40–43. Population growth


A culture of bacteria in a Petri dish has an initial population of 1500 cells and grows at a rate (in cells/day) of N′(t) = 100e^−0.25t. Assume t is measured in days.


a. What is the population after 20 days? After 40 days?

Verified step by step guidance

1. Identify the given rate of change of the population, which is the derivative of the population function: \(N\'(t) = 100e^{-0.25t}\), where \(t\) is in days.
2. Recognize that to find the population function \(N(t)\), you need to integrate the rate function \(N\'(t)\) with respect to \(t\): \(N(t) = \int 100e^{-0.25t} \, dt + C\).
3. Perform the integration of \$100e^{-0.25t}\(: use the integral formula \)\int e^{kt} dt = \frac{1}{k} e^{kt} + C\(, keeping track of the constant of integration \)C$.
4. Use the initial condition given, \(N(0) = 1500\), to solve for the constant \(C\) by substituting \(t=0\) into the integrated function and setting it equal to 1500.
5. Once you have the explicit formula for \(N(t)\), substitute \(t=20\) and \(t=40\) into this formula to find the population after 20 days and 40 days, respectively.

Key Concepts

Derivative as Rate of Change

The derivative N′(t) represents the instantaneous rate of change of the population with respect to time. In this problem, N′(t) = 100e^−0.25t gives the growth rate in cells per day at any time t, which is essential for determining how the population changes over time.

Definite Integral for Accumulated Change

To find the total change in population over a time interval, we integrate the rate function N′(t) from the start to the end time. The definite integral of N′(t) from 0 to t gives the net increase in population during that period.

Initial Condition and Population Function

The initial population N(0) = 1500 serves as the starting value for the population function N(t). After integrating the rate function, adding this initial value gives the total population at any time t, allowing calculation of population after 20 and 40 days.