Question 1003940: 1. given that a quantity Q(t) exhibiting exponential decay is described by the function Q(t)=2000e^0.06t , where t measured in years, answer the following questions :
a) what is the decay constant?
b) what quantity is present initially?
c) complete the following table of values?
t= 0, 5, 10, 20, 100
Q = ?, ?, ?, ?, ?
The growth rate of the bacterium Escherichia Coli a common bacterium in the human intestine, is proportional to its size. Under ideal laboratory conditions, when this bacterium is grown in a nutrient broth medium, the number of cells in a culture doubles approximately every 20 min.
a. If the initial cell population is 100, determine the function Q(t) that expresses the exponential growth of the number of cells of this bacterium of time t (in minutes)
b. How long will it take for a colony of 100 cells to increase to a population of 1 million?
c. If the initial cell population were 1000, how would this alter our model?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 1. given that a quantity Q(t) exhibiting exponential decay is described by the function Q(t) =2000 * e^0.06t , where t measured in years, answer the following questions :
a) what is the decay constant?
the decay constant shown is 0.06
this, however, doesn't look like a decay constant.
it would, if it were negative.
since it is positive, then it is a growth constant.
if you are talking about a decay formula, then the growth constant should have been negative.
Q(t) = 2000 * e^-.06t would have been more like it.
b) what quantity is present initially?
the quantity that is present initially is 2000.
c) complete the following table of values?
t= 0, 5, 10, 20, 100
Q = ?, ?, ?, ?, ?
replace t in the formula f = 2000 * e^(.06t) with t = 0,5,10,20,100, and then solve.
your values are:
t = 0,5,10,20,100
Q = 2000,2699.7,3644.2,6640.2,806858
as you can see, there is growth, not decay.
i change the sign of the decay constant from .06 to -.06 and these are the results i got.
t = 0,5,10,20,100
Q = 2000,1481.6,1097.6,602.39,4.9575
The growth rate of the bacterium Escherichia Coli a common bacterium in the human intestine, is proportional to its size. Under ideal laboratory conditions, when this bacterium is grown in a nutrient broth medium, the number of cells in a culture doubles approximately every 20 min.
a. If the initial cell population is 100, determine the function Q(t) that expresses the exponential growth of the number of cells of this bacterium of time t (in minutes)
first you want to find the constant of growth.
the formula for exponential growth is f = p * e^(kt)
f is the future value
p is the present value
e is the scientific constant of 2.718281828...
k is the constant of growth per minute
t is the number of minutes
set f = 200 and p = 100 to get double growth.
set t = 20 minutes.
formula becomes 200 = 100 * e^(20k).
divide both sides of this formula by 100 to get:
2 = e^(20k)
take natural log of both sides of this equation to get:
ln(2) = ln(e^20k) which becomes:
ln(2) = 20k*ln(e) which becomes:
ln(2) = 20k*1 which becomes:
ln(2) = 20k
divide both sides of this equation by 20 to get:
ln(2)/20 = k
solve for k to get:
k = .034657359
that's your growth constant.
replace in your formula to confirm it's good.
200 = 100 * e^(20k) becomes 200 = 100 * e^(20 * .034657359) which becomes:
200 = 200, confirming the value of k is good.
since 200 is two times 200, your number of cells have doubled in 20 minutes.
b. How long will it take for a colony of 100 cells to increase to a population of 1 million?
same formula of f = p * e^(kt)
same constant of growth of k = .034657359
p = 100
f = 1000000
formula becomes:
1000000 = 100 * e^(.034657359*t)
divide both sides of this formula by 100 to get:
10000 = e^(.034657359*t)
take the natural log of both sides of this equation to get:
ln(10000) = ln(e^(.034657359*t)) which becomes:
ln(10000) = .034657359*t*ln(e) which becomes:
ln(10000) = .034657359*t.
divide both sides of this equation by .034657359 to and solve for t to get:
t = 265.7542476 minutes.
it will take 265.7542476 minutes for the solution to increase to 1000000 cells.
replace t in the original formula to confirm the solution is good.
1000000 = 100 * e^(.034657359*t) becomes:
1000000 = 100 * e^(.034657359*265.7542476) which becomes:
1000000 = 1000000, confirming the solution is good.
c. If the initial cell population were 1000, how would this alter our model?
the constant of growth would be the same because 2000 would then be twice 1000 and 2000 / 1000 = 2 which is the same as we got before.
100 doubles to 200
1000 doubles to 2000
growing to 1000000 would now be from 1000 to 1000000 rather than from 100 to 1000000 which would then be equal to a growth of 1000 times rather than a growth of 10000 times.
the formula, when the initial quantity was 100, was:
10000 = e^(kt)
the formula, when the initial quantity is now 1000, becomes:
1000 = e^(kt)
the original formula when p = 100 gave a time of 265.7542476 minutes.
the new formula whan p = 1000 give a time of 199.3156857.
it will take less time to reach a million when you're starting from 1000 than when you're starting from 100.
when p = 100, the formula became ln(10000) / .034657359 = t.
when p = 1000, the formula becomes ln(1000) / .034657359 = t.
since ln(1000) is smaller than ln(10000), the value of t will be less.
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