The population of the world was about $5.3$ billion in 1990 . Birth rates in the 1990 s range from 35 to 40 million per year and death rates range from 15 to 20 million per year. Let's assume that the carrying capacity for world population is 100 billion. Use the logistic model to predict the world population in the 2,450 year. Calculate your answer in billions to one decimal place. (Because the initial population is small compared to the carrying capacity, you can take $k$ to be an estimate of the initial relative growth rate.)
A) $78.3$ billion
B) $27.0$ billion
C) $17.1$ billion
D) $59.2$ billion
E) $32.9$ billion

Question

Quizplus · Accepted Answer

Let's first find the value of $k$:
$$\frac{dP}{dt}=kP\left(1-\frac{P}{100}\right)$$
When $P=5.3\text{ billion}$, birth rate $B$ is about $35$ million per year and death rate $D$ is about $15$ million per year. So, we have:
$$B-D=(35-15)\text{ million}=20\text{ million}$$
Therefore, we can estimate the initial relative growth rate as:
$$k=\frac{B-D}{P}= \frac{20\text{ million}}{5.3\text{ billion}}\approx 0.00377$$
Now, we can use the logistic model to predict the world population in the year 2450:
\begin{align*}
\frac{dP}{dt}&=0.00377 P\left(1-\frac{P}{100}\right) \
\frac{dP}{P(1-P/100)}&=0.00377 dt \
\int \frac{dP}{P(1-P/100)}&=\int 0.00377 dt \
\ln\left|\frac{P}{100-P}\right|&=0.00377 t+C\
\frac{P}{100-P}&=e^{0.00377 t+C}=e^{0.00377 t}k_1 &(k_1>0)\
P&=100\frac{k_1}{1+k_1e^{-0.00377 t}}
\end{align*}
Using the initial condition $P(0)=5.3$, we get $k_1=0.0526$. Thus, the world population in 2450 will be:
$$P(2450)=100\frac{0.0526}{1+0.0526e^{-0.00377\times 560}} \approx 27.0\text{ billion}$$
So, the best choice is B.

The Population of the World Was About 5.35.35.3 Billion in 1990 kkk

The Population of the World Was About $5.3$ Billion in 1990 $k$