Question

A culture of yeast grows at a rate proportional to its size. If the initial population is 3000 cells and it doubles after 2 hours, answer the following questions.
(a) Write an expression for the number of yeast cells after t hours.
(b) Find the number of yeast cells after 8 hours.
(c) Find the rate at which the population of yeast cells is increasing at 8 hours.

Answer 1

Answer:

a) C(t) = 3000e^(rt)

b) C(8) = 47400 yeast cells.

c) 13870 yeast cells a hour.

Step-by-step explanation:

The growth of the culture of yeast in hours can be modeled by the following differential equation:

1) dC/dt = rC,

where C is the number of cells and r is the growth rate.

For question a), to write an expression for the number of yeast cells after t hours, we need to solve the differential equation 1). I am going to solve it by the variable separation method.

dC/C = rdt

Integrating both sides, we have:

ln C = rt + C0

where C0 is the initial population of cells.

We need to isolate C in this equation, so we do this

e^(ln C) = e^(rt + C0)

So

C(t) = C0e^(rt)

The initial population of cells is given as 3000, so:

C(t) = 3000e^(rt)

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b)

After two hours, the number of cells grows. So C(2) = 6000. This is helpful so we can find the growth rate r.

6000 = 3000e^(2r)

e^(2r) = 2

ln(e^(2r)) = ln 2

2r = 0.69

r = 0.345

Now we have C(t) = 3000e^(0.345t), so

C(8) = 3000e^(0.345*8) = 47400 yeast cells.

c)

C(7) = 3000e^(0.345*7) = 33570 yeast cells.

C(8) - C(7) = 47400 - 33570 = 13870 yeast cells. So, at 8 hours, the population of yeast cells is increasing at the rate of 13870 yeast cells a hour.