Answer:
Explanation:
Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m
a) By definition: [tex]m'=dm/dt[/tex]
b) Given: [tex]rate=0.2m[/tex]
c) By substitution: [tex]dm/dt=0.2m[/tex]
Question 2: Find the general solution of this equation. Use A as a constant of integration.
a) Separate variables
[tex]dm/m=0.2dt[/tex]
b) Integrate
[tex]\int dm/m=\int 0.2dt[/tex]
[tex]ln(m)=0.2t+C[/tex]
c) Antilogarithm
[tex]m=e^{0.2t+C}[/tex]
[tex]m=e^{0.2t}\cdot e^C[/tex]
[tex]e^C=A\\\\m=Ae^{(0.2t)}[/tex]
Question 3. Which particular solution matches the additional information?
Use the measured rate of 4 grams per hour after 3 hours
[tex]t=3hours,dm/dt=4g/h[/tex]
First, find the mass at t = 3 hours
[tex]dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g[/tex]
Now substitute in the general solution of the differential equation, to find A:
[tex]m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976[/tex]
Round A to 1 significant figure:
Particular solution:
[tex]m=10e^{(0.2t)}[/tex]
Question 4. What was the mass of the bacteria at time =0?
Substitute t = 0 in the equation of the particular solution:
[tex]m=10e^{0}\\\\m=10g[/tex]
