Chapter 3.8, Problem 26E

The table shows how the average age of first marriage of Japanese women has varied since 1950.

(a) Use a graphing calculator or computer to model these data with a fourth-degree polynomial.

(b) Use part (a) to find a model for A′(t).

(c) Estimate the rate of change of marriage age for women in 1990.

(d) Graph the data points and the models for A and A′.

To determine

To find: To model the given data with a fourth-degree polynomial.

Answer to Problem 26E

The fourth-degree polynomial is $y=-1.2\times {10}^{-6}{t}^4+9545.853t^3-28.48t^2+37757.11t-1.88\times {10}^7$.

Explanation of Solution

Given:

The average age of first marriage of Japanese women has varied since 1950 is shown in table (1).

$t$	$A\left(t\right)$
1950	23.00
1955	23.80
1960	24.40
1965	24.50
1970	24.20
1975	24.70
1980	25.20
1985	25.50
1990	25.90
1995	26.30
2000	27.00
2005	28.00

Calculation:

Model the given data with a fourth-degree polynomial.

$A\left(t\right)=a{t}^4+b{t}^3+c{t}^2+dt+e$ (1)

Plot the graph between $t$ and $A\left(t\right)$ using table (1) as shown in figure (1).

Refer the figure (1).

The fourth-degree polynomial for the curve is as below.

$A\left(t\right)=-1.2\times {10}^{-6}{t}^4+0.0095t^3-28.48t^2+37757.11t-1.88\times {10}^7$ (2)

Equate the equation (1) with equation (2).

$a{t}^4+b{t}^3+c{t}^2+dt+e=-1.2\times {10}^{-6}{t}^4+0.0095t^3-28.48t^2+37757.11t-1.88\times {10}^7$

Compare the above equation.

$a=-1.2\times {10}^{-6}$, $b=0.0095$, $c=-28.48$, $d=37757.11$ and $e=-1.88\times {10}^7$.

To determine

To find: The model for $A\text{'}\left(t\right)$ by using part (a).

Answer to Problem 26E

The value of $A\text{'}\left(t\right)$ is $-0.000005t^3+28637.55t^2-56.96t+37757.11$.

Explanation of Solution

Calculate the value of $A\text{'}\left(t\right)$.

Differentiate equation (1) with respect to $t$.

$A\text{'}\left(t\right)=4a{t}^3+3b{t}^2+2ct+d$ (3)

Substitute $-1.2\times {10}^{-6}$ for the expression $a$, $9545.853$ for the expression $b$, $-28.48$ for the expression $c$ and $37757.11$ for the expression $d$.

$A\text{'}\left(t\right)=4\left(-1.2\times {10}^{-6}\right)t^3+3\left(0.0095\right)t^2+2\left(-28.48\right)t+37757.11$

$A\text{'}\left(t\right)=-0.000005t^3+0.0285t^2-56.96t+37757.11$ (4)

Thus, the value of $A\text{'}\left(t\right)$ is $-0.000005t^3+0.0285t^2-56.96t+37757.11$.

To determine

To estimate: The rate of change of marriage age for women in 1990.

Answer to Problem 26E

The rate of change of marriage age for women in 1990 is $-2,133.44$.

Explanation of Solution

Compute the rate of change of marriage age for women in 1990 using the formula below.

$A^{\text{'}}\left(t\right)=-0.000005t^3+0.0285t^2-56.96t+37757.11$

Substitute 1990 for t.

$\begin{array}{c}{A}^{\text{'}}\left(t\right)=-0.000005\times 1,{990}^3+0.0285\times 1,{990}^2-56.96\times 1,990+37,757.11\\ =-2,133.44\end{array}$

To determine

To sketch: The graph of data points and models for A and $A\text{'}$.

Explanation of Solution

Plot the graph between $t$ and $A\left(t\right)$ as shown in Figure (2).

Calculate the difference between the year 1950 and 1955.

$\begin{array}{c}Difference=1950-1945\\ =5\end{array}$

Substitute the value 5 for the expression $t$ in the equation (2).

$\begin{array}{c}A\text{'}\left(t\right)=-0.000005{\left(5\right)}^3+28637.55{\left(5\right)}^2-56.96\left(5\right)+37757.11\\ =0.114614\end{array}$

Similarly, calculate the value of $A\text{'}\left(t\right)$ for the following values of $t$ in table (1).

Tabulate the value of $A\text{'}\left(t\right)$ as shown in table (2).

Difference	$A\text{'}\left(t\right)$
0	0.221834
5.0	0.114614
10	0.052734
15.0	0.027194
20	0.028994
25.0	0.049134
30	0.078614
35.0	0.108434
40	0.129594
45.0	0.133094
50	0.109934
55.0	0.051114
60	-0.052366

Plot the graph between $t$ and $A\text{'}\left(t\right)$ using the table (2) as shown in Figure (3).