Solution :
Given initial population = 300
Final population after 1 day = 800
Number of days = 6
∴ [tex]$\frac{dP}{dt} =kt^{1/2} $[/tex]
P(0) = 300 P(1) = 300
We need to find P(8).
[tex]$dP = kt^{1/2} dt$[/tex]
[tex]$ \int 1 dP = \int kt^{1/2} dt$[/tex]
[tex]$P(t) = k \left(\frac{t^{3/2}}{3/2}\right)+c$[/tex]
[tex]$P(t)= \frac{2k}{3}t^{3/2} + c$[/tex]
When P(0) = 300
[tex]$300 = \frac{2k}{3} (0)^{3/2} + c$[/tex]
∴ c = 300
∴ [tex]$P(t)= \frac{2k}{3}t^{3/2} + 300$[/tex]
When P(1) = 800
[tex]$800 = \frac{2k}{3} (1)^{3/2} + 300$[/tex]
[tex]$500 = \frac{2k}{3}$[/tex]
∴ k = 750
[tex]$P(t)= 500t^{3/2} + 300$[/tex]
So, P(8) is
[tex]$P(t)= 500(8)^{3/2} + 300$[/tex]
= 11,614
So the population becomes 11,614 after 8 days.
