# Exponential growth and decay

1. Oct 31, 2009

### hover

1. The problem statement, all variables and given/known data
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A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 63 cells.
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(a) Find the relative growth rate.

2. Relevant equations
Well this has to do with exponential growth and decay. So equations that could apply are

$$\frac{dy}{dt}=ky$$
where y is some function, k is a constant and dy/dt is a change in that function

$$y(t)=y_0e^k^t$$

where y of t is a function, y of 0 is an initial value, k is a constant and t is time

$$y^-^1* \frac{dy}{dt}=k$$

3. The attempt at a solution

(a) Find the relative growth rate in cells per hour.

well I thought since I know the change in the change of the number of bacteria per minute and the starting number of bacteria, I thought i could do this-

$$\frac{1}{63}*\frac{2}{20}$$

$$\frac{1}{63}*\frac{1}{10}$$

$$\frac{1}{630}$$ cells per minute

$$\frac{60}{630}$$cells per hour

$$.095$$cells per hour

that answer is wrong. I don't know how to go about this as you can clearly tell. can someone give me a kick start?

thanks :)

2. Oct 31, 2009

### futurebird

Plug in the values to:

$$y(t) = y_0e^{kt}$$

After 20 min you have twice as many...

$$126 = 63 e^{k(20)}$$

Now solve for k. Then you will have the equation for how many there are after t min.

3. Oct 31, 2009

### arildno

$$y(t)=y_{0}e^{kt}$$
Since we have:
$$63=y(0)=y_{0}$$, we ave determined ONE of the two constants, y_0.

The relative growt rate is, indeed, k.
We know that after t=20 minutes, the population has doubled.
Thus, we have:
$$2y_{0}=y(20)=y_{0}e^{20k}$$,
which means:
$$e^{20k}=2\to{k}=\frac{\ln(2)}{20}$$
Thus, we get:
$$y(t)=63e^{\frac{t\ln(2)}{20}}=63*2^{\frac{t}{20}}$$
if you want to make the doubling time explicit, t being understood to be measured in minutes.

4. Oct 31, 2009

### hover

Thanks guys :). I knew I kind of overlooked this question. I should have known better. k=2.07 cells per hour. I got part a right and once I got that, all the other parts of the question (a-e) fell right out. I thank all of you.

-Hover