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
(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?
1 answer
1 answer