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.
Write down the differential equation the mass of the bacteria, m, satisfies: ′= .2m
Find the general solution of this equation. Use A as a constant of integration.
() = Ae^(.2t)
Which particular solution matches the additional information? () =
What was the mass of the bacteria at time =0? mass =

Answer 1

Answer:

• Question 1:      $dm/dt=0.2m$

• Question 2:     $m=Ae^{(0.2t)}$

• Question 3:      $m=10e^{(0.2t)}$

• Question 4:      $m=10g$

Explanation:

Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m

a) By definition:  $m'=dm/dt$

b)  Given:  $rate=0.2m$

c) By substitution:  $dm/dt=0.2m$

Question 2: Find the general solution of this equation. Use A as a constant of integration.

a) Separate variables

     $dm/m=0.2dt$

b) Integrate

           $\int dm/m=\int 0.2dt$

            $ln(m)=0.2t+C$

c) Antilogarithm

       $m=e^{0.2t+C}$

       $m=e^{0.2t}\cdot e^C$

         $e^C=A\\\\m=Ae^{(0.2t)}$

Question 3. Which particular solution matches the additional information?

Use the measured rate of 4 grams per hour after 3 hours

            $t=3hours,dm/dt=4g/h$

First, find the mass at t = 3 hours

            $dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g$

Now substitute in the general solution of the differential equation, to find A:

          $m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976$

Round A to 1 significant figure:

• A = 10.

Particular solution:

           

             $m=10e^{(0.2t)}$

Question 4. What was the mass of the bacteria at time =0?

Substitute t = 0 in the equation of the particular solution:

         $m=10e^{0}\\\\m=10g$