The population of a colony of mosquitoes obeys the law of uninhibited growth: Use this information to answer parts (a) through (c): (a) If N is the population of the colony and t is the time in days, express N as a function of t. Consider No is the original amount at t = 0 and k # 0 is a constant that represents the growth rate. N(t) = (Type an expression using t as the variable and in terms of e.) (b) The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there are 1500 after day; what is the size of the colony after 4 days? Approximately mosquitoes (Do not round until the final answer: Then round to the nearest whole number as needed ) (c) How long is it until there are 40,000 mosquitoes? About days (Do not round until the final answer: Then round to the nearest tenth as needed )
The Correct Answer and Explanation is:
Part (a): Express N(t)N(t)N(t) as a function of ttt
The law of uninhibited growth for a population N(t)N(t)N(t) is modeled by the exponential growth equation:N(t)=N0ektN(t) = N_0 e^{kt}N(t)=N0ekt
Where:
- N(t)N(t)N(t) is the population at time ttt,
- N0N_0N0 is the initial population at t=0t = 0t=0,
- kkk is the growth rate constant,
- eee is the base of the natural logarithm.
This equation describes how the population grows over time if there are no restrictions on growth.
Part (b): Population after 4 days
We are given:
- Initial population N0=1000N_0 = 1000N0=1000,
- After 1 day, the population is 1500.
We need to find the value of kkk and use that to determine the population after 4 days.
- Step 1: Find the growth rate constant kkk:
Using the information that the population is 1500 after 1 day, we substitute into the formula:1500=1000ek⋅11500 = 1000 e^{k \cdot 1}1500=1000ek⋅1
Solving for kkk:15001000=ek\frac{1500}{1000} = e^k10001500=ek1.5=ek1.5 = e^k1.5=ek
Taking the natural logarithm of both sides:ln(1.5)=k\ln(1.5) = kln(1.5)=kk≈0.4055k \approx 0.4055k≈0.4055
- Step 2: Find the population after 4 days:
Now that we know k≈0.4055k \approx 0.4055k≈0.4055, we can use the formula to find the population after 4 days:N(4)=1000e0.4055⋅4N(4) = 1000 e^{0.4055 \cdot 4}N(4)=1000e0.4055⋅4N(4)=1000e1.6222N(4) = 1000 e^{1.6222}N(4)=1000e1.6222N(4)≈1000×5.080N(4) \approx 1000 \times 5.080N(4)≈1000×5.080N(4)≈5080 mosquitoesN(4) \approx 5080 \text{ mosquitoes}N(4)≈5080 mosquitoes
Part (c): Time until the population reaches 40,000 mosquitoes
We are asked to find the time ttt when the population reaches 40,000. Using the same formula:40,000=1000e0.4055⋅t40,000 = 1000 e^{0.4055 \cdot t}40,000=1000e0.4055⋅t
- Step 1: Solve for ttt:
40,0001000=e0.4055⋅t\frac{40,000}{1000} = e^{0.4055 \cdot t}100040,000=e0.4055⋅t40=e0.4055⋅t40 = e^{0.4055 \cdot t}40=e0.4055⋅t
Taking the natural logarithm of both sides:ln(40)=0.4055⋅t\ln(40) = 0.4055 \cdot tln(40)=0.4055⋅tt=ln(40)0.4055t = \frac{\ln(40)}{0.4055}t=0.4055ln(40)t≈3.68890.4055t \approx \frac{3.6889}{0.4055}t≈0.40553.6889t≈9.1 dayst \approx 9.1 \text{ days}t≈9.1 days
Final Answers:
- (a) The population function is N(t)=N0ektN(t) = N_0 e^{kt}N(t)=N0ekt.
- (b) The population after 4 days is approximately 5080 mosquitoes.
- (c) It will take approximately 9.1 days for the population to reach 40,000 mosquitoes.
