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The population of a southem city follows the exponential law. Use this information to answer parts a and b . (a) If N is the population of the city and t is the time in years, express N as a function of t . \( \mathrm{N}(\mathrm{t})=\mathrm{N}_{0 \mathrm{e}} \frac{12 \ln (2)}{13} \mathrm{t} \) (Type an expression using t as the variable and in terms of \( e \).) (b) If the population doubled in size over 13 months and the current population is 40,000 , what will the population be 5 years from now? The population will be approximately 320,000 people. (Do not round until the final answer. Then round to the nearest whole number as needed.)

Ask by Goodwin Erickson. in the United States
Jan 22,2025

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The population 5 years from now will be approximately 980,000 people.

Solución

Let's solve each part of the problem step-by-step. ## Part (a) **Expressing Population \( N(t) \) as a Function of Time \( t \):** Given that the population follows an exponential growth model, the general form of the population function is: \[ \mathrm{N}(\mathrm{t}) = \mathrm{N}_0 \times e^{kt} \] Where: - \( \mathrm{N}_0 \) is the initial population. - \( k \) is the growth rate. - \( t \) is the time in years. **Determining the Growth Rate \( k \):** We know that the population doubles in 13 months. To find \( k \), follow these steps: 1. **Convert 13 Months to Years:** \[ 13 \text{ months} = \frac{13}{12} \text{ years} \] 2. **Set Up the Doubling Equation:** \[ 2\mathrm{N}_0 = \mathrm{N}_0 \times e^{k \times \frac{13}{12}} \] 3. **Solve for \( k \):** \[ 2 = e^{k \times \frac{13}{12}} \\ \ln(2) = k \times \frac{13}{12} \\ k = \frac{12 \ln(2)}{13} \] **Final Expression for \( N(t) \):** \[ \mathrm{N}(\mathrm{t}) = \mathrm{N}_0 \times e^{\left(\frac{12 \ln(2)}{13}\right) t} \] This matches the expression you provided: \[ \mathrm{N}(\mathrm{t}) = \mathrm{N}_0 \, e^{\frac{12 \ln(2)}{13} \, t} \] ## Part (b) **Calculating the Population 5 Years from Now:** Given: - **Current Population (\( \mathrm{N}_0 \)):** 40,000 - **Time (\( t \)):** 5 years Using the population model from part (a): \[ \mathrm{N}(5) = 40{,}000 \times e^{\left(\frac{12 \ln(2)}{13}\right) \times 5} \] **Simplifying the Exponent:** \[ \frac{12 \ln(2)}{13} \times 5 = \frac{60 \ln(2)}{13} \approx \frac{60 \times 0.6931}{13} \approx 3.198 \] \[ \mathrm{N}(5) \approx 40{,}000 \times e^{3.198} \approx 40{,}000 \times 24.5 \approx 980{,}000 \] **Final Answer:** The population 5 years from now will be approximately **980,000** people. --- **Note:** It appears there was a miscalculation in your initial estimate of 320,000 people. The correct population, based on the exponential growth model and the given doubling time, is approximately **980,000**.

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Beyond the Answer

When we talk about populations following exponential growth, we often see the classic formula \( N(t) = N_0 e^{kt} \), where \( N_0 \) is the initial population and \( k \) is the growth constant. In this city’s case, since the population doubles in 13 months (which is approximately 1.08 years), we can calculate the growth constant \( k \) using the formula \( k = \frac{\ln(2)}{t_{double}} \). Now for any community experiencing exponential growth, a critical tip is to remember that time frames must be consistent. In the example, you converted months to years for calculations; however, in other scenarios, confusion could arise if you switch between units. Always take a moment to double-check that the time you’re using aligns with the growth constant you're working with!

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