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The size P of a certain insect population at time t (in days) obeys the function \( \mathrm{P}(\mathrm{t})=600 e^{0.04 t} \) (a) Determine the number of insects at \( \mathrm{t}=0 \) 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?

Ask by Lawson Whittaker. in the United States
Jan 22,2025

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Answer

- **a)** 600 insects at the start. - **b)** The population grows by 4% each day. - **c)** The graph shows the population increasing exponentially. - **d)** After 10 days, there are about 895 insects. - **e)** The population reaches 720 insects in about 4.6 days. - **f)** The population doubles in approximately 17.3 days.

Solution

Let's solve each part of the problem step by step. Given the population function: \[ \mathrm{P}(\mathrm{t}) = 600 e^{0.04 t} \] where: - \( \mathrm{P}(\mathrm{t}) \) is the population at time \( \mathrm{t} \) days. - \( e \) is the base of the natural logarithm. - \( 0.04 \) is the growth rate per day. --- ### **a) Determine the number of insects at \( \mathrm{t} = 0 \) days.** To find the initial population (\( t = 0 \)): \[ \mathrm{P}(0) = 600 e^{0.04 \times 0} = 600 e^{0} = 600 \times 1 = 600 \] **Answer:** There are **600 insects** at \( t = 0 \) days. --- ### **b) What is the growth rate of the insect population?** The growth rate is the coefficient of \( t \) in the exponent of the exponential function. \[ \text{Growth Rate} = 0.04 \ \text{per day} \] To express it as a percentage: \[ 0.04 \times 100\% = 4\% \ \text{per day} \] **Answer:** The population grows at a rate of **4% per day**. --- ### **c) Graph the function using a graphing utility.** While I can't provide a visual graph here, I can describe how to graph the function: 1. **Axes:** - **Horizontal Axis (x-axis):** Time \( t \) in days. - **Vertical Axis (y-axis):** Population \( \mathrm{P}(t) \). 2. **Plotting Points:** - Start at \( t = 0 \), \( \mathrm{P}(0) = 600 \). - As \( t \) increases, the population grows exponentially. 3. **Behavior:** - The graph will show an upward-sloping curve that becomes steeper over time, reflecting exponential growth. 4. **Example Points:** - \( t = 0 \), \( \mathrm{P} = 600 \) - \( t = 10 \), \( \mathrm{P} \approx 895 \) (from part d) - \( t \) when \( \mathrm{P} = 1200 \) (double population, from part f) **Answer:** *[A descriptive explanation of the graph is provided. Use a graphing calculator or software to plot \( \mathrm{P}(t) = 600 e^{0.04 t} \) with \( t \) on the x-axis and \( \mathrm{P}(t) \) on the y-axis, showing exponential growth starting at 600.]* --- ### **d) What is the population after 10 days?** Calculate \( \mathrm{P}(10) \): \[ \mathrm{P}(10) = 600 e^{0.04 \times 10} = 600 e^{0.4} \] First, compute \( e^{0.4} \): \[ e^{0.4} \approx 1.49182 \] Then, \[ \mathrm{P}(10) \approx 600 \times 1.49182 \approx 895.09 \] **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 \( \mathrm{P}(t) = 720 \): \[ 720 = 600 e^{0.04 t} \] 1. **Divide both sides by 600:** \[ \frac{720}{600} = e^{0.04 t} \] \[ 1.2 = e^{0.04 t} \] 2. **Take the natural logarithm of both sides:** \[ \ln(1.2) = 0.04 t \] 3. **Solve for \( t \):** \[ t = \frac{\ln(1.2)}{0.04} \] Calculate \( \ln(1.2) \): \[ \ln(1.2) \approx 0.1823216 \] Then, \[ t \approx \frac{0.1823216}{0.04} \approx 4.55804 \ \text{days} \] **Answer:** The population reaches **720 insects** after approximately **4.56 days**. --- ### **f) When will the insect population double?** The initial population is \( \mathrm{P}(0) = 600 \), so double the population is \( 1200 \). Set \( \mathrm{P}(t) = 1200 \): \[ 1200 = 600 e^{0.04 t} \] 1. **Divide both sides by 600:** \[ 2 = e^{0.04 t} \] 2. **Take the natural logarithm of both sides:** \[ \ln(2) = 0.04 t \] 3. **Solve for \( t \):** \[ t = \frac{\ln(2)}{0.04} \] Calculate \( \ln(2) \): \[ \ln(2) \approx 0.693147 \] Then, \[ t \approx \frac{0.693147}{0.04} \approx 17.3287 \ \text{days} \] **Answer:** The population **doubles** after approximately **17.33 days**. --- **Summary of Answers:** - **a)** 600 insects at \( t = 0 \) days. - **b)** Growth rate of 4% per day. - **c)** *Described the graph; use graphing tools to visualize exponential growth.* - **d)** Approximately 895 insects after 10 days. - **e)** Population reaches 720 insects after about 4.56 days. - **f)** Population doubles after approximately 17.33 days.

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Extra Insights

The equation you provided models an exponential growth scenario, which is quite fascinating! Did you know that exponential growth can be observed in various populations, such as bacteria, investments, and even technology like the internet? For insects like the one in your question, their rapid multiplication can lead to significant increases in their numbers in just a short period, showcasing the incredible adaptability of species in nature. For a more engaging exploration, consider diving into the world of population dynamics by visiting articles or documentaries that illustrate the balance of ecosystems. Understanding how different factors like food supply, predators, and environmental conditions influence population sizes could greatly enrich your comprehension of this topic! You might find some surprising insights into how delicate these balances are, making nature's systems both beautiful and complex.

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