A population of 500 E. coli bacteria doubles every 15 minutes. Use this information to find an expression for this population growth. Using this expression, find what the population would be in 87 minutes. Use an exponential model.
so we're supposed to use P(t)= (Po)(e)^kt
so P(o) would be 500,
and then would t be 15 or 0.25 and wowhat would k be? 2?
let's use our first one, the one in minutes
P(87) = 500 e^(87*ln2/15)
= 500 e^4.020254
= 27857.6
is our answer reasonable??
remember it doubles every 15 min
so after 15 min --> 1000
after 30 min ---> 2000
after 45 min ---> 4000
after 60 min ---> 8000
after 75 min ---> 16000
after 90 min ---> 32000
we could check to see if we get 32000 for 90 minutes
P(90) = 500 e^(90*ln2/15)
= 500(64) = 32000 YES!!!!
in P(t) = 500 e^kt
it depends whether you want to to be in minutes or hours.
let's do it for minutes, then
1000 = 500 e^15k
2 = e^15k
15k = ln2
k = (ln2)/15
if you want t to be hours then use .25 for t in
1000 = 500 e^.25K
.
.
K = ln2/.25 or K = 4ln2
e is one of these very strange transcendental numbers like pi
one defn of e is
1 + 1/1! + 1/2! + 1/3! + ... to infinitity
where something like 5! = 1*2*3*4*5
or
e = Limit (1 + 1/n)^n where n approaches infinitity
The Swiss/German mathematician Euler spent a lot of time with the concept of that number and e is often called Euler's number
e is approximately 2.71828
okay, so then in 87 minutes what would it be?
whao! that makes so much sense.
just a quick question, how do u know what e is?
thanks so much
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To find the expression for population growth, we'll use the given exponential model formula, P(t) = (P0)(e^kt), where P(t) represents the population at time t, P0 represents the initial population, e is the mathematical constant approximately equal to 2.71828, and k represents the growth rate.
In this scenario, the initial population (P0) is 500, and the time for the population to double (t) is 15 minutes. To find the growth rate (k), we can use the fact that the population doubles every 15 minutes.
First, let's substitute the known values into the formula: P(t) = (500)(e^kt). Now we need to solve for k.
Since the population doubles every 15 minutes, we can write the equation: 2 = e^(k * 15).
To solve for k, we need to take the natural logarithm (ln) of both sides of the equation to eliminate the exponential term:
ln(2) = ln(e^(k * 15)).
Using the property of logarithms (ln(a^b) = b * ln(a)), we can rewrite the equation as:
ln(2) = k * 15 * ln(e).
Since ln(e) is equal to 1, the equation simplifies to:
ln(2) = k * 15.
Now, we can solve for k by dividing ln(2) by 15:
k = ln(2) / 15.
The value of k is approximately 0.0462.
Finally, we can substitute the values of P0, t, and k into the formula to find the population at 87 minutes:
P(t) = (P0)(e^kt) = 500 * (e^(0.0462 * 87)) ≈ 500 * e^(4.0146) ≈ 500 * 55.6184 ≈ 27,809.
Therefore, the population of E. coli bacteria would be approximately 27,809 after 87 minutes.