Question

A culture of bacteria is growing at a rate of 4e^0.5t per hour, with t in hours and 0 ≤ t ≤ 20.
A) How many new bacteria will be in the culture after the first six hours?

B) How many new bacteria are introduced from the end of the fifth hour through the fifteenth hour?

C) At approximately how many hours after t=0 will the culture contain 200 new bacteria?

Answer 1

(A) if you mean how many total bacteria have grown during the first 6 hours, then that is of course ∫[0,6] 4e^(t/2) dt = 8(e^3 - 1)

(B) ∫[8,15] 4e^(t/2) dt
(C) ∫[0,x] 4e^(t/2) dt = 200
8(e^(x/2) - 1) = 200
e^(x/2) - 1 = 25
e^(x/2) = 26
x/2 = ln26
x = 2 ln26 = 6.516

Answer 2

dB/dt = 4 e^ (t/2)
B = 8 e^(t/2) + C
at t = 0, say there are -8 so B is only the new ones
B = 8 e^(t/2)

A) B = 8 e^3 = 160
B) B = 8e^7.5 - 8 e^2.5 = 8 (1808 - 12) = 14,368
C) 200 = 8 e^t/2
e^t/2 = 25
ln e^t/2 = t/2 = ln 25
so t = 2 ln 25 = 6.43 hours