Answer:
a) [tex]P(t) = 413e^{1.05t}[/tex]
b) [tex]P(15) \approx 2.86 \cdot 10^9[/tex]
c) year 1 and year 2
Step-by-step explanation:
a) We can model this situation using the relative growth model:
[tex]P(t) = Ae^{kt}[/tex]
where [tex]A[/tex] is the initial population and [tex]k[/tex] is the relative growth rate.
Note that [tex]e[/tex] is not a variable. It is a constant (Euler's number) that represents the limit for relative growth. An example of this would be continually compounding interest.
- [tex]e\approx 2.71828[/tex]
We are given the initial population and growth rate to be:
- [tex]A=413[/tex]
- [tex]k = (100 + 5)\% = 1.05[/tex]
Plugging these values into the above equation, we get:
[tex]\boxed{P(t) = 413e^{1.05t}}[/tex]
b) To find the rabbit population in 15 years, we can plug in [tex]t = 15[/tex]:
[tex]P(15) = 413e^{1.05(15)}[/tex]
[tex]P(15) = 413e^{15.75}[/tex]
[tex]\boxed{P(15) \approx 2.86 \cdot 10^9}[/tex]
c) We can plug in [tex]P(t) = 1921[/tex] and solve for t:
[tex]1921 = 413e^{1.05t}[/tex]
↓ dividing both sides by 413
[tex]\dfrac{1921}{413} = e^{1.05t}[/tex]
↓ taking the natural log of both sides
[tex]\ln\!\left(\dfrac{1921}{413}\right) = 1.05t[/tex]
↓ dividing both sides by 1.05
[tex]\dfrac{\ln\!\left(\dfrac{1921}{413}\right)}{1.05} = t[/tex]
↓ evaluating using a calculator
[tex]t \approx 1.46[/tex]
So, the population will reach 1921 between year 1 and year 2.
