A bacterial colony undergoes a two-phase process. In the first phase, the population grows exponentially at a rate of 5% per minute for a duration of \(t_1\) minutes. In the second phase, due to limited resources, the population decays exponentially at a rate of 2% per minute.
(a) Formulate the mathematical expressions for the two phases. For the growth phase, derive the expression for \(P(t)\) when \(0 \leq t \leq t_1\). For the decay phase, express \(P(t)\) for \(t > t_1\) in terms of \(P(t_1)\) and the decay rate.
(b) Given \(P_0 = 100\), \(t_1 = 10\) minutes, and a decay phase duration of 5 minutes, calculate the population at the end of the decay phase.
(c) Interpret the overall change in population from the initial value \(P_0\) to the final value after both phases, expressing the overall percentage change.
Question
Formulate the mathematical expressions for the two phases. For the growth phase, derive the expression for \(P(t)\) when \(0 \leq t \leq t_1\). For the decay phase, express \(P(t)\) for \(t > t_1\) in terms of \(P(t_1)\) and the decay rate.
Hint
Hint
Hint
Write the formula for exponential growth and then for decay starting from the population at \(t_1\).
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Complete Answer
Growth phase: \(P(t) = P_0 e^{0.05t}\) for \(0 \leq t \leq t_1\); Decay phase: \(P(t) = P(t_1) e^{-0.02(t – t_1)}\) for \(t > t_1\).
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Question
Given \(P_0 = 100\), \(t_1 = 10\) minutes, and a decay phase lasting 5 minutes, calculate the population at the end of the decay phase.
Hint
Hint
Hint
First calculate \(P(10)\) using the growth expression, then apply the decay factor for 5 minutes.
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At \(t=10\): \(P(10)=100e^{0.05×10}=100e^{0.5}\).
After 5 minutes of decay: \(P = 100e^{0.5}e^{-0.02×5}=100e^{0.5-0.1}=100e^{0.4}\) which is approximately 149.18.
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Question
Interpret the overall change in the population from the initial value \(P_0\) to the final value after both phases. Express the overall percentage change.
Hint
Hint
Hint
Determine the overall multiplier on \(P_0\) then convert that into a percentage change.
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Complete Answer
The final population is \(100e^{0.4}\) (approximately 149.18), indicating an overall increase of about 49.18% from the initial population of 100.
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