Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 23

The table below gives the population of the world in the 20th century.

$
\begin{array}{|c|c|c|c|}
\hline\\
\text{Year} & \text{Populations} && \text{Year} & \text{Populations}\\
& \text{(in millions)} && & \text{(in millions)}\\
\hline\\
1900 &1650 &&1960 &3040\\
1910 &1750 &&1970 &3710\\
1920 &1860 &&1980 &4450\\
1930 &2070 &&1990 &5280\\
1940 &2300 &&2000 &6080\\
1950 &2560 &&\\
\hline
\end{array}
$


For the year 1920,

$
\begin{equation}
\begin{aligned}
m_1 &= \frac{P(1920) - P(1910)}{1920 - 1910} = \frac{1860-1750}{1920-1910} = 11\\
\\
m_2 &= \frac{P(1930) - P(1920)}{1930-1920} = \frac{2070-1860}{1930-1920} = 21
\end{aligned}
\end{equation}
$

The rate of population growth in 1920 is $\displaystyle \frac{m_1+m_2}{2} = \frac{11+21}{2} = 16 \frac{\text{million}}{\text{years}}$
For the year 1980,

$
\begin{equation}
\begin{aligned}
m_1 &= \frac{P(1980) - P(1970)}{1980 - 1970} = \frac{4450-3710}{1980-1970} = 74\\
\\
m_2 &= \frac{P(1990) - P(1980)}{1990-1980} = \frac{5280-4450}{1990-1980} = 83
\end{aligned}
\end{equation}
$

The rate of population growth in 1980 is $\displaystyle \frac{m_1+m_2}{2} = \frac{74+83}{2} = 78.5 \frac{\text{million}}{\text{years}}$
b.)


Based from the graph, the model for the cubic function is
$P(x) = 0.0013x^3 + 7.0614x^2 + 12823 x - 8\times 10^6$


$
\begin{equation}
\begin{aligned}
\text{c.) } P'(x) &= 3(0.0013)x^2 - 2(7.0614) x + 12823(1)\\
\\
P'(x) &= 0.0039x^2 - 14.1228 x + 12823
\end{aligned}
\end{equation}
$

when $x= 2000$,

$
\begin{equation}
\begin{aligned}
P'(2000) &= 0.0039(2000)^2 - 14.1228(2000) + 12823\\
\\
P'(2000) &= 177.4 \frac{\text{million}}{\text{year}}
\end{aligned}
\end{equation}
$


d.) when $x = 1920$

$
\begin{equation}
\begin{aligned}
P'(1920) &= 0.0039(1920)^2 - 14.1228(1920) + 12823\\
\\
P'(2000) &= 84.184 \frac{\text{million}}{\text{year}}
\end{aligned}
\end{equation}
$


when $x = 1980$,

$
\begin{equation}
\begin{aligned}
P'(2000) &= 0.0039(1980)^2-14.1228(1980) + 12823\\
\\
P'(2000) &= 149.416 \frac{\text{million}}{\text{year}}
\end{aligned}
\end{equation}
$

Our estimated values in part(a) is very much different with our answer in part(d).