The population of Village increases continuously at the rate proportional to the number of its inhabitants present at any time . if the population of village was 20000 in 1999 and 25000 in the year 2003 , what will be the population of the village in 2009
Answers
- Hii
Step-by-step explanation:
Let the population of the village is y in 2009
since, population in 1999
dy/dt =ky .(K is proportionality constant )
dy/y = K dt
logy = Kt + c
logy = kt+c
log (20000) = k(0) + C at time t = 0
Log20000 = C _________(1)
Again , population in 2003 at time t = 5year
dy/dt= kt + c
Log (25000) = k(5)+ log20000
Log(5/4) = k* 5
K,= log(5/4)/5 _______(2)
Now in 2009
At time t = 10 year
dy/dt= Ky
logy = kt+c
Logy = log(5/4)/5* 10 + log20000
logy = log (25/16*20000)
y = 31250 Answer !
_______________________________
Hope it helps you !!!
@Rajukumar111
Answer:
31250
Step-by-step explanation:
Let the population at time t be x
Given the population of the village increases at the rate proportional to its number of its inhabitants.
⇒ (dx/dt) ∝ x
⇒ (dx/dt) = kx
⇒ (dx/x) = k dx
On integrating both sides, we get
∫ (dx/x) = k ∫ dt
log x = kt + C ------ (1)
In 1999 :
Population of village was 20000 in 1999.
log 20000 = k(0) + C
C = log 20000
Place value in (1), we get
log x = kt + C
log x = kt + log 20000 ------ (2)
In 2004 :
t = 5, y = 25000.
Place value in (2), we get
log 25000 = 5k + log 20000
⇒ log 25000 - log 20000 = 5k
⇒ log 25000/20000 = 5k
⇒ (1/5) log 5/4 = k
⇒ k = (1/5) log (5/4)
Place value of k in (2), we get
log x = kt + log 20000
log x = (t/5) log (5/4) + log 20000 ---- (3)
In 2009 :
In t = 2009, t = 10, we need to find the value of x.
log x = (10/5) log (5/4) + log 20000
⇒ log x - log 20000 = 2 log(5/4)
⇒ log(x/20000) = log(5²/4²)
⇒ log(x/20000) = log(25/16)
⇒ x = (25/16) * 20000
⇒ x = 31250
Therefore,
Population of village = 31250
Hope it helps!