Question:
The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village wa S 20, 000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?
Answer:
Let the population at any instant t is y.
Now it is given that the rate of increase of population is proportional to the number of inhabitants at any instant.
So dy/dt ∝ y
=> dy/dt = ky (k is a constant)
=> dy/y = kdt
Integrate on both side, we get
∫dy/y = ∫kdt
=> log y = kt + C .........1 (C is integral constant)
In the year 1990
t = 0 and y = 20000
from equation 1
log 20000 = 0 + C
=> C = log 20000
Again in the year 2004,
t = 5 and y = 25000
From equation 1
log 25000 = 5k + C
=> log 25000 = 5k + log 20000
=> log 25000 - log 20000 = 5k
=> 5k = log(25000/20000)
=> 5k = log(5/4)
=> k = 1/5 *log(5/4)
Now in the year 2009, t = 10
Put the value of t, k and C in equation 1, we get
log y = 10 * 1/5 *log(5/4) + log 20000
=> log y - log 20000 = 2*log(5/4)
=> log(y/20000) = log(5/4)2
=> y/20000 = (5/4)2
=> y/20000 = 25/16
=> y = (20000*25)/16
=> y = (5000*25)/4 (20000 and 16 is divided by 4)
=> y = 1250*25 (5000 and 4 is divided by 4)
=> y = 31250
So population of the village in the year 2009 will be 31250.
Not what you are looking for? Go ahead and submit the question, we will get back to you.
