Page 179 - How To Solve Word Problems In Calculus

t ln 2
100e 2 = 5000
t ln 2
e 2 = 50
t
ln 2 = ln 50
2
t ln 50
=
2 ln 2

2ln50
t = ≈ 11.29
ln 2
It takes approximately 11.29 hours for the population to reach 5000.
2. Let y(t) be the population of the colony t hours after the colony
starts growing. Assuming the rate at which the colony grows is
dy
proportional to the size of the colony, we have = ky. This leads
dt
kt
to the solution y = y 0 e . We wish to determine y 0 , the value of y
when t = 0. When t = 1, y = 9000 and when t = 2, y = 12,000.
Substituting, we obtain
9000 = y 0 e k

12,000 = y 0 e 2k

Dividing the second equation by the first, we obtain
12,000 y 0 e 2k
=
9000 y 0 e k
4
= e k
3

4
k = ln
3
k
Since 9000 = y 0 e , it follows that
9000 = y 0 e ln 4 3

4
9000 = y 0
3
3
y 0 = 9000 × = 6750 bacteria
4
166