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.
Reviewed and approved by the UpStudy tutoring team
Explain
Simplify this solution