Question: Math-014 Homework Assignment #5 Growth and Decay Exponential Growth The exponential growth…

Math-014 Homework Assignment #5 Growth and Decay

Exponential Growth

The exponential growth formula is P(t) = P_0e^{kt}, where P_0 is the initial amount, k is the growth rate (as a decimal) and t is the time.

Fill in the growth formula for the problems below:

1. The population of Haiti has a growth rate of 1.08% per year. In 2015, the population was approximately 9,996,000. Fill in the growth rate formula for this problem.

P(t) = _________________

Doubling and Half-Life

The formula ln 2 = kT describes the relationship between the rate of growth or decay (k) and the time passed (T) for both doubling and half-rate problems.

2. Use the formula to fill in the doubling time table below:

Country	Growth rate per year, k	Doubling time, T
A	1.9%
B		98 years

3. Use the formula to fill in the half-life table below:

Substance	Decay rate, k	Half-life, T
A	0.012% per year
B		1380 years

Solution

Exponential Growth For the population of Haiti, we need to fill in the growth formula using the given information: The growth rate \( k \) is 1.08% per year, which we convert to a decimal: \( k = 0.0108 \). The initial amount \( P_0 \) is 9,996,000. Using the formula \( P(t) = P_0 e^{kt} \), we have: \[ P(t) = 9,996,000 \cdot e^{0.0108t} \] Doubling and Half-Life 1. Use the formula \( \ln 2 = kT \) to fill in the doubling time table: For Country A: The growth rate \( k \) is 1.9% per year, so \( k = 0.019 \). We find \( T \) (doubling time) using \( \ln 2 = 0.019 \cdot T \). \[ T = \frac{\ln 2}{0.019} \] Calculating \( T \): \[ T \approx 36.46 \text{ years} \] For Country B, given that \( T = 98 \) years: We find \( k \) using \( \ln 2 = k \cdot 98 \). \[ k = \frac{\ln 2}{98} \] Calculating \( k \): \[ k \approx 0.00708 \text{ or } 0.708\% \] 2. Use the formula \( \ln 2 = kT \) to fill in the half-life table: For Substance A: The decay rate \( k \) is 0.012% per year, so \( k = 0.00012 \). To find \( T \) (half-life) using \( \ln 2 = 0.00012 \cdot T \). \[ T = \frac{\ln 2}{0.00012} \] Calculating \( T \): \[ T \approx 5777.23 \text{ years} \] For Substance B, given that \( T = 1380 \) years: We find \( k \) using \( \ln 2 = k \cdot 1380 \). \[ k = \frac{\ln 2}{1380} \] Calculating \( k \): \[ k \approx 0.0005024 \text{ or } 0.05024\% \text{ per year} \]