Question

A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of the bacteria is growing at a rate of 4 grams per hour after 3 hours.
a.Write down the differential equation the mass of the bacteria, m, satisfies:
b.Find the general solution of this equation. Use A as a constant of integration.
c.Which particular solution matches the additional information? m(t) =
d.What was the mass of the bacteria at time t=0? mass =

Answer 1

Answer:

a) dm/dt = 0.2m

b) ln(m) = 0.2t + A

c) ln(m) = 0.2t + ln(20) - 0.6

d) 11 grams

Step-by-step explanation:

a) dm/dt = 0.2m

b) dm/m = 0.2dt

ln(m) = 0.2t + A

c) At t = 3, dm/dt = 4

dm/dt = 0.2m

4 = 0.2m

m = 4/0.2 = 20

So, when t = 3, m = 20

ln(20) =0.2(3) + A

A = ln(20) - 0.6

ln(m) = 0.2t + ln(20) - 0.6

d) find m when t=0

ln(m) = 0.2(0) + ln(20) - 0.6

ln(m) = ln(20) - 0.6

ln(m) = 2.3957322736

m = e^2.3957322736

m = 10.9762327224

m = 11