \times 2.403 = 3965 \, \text{million}$$ Actual: $6080$ million --- **Final answers:** - (a) Relative growth rate: $86.64\%$ per hour - (b) Initial size: $90$ bacteria - (c) Expression: $y(t) = 90 e^{0.866434 t}$ - (d) Bacteria after $7.5$ hours: $59,805$ - (e) Growth rate at $7.5$ hours: $51,800$ bacteria/hr - (f) Time to reach $117,000$: $8.27$ hours - World population predictions: - (a) 1900: $1,505$ million, 1950: $1,867$ million - (b) 1950: $2,082$ million - (c) 2000: $3,965$ million