Question

The rabbit population at the city park increases by 5% per year. If there are intially 413 rabbits in the city park.

a) Write a model for the population (y) in terms of years (t).



b) Find the rabbit population in 15 years. (Round to the nearest whole rabbit)


c) [Estimate] When the rabbit population reaches 1921. It will happen between year
and year.

Answer 1

Answer:

a) $P(t) = 413e^{1.05t}$

b) $P(15) \approx 2.86 \cdot 10^9$

c) year 1 and year 2

Step-by-step explanation:

a) We can model this situation using the relative growth model:

$P(t) = Ae^{kt}$

where $A$ is the initial population and $k$ is the relative growth rate.

Note that $e$ 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.

• $e\approx 2.71828$

We are given the initial population and growth rate to be:

• $A=413$
• $k = (100 + 5)\% = 1.05$

Plugging these values into the above equation, we get:

$\boxed{P(t) = 413e^{1.05t}}$

b) To find the rabbit population in 15 years, we can plug in $t = 15$:

$P(15) = 413e^{1.05(15)}$

$P(15) = 413e^{15.75}$

$\boxed{P(15) \approx 2.86 \cdot 10^9}$

c) We can plug in $P(t) = 1921$ and solve for t:

$1921 = 413e^{1.05t}$

↓ dividing both sides by 413

$\dfrac{1921}{413} = e^{1.05t}$

↓ taking the natural log of both sides

$\ln\!\left(\dfrac{1921}{413}\right) = 1.05t$

↓ dividing both sides by 1.05

$\dfrac{\ln\!\left(\dfrac{1921}{413}\right)}{1.05} = t$

↓ evaluating using a calculator

$t \approx 1.46$

So, the population will reach 1921 between year 1 and year 2.