Barrons AP Calculus - David Bock

Q(t) = Q 0 e−0.00012t

(where Q 0 is the original amount). We are asked to find t when Q(t) = 0.25Q 0.

Rounding to the nearest 500 yr, we see that the animal died approximately 11,500yr ago.

EXAMPLE 17

In 1970 the world population was approximately 3.5 billion. Since then it has been growing at a

rate proportional to the population, and the factor of proportionality has been 1.9% per year. At

that rate, in how many years would there be one person per square foot of land? (The land area

of Earth is approximately 200,000,000 mi^2 , or about 5.5 × 10^15 ft^2 .)

SOLUTION: If P(t) is the population at time t, the problem tells us that P satisfies the equation

Its solution is the exponential growth equation

P(t) = P 0 e0.019t,

where P 0 is the initial population. Letting t = 0 for 1970, we have

3.5 × 10^9 = P(0) = P 0 e^0 = P 0.

Then

P(t) = (3.5 × 10^9 )e0.019t.

The question is: for what t does P(t) equal 5.5 × 10^15? We solve

Taking the logarithm of each side yields

where it seems reasonable to round off as we have. Thus, if the human population continued to

grow at the present rate, there would be one person for every square foot of land in the year

2720.

Case II: Restricted Growth
The rate of change of a quantity y = f (t) may be proportional, not to the amount present, but to a
difference between that amount and a fixed constant. Two situations are to be distinguished: The rate
of change is proportional to

(a) a fixed constant A minus the amount of the

quantity present:

(b) the amount of the quantity present minus a

fixed constant A:

f ′(t) = k[A − f (t)] f ′(t) = −k[f (t) − A]