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.

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)