Quiz - Chapter 7 & 10
April 14, 2000 - 25 points

1. Exponential Growth (10 points)
According to my old geography book, Mexico had a population of 97.8 million people in 1997, with an annual growth rate of 2.2% per year.
a) Using the formula for exponential growth (the one similar to the compound interest formula), estimate the population of Mexico in 2010 and 2030.

For 2010, t=13 years, r=.022, and $Q_{0}$=97.8 million

$\nonumber Q = Q_{0}(1+r)^{t} = 97.8(1+.022)^{13}=129.77 \mbox{ million.}$

For 2030, t=33 years, r=.022, and $Q_{0}$=97.8 million

$\nonumber Q = Q_{0}(1+r)^{t} = 97.8(1+.022)^{33}=200.54 \mbox{ million.}$

b) Using the rule of 70, estimate the doubling time for Mexico.

Rule of 70 is that $T_{D} \approx \frac{70}{P}$
P = 2.2 so therefore $T_{D} \approx \frac{70}{2.2}=31.82$ years

c) Using the exponential growth formula in terms of doubling time (use the doubling time from part b), make two more estimates for the population of Mexico in 2010 and 2030.
Does this match your estimates from (a)?

For 2010, t=13 years, $T_{D}=31.82$, and $Q_{0}$=97.8 million

$\nonumber Q = Q_{0}2^{\frac{t}{T_{D}}} = 97.8(2)^{\frac{13}{31.82}}=129.81 \mbox{ million.}$

For 2030, t=33 years, $T_{D}=31.82$, and $Q_{0}$=97.8 million

$\nonumber Q = Q_{0}2^{\frac{t}{T_{D}}} = 97.8(2)^{\frac{33}{31.82}}=200.70 \mbox{ million.}$

2. (8 points) The population of a certain city is fdropping at a rate of 1.2% per month due to certain economic factors. If the city started out with 50,000 people at the beginning of 1999,
a) How many will there be left in this city at the end of the year 2001 (3 years)

t = 36 months, r=0.012, $Q_{0}=50,000$

$\nonumber Q = Q_{0}(1-r)^{t} = 50000(1-0.012)^{36}=32376 \mbox{ people.}$

b) How long will it take toreach a population of 25,000? (Hint: 25,000 is half of 50,000

This is just a half life problem. Use either the rule of 70 or the exact half- life formula.

Using rule of 70

$T_{H} \approx \frac{70}{1.2}=58.83$ months, or 4 years, 11 months

Using exact Half-Life formula,

$T_{H}= \frac{log(\frac{1}{2})}{log(1-r)} =\frac{log(\frac{1}{2})}{log(1-0.012)}=57.41$ months, or 4 years, 9 $\frac{1}{2}$ months

3. (7 points) Find the area and perimeter of the following figure.

(figure to appear later)
rectangle with horizontal sides having length 4 ft, and vertical sides having length 2 ft. On the right side, there is a half circle on the end of the rectangle whose diameter is the end of the rectangle.

Area of Rectangle = 2 ft $\times$ 4 ft = 8 $ft^{2}$

Area of the half circle = $\frac{\pi r^{2} }{2}=\frac{\pi}{2}= 1.57 ft^{2}$

Total Area = 8 +1.57 = 9.57 $ft^{2}$

Perimeter is length around outside.

Notice that one end of the rectangle is NOT on the outside.

Perimeter of half circle = $\pi$ r = 3.14 ft

Total Perimeter = 2 ft + 4 ft + 4 ft + 3.14 ft = 13.14 ft