#### Answer

(a) The population density in 1800 was $6.1~people/mi^2$
The population density in 2010 was $87.4~people/mi^2$
(b) The percent increase in population density from 1800 to 2010 was $1332.8\%$

#### Work Step by Step

(a) Let $P_1$ be the population in 1800.
Let $A_1$ be the land area in 1800.
We can find the population density in 1800.
$\frac{P_1}{A_1} = \frac{5,308,483~people}{864,746~mi^2} = 6.1~people/mi^2$
The population density in 1800 was $6.1~people/mi^2$
Let $P_2$ be the population in 2010.
Let $A_2$ be the land area in 2010.
We can find the population density in 2010.
$\frac{P_2}{A_2} = \frac{308,745,538~people}{3,531,905~mi^2} = 87.4~people/mi^2$
The population density in 2010 was $87.4~people/mi^2$
(b) We can find the percent increase in population density from 1800 to 2010.
$\frac{87.4~people/mi^2-6.1~people/mi^2}{6.1~people/mi^2}\times 100\% = 1332.8\%$
The percent increase in population density from 1800 to 2010 was $1332.8\%$