# Calculus

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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?

• Calculus - ,

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

• Calculus - ,

okay, so then in 87 minutes what would it be?

• Calculus - ,

let's use our first one, the one in minutes
P(87) = 500 e^(87*ln2/15)
= 500 e^4.020254
= 27857.6

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!!!!

• Calculus - ,

whao! that makes so much sense.
just a quick question, how do u know what e is?

thanks so much
<3333333333

• Calculus - ,

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

• Calculus - ,

e is approximately 2.71828