Question

The rate of change of the population of a small town is dPdt=kP, where P is the population, t is time in years and k is the growth rate. If P=30000 when t=2 and P=40000 when t=4,
what is the population when t=10?
Round your answer to the nearest integer.

Answer 1

Answer:

P  =  94800  inh   (town population at t = 10)

Step-by-step explanation: We get:

dP/dt  =  kP   (1)

Population in year   2        is      30000

Population in year   4        is      40000

From equation (1)

dP/P    =  kdt

Then

∫ 1/P dp   =   k ∫dt

Ln (P)   =  kt   + c       (2)

Now we know populations in years 2 and 4 then by subtitution

Ln (40000)   = 4k  + c

Ln ( 30000)  = 2k  + c

Subtracting these equation we have

Ln (40000) - Ln ( 30000)  =  2k

Then we can get k ??

ln ( 40000/30000)  = 2k          k  =  1/2 Ln  ( 4/3)

Now weareable to find population when t = 10

Equation (2)    

Ln (P)   =  kt   + c     and from previous data    

Ln (40000)   = 4k  + c

ln  (P)    =  10* k  + c

Ln (40000)   = 4k  + c

The same procedure

Ln  (P)  -  Ln ( 40000) =  6k

Ln  (P)  -  Ln ( 40000) =   6*  1/2 Ln  ( 4/3)

Ln  (P)  -  Ln ( 40000) = 3*Ln (4/3)

Ln  [  P/ 40000 ]  = 3*Ln (4/3)

Eliminating Ln on both sides of the equation

P /  40000  = (4/3)³

P    =   40000 *  (4/3)³

P  =  40000 *  2.37

P  =  94800  inh   (town population at t = 10)