If the population was 200,000 bacteria when observations began, what will the population be 12 hours later?

Question

Answer the given problem.

The population P(t) of a bacterial colony t hours after observation begins is

found to be changing at the rate dp/dt=200e^0.1t+150e^-0.03t. If the population

was 200,000 bacteria when observations began, what will the population be

12 hours later?

Include:

Important Notations

Useful Given

Solution:

Initial function

The population

function

value of C

Final Answer

Interpretation

Answer 1

∫Separate dp and dt onto opposite sides of the equation in order to integrate.

dp = dt * 200e^0.1t+ dt *150e^-0.03t ⇒ P = ∫(dt * 200e^0.1t+ dt *150e^-0.03t)

The integral goes from 0 hours to 12 hours, so

∫₀¹²(dt * 200e^0.1t+ dt *150e^-0.03t) = P(12) - P(0)

Then you use two u-substitutions for 0.1t and -0.03t so solve the integral. Which gives you a final equation of

(2000e^0.1t - 5000e^-0.03t) ]₀¹² = P(12) - 200,000

(2000e^1.2 - 5000e^-0.36) - (2000 - 5000) = P(12) - 200,000

3151.85 + 3000 = P(12) - 200,000

P(12) = 206,151 Bacteria at 12 hours