A culture of bacteria in a Petri dish has an initial...

A culture of bacteria in a Petri dish has an initial population of 1500 cells and grows at a rate (in cells/day) of $N^{\prime}(t)=100 e^{-0.25 t}.$
a. What is the population after 20 days? After 40 days?
b. Find the population $N(t)$ at any time $t \geq 0.$

Key Concepts

Exponential Functions

Exponential functions are common in growth and decay models due to their natural appearance in processes with continuous proportional change. Their properties and rules of integration are central to solving differential equations related to phenomena like population growth.

Integration of Rate Functions

Integration is the process used to recover the original function from its derivative, especially when the derivative represents a growth rate. By integrating the rate at which the population increases, one can determine the total population at any given time.

Initial Value Problems

Initial value problems incorporate a specified condition at the beginning of observation. This condition is used to determine the constant of integration when solving the differential equation, ensuring that the resulting function accurately reflects the system's starting state.

Differential Equations

Differential equations express the relationship between a function and its rate of change. In the context of population growth, they model how the population evolves over time based on its instantaneous rate of change.

Transcript

00:02 Okay, so here we have a culture of bacteria that has an initial population of 1 ,500, as we can see, n of 0 is equal to 1 ,500, and we have the function n prime of t equals 100 e to the power of negative 0 .25t.

00:19 And this is supposed to represent the growth rate of our bacteria culture.

00:25 So we're going to figure out how many bacteria have spawned at 20 days.

00:31 So to do that, we're going to be using 100 negative 2 .5.

00:36 Oops, not 2 .5, 0 .25 t.

00:44 And we're going to take the integral of that from 0 to 20.

00:47 If we do that, we get 100, negative 400, e to the power of negative 0 .25t.

01:00 The reason why i have negative 400 here is because we divided by negative 0 .25, and we got 400.

01:10 So this is going to be equal, and we're going to do this from 0 to 20.

01:19 If we do that, we get that's equal to negative 400, e to the power of negative 0 .25 times 20 minus negative 400, e to the power of negative 0 .25 times 0 .0.

01:40 Note this becomes 1 so this is just plus 400 at the end and we need to figure out what eats the power of negative 0 .25 times 20 is if you do that using a calculator you will get so that be negative 5 if you do that you get that we get 0 .07 times 400 so this is approximately negative 2 .7 plus the 400.

02:29 If we add those up, we get, oh, and it's in cells.

02:37 So because it's in cells, what i'm going to do here is i'm just going to round up to 3.

02:41 It'll make that easy.

02:43 So here we get 397.

02:46 Don't forget, we had 1 ,500 cells in the first place.

02:49 So we're going to add that 1 ,500 to our added new cells, and we get 1 ,897 cells...

Please give Ace some feedback