# Calculus population questions?

Okay, so I have no idea how to do this. So, if anyone could help me out and breifly describe how to get these answers, it would be greatly appreciated.

1. A cell of bacteria divides in two every 20 minutes. The initial population is 60 cells.

a. Find the relative growth rate.

b. Find an expression of the numer of cells after t hours.

c. Find the number of cells after 8 hours.

d. Find the rate of growth after 8 hours.

e. When will the population reach 20,000 cells?

(I really only need help on "a" of number 1, then I should be able to find the others)

2. A bacteria culture grows with constant relative growth rate. After 2 hours, there are 600 bacteria and after 8 hours, there are 75,000 bacteria.

a. Find the initial population.

b. Find an expression for the population after t hours.

c. Find the number of cells after 5 hours.

d. Find the rate of growth after 5 hours.

e. When will the population reach 200,000?

Relevance
• cidyah
Lv 7
10 years ago

2)

An error was pointed out by another user. Please make the correction and redo the problem.

a)

Let N(t) = N(0) e^kt , where N(0) is the initial population and k the growth rate

When t=2, N(t)=600

600 = N(0) e^2k ----(1)

When t=8, N(t) =75,000

75,000 = N(0) e^8k -----(2)

subtract (1) from (2)

divide (2) by (1)

e^6k = 75,000/600

e^6k =750

6k = ln(750)

k = ln(750)/6 = 1.103345534

600 = N(0) e^1.103345534(2)

600 = N(0) e^2.206691068

N(0) = 600 / e^2.206691068 = 66.038

Initial population is 66

b)

N(t) = 66 e^1.103345534t

c)

substitute t=5 into N(t)

N(5) = 66 e^1.103345534(5) = 16422.08 = 16422

d)

16422 = 66 e^5k

5k = ln(16422/66)

k = ln(16422/66) /5 = 1.103344487

e)

200,000 = 66 e^1.103345534t

solve for t

200,000 /66 = e^1.103345534t

1.103345534t = ln[200,000 /66]

t = ln[200,000 /66] / 1.103345534

t=4.802

t= 4.8 hours

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For problem (1), I have attached a yahoo-answer reference.