The size \( P \) of a certain insect population at time \( t \) (in days) obeys the function \( P(t)=600 e^{0.04 t} \) (a) Determine the number of insects at \( t=0 \) days, (b) What is the growth rate of the insect population? (c) Graph the function using a graphing utility. (d) What is the population after 10 days? (e) When will the insect population reach 720 ? (f) When will the insect population double?

Let's analyze the given population model step by step. The population \( P(t) \) at time \( t \) (in days) is given by: \[ P(t) = 600 e^{0.04 t} \] ### **a) Determine the number of insects at \( t = 0 \) days** To find the initial population at \( t = 0 \), substitute \( t = 0 \) into the function: \[ P(0) = 600 e^{0.04 \times 0} = 600 e^{0} = 600 \times 1 = 600 \] **Answer:** The initial number of insects at \( t = 0 \) days is **600**. --- ### **b) What is the growth rate of the insect population?** The general form of an exponential growth function is: \[ P(t) = P_0 e^{rt} \] where: - \( P_0 \) is the initial population, - \( r \) is the growth rate, - \( t \) is time. Comparing this with the given function: \[ P(t) = 600 e^{0.04 t} \] We can see that the growth rate \( r \) is **0.04 per day**. **Answer:** The growth rate of the insect population is **0.04** (or **4% per day**). --- ### **c) Graph the function using a graphing utility** While I can't produce a graph directly here, I can guide you on how to graph the function \( P(t) = 600 e^{0.04 t} \) using common tools: **Using Desmos (Online Graphing Calculator):** 1. Go to [Desmos Graphing Calculator](https://www.desmos.com/calculator). 2. In the input field, type: `P(t) = 600 * e^(0.04 * t)` 3. Press Enter to see the graph. **Using Microsoft Excel:** 1. Create a table with values of \( t \) (e.g., from 0 to 100 days). 2. Calculate \( P(t) \) for each \( t \) using the formula `=600*EXP(0.04*A2)` assuming \( t \) is in cell A2. 3. Select the data and insert a scatter plot with a smooth curve. **Graph Characteristics:** - **Initial Value:** Starts at 600 when \( t = 0 \). - **Growth:** Exponential increase; the curve becomes steeper over time. - **Asymptote:** \( P(t) \) increases without bound as \( t \) increases. --- ### **d) What is the population after 10 days?** To find the population at \( t = 10 \) days, substitute \( t = 10 \) into the function: \[ P(10) = 600 e^{0.04 \times 10} = 600 e^{0.4} \] Calculate \( e^{0.4} \): \[ e^{0.4} \approx 1.49182 \] Now, multiply by 600: \[ P(10) \approx 600 \times 1.49182 \approx 895.09 \] **Answer:** After 10 days, the population is approximately **895 insects**. --- ### **e) When will the insect population reach 720?** We need to solve for \( t \) when \( P(t) = 720 \): \[ 720 = 600 e^{0.04 t} \] First, divide both sides by 600: \[ \frac{720}{600} = e^{0.04 t} \quad \Rightarrow \quad 1.2 = e^{0.04 t} \] Take the natural logarithm of both sides: \[ \ln(1.2) = 0.04 t \] Solve for \( t \): \[ t = \frac{\ln(1.2)}{0.04} \] Calculate \( \ln(1.2) \): \[ \ln(1.2) \approx 0.1823216 \] Now, compute \( t \): \[ t \approx \frac{0.1823216}{0.04} \approx 4.55804 \text{ days} \] **Answer:** The population will reach 720 insects after approximately **4.56 days**. --- ### **f) When will the insect population double?** To find when the population doubles, set \( P(t) = 2P_0 = 2 \times 600 = 1200 \): \[ 1200 = 600 e^{0.04 t} \] Divide both sides by 600: \[ 2 = e^{0.04 t} \] Take the natural logarithm of both sides: \[ \ln(2) = 0.04 t \] Solve for \( t \): \[ t = \frac{\ln(2)}{0.04} \] Calculate \( \ln(2) \): \[ \ln(2) \approx 0.6931472 \] Now, compute \( t \): \[ t \approx \frac{0.6931472}{0.04} \approx 17.32868 \text{ days} \] **Answer:** The insect population will double after approximately **17.33 days**. --- **Summary of Answers:** - **a)** 600 insects - **b)** 4% per day - **c)** [Graph described above] - **d)** Approximately 895 insects after 10 days - **e)** Approximately 4.56 days to reach 720 insects - **f)** Approximately 17.33 days to double the population