Single Variable Calculus Questions and Answers

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

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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).