Insects-Problem


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The number of bees in a beehive is now about 5000 . In 6 hours it grows to about 6000 bees.
About how long will it take for the beehive to grow to 12000 bees?
The rate of the growth is proportional to the initial number of bees.
Enter-01: Enter initial number of bees (Q1 = 5000)
Enter-02: Enter known time interval of known growth (T1 = 6 Hours)
Enter-03: Enter number of bees after known time interval (Q2 = 6000)
Enter-04: Enter number of bees you are looking for (Q3 = 12000)

Result-01: First we must calculate the proportionality constant K = (ln-6000 - ln-5000)/6
Result-02:
Here we calculate the time for the proportional growth rate to result in 12000 bees.
T = (ln-12000 - ln-6000)/K = 22.810704101543774 Hours
This also solves these problems:
A certain mold doubles in size every 6 hours. How long does it take to get 100 times bigger?
In this case Q1 = .5 and Q2 = 1 and T1 = 6 hrs and Q3 = 100
=> K = 0.11552453009332421 and T2 = 39.863137138648355 hours
A certain ant killer reduces the ant bed population to half as many in 3 hours.
How long before the ant population is reduced to one tenth its size?
In this case Q1 = .5 and Q2 = 1 and T1 = 3 hours and Q3 = .1
=> K = 0.23104906018664842 and T2 = 9.965784284662087 hours
=> After 9.9657*2 = 19.9314 hours only 1/100 of the ant population would be left.
The Half-Life of Sodium-24 is 14.9 hours.
How many hours will it take for
a 40 gram sample to reduce down to 3 grams?
In this case Q1 = .5
and Q2 = 1 and T1 = 14.9 hours
Q3 = 3/40 = .075
=> K = 0.04651994500402317
=> T2 = -55.68078735307648 Hours

Find the half-life of a substance.
There is initially 100 mg of the substance.
Two hours later there is 90 mg.
=> Q1 = 100 mg and T1 = 2 hours
=> Q2 = 90 mg and Q3 = 45 mg
K = (ln-90 - ln-100)/2 = -.052680257
T = (ln-45 - ln-90)/K = 13.157 hours = Half-Life of substance

Carbon 14 Dating Problem (Carbon-14 Half-Life = 5730 Years)
The amount of Carbon-14 in the sample at T2 = 0 is the unknown
=> Q1 = 1 and T1 = 5730
and Q2 = .5 => Half for Half-Life
Q3 = Amount of Carbon-14 still in the sample = 5 grams
=> K = - .000120968
And T2 = - 19034.64798 Years


*** This is a different question ***
The Half-Life of Sodium-24 is 14.9 hours.
How many grams of a 40 gram sample of
Sodium-24 will be left after 24 hours?
Enter-01: Enter .5 Here for half life (Q1 = .5)
Enter-02: Enter known time interval of half-life (T1 = 14.9 Hours)
Enter-03: Enter 1 here for half life (Q2 = 1)
Enter-04: Enter elapsed time (T2 = 24 hours)
Enter-05: Enter initial Quantity (Q3 = 40 grams)

Result-01: First we must calculate the proportionality constant K = (ln-.5)/14.9 = -0.04651994500402317
Result-02: Here we calculate the grams left after the elapsed time.
Q4 = Quantity after 24 hours => Q4 = Q3*(e^(K*T2)) = 13.097230216768821 grams

The formula => f(t) = K*((e^Kt))
K = Rate of change and e = 2.719 Euler's Constant