Posted by **Mitch** on Monday, June 29, 2009 at 10:45pm.

(Growth of Cells) Suppose that after t hrs, there are p (t) cells present in a culture, where P(t) = 5000e*^2t

a) How many cells were present initially?

b) Give a differential equation satisfied by P(t)

c) When will the population double?

d) When will 20,000 cells be present?

- Calculus -
**Marth**, Tuesday, June 30, 2009 at 10:01am
a. "initially" means time = 0

P(0) = 5000e^2(0) = 5000e^0 = 5000

b. P(t) = 5000*e^(2t)

[d/du]e^u = e^u du

Therefore, P'(t) = 5000*2e^(2t)

P'(t) = 10000e^(2t)

c. Set e^(2t) = 2.

2t = ln2

t = (ln2)/2

The population will double every (ln2)/2 hours.

d. set P = 20000 and solve for t.

20000 = 5000e^(2t)

4 = e^(2t)

ln4 = 2t

2ln2 = 2t

ln2 = t

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