Answer
$P(7)\approx2352$
Work Step by Step
The bacteria population growth rate is modeled by the given different equation where:
$\frac{dP}{dt}=k\sqrt{t}$
The particular solution of this differential equation can be solved for to result in a function that represents the population of the bacteria at any moment in time.
Multiply both sides of the equation by the differential $dt$:
$dP=k\sqrt{t}dt$
Integrating both sides now each with respect to the differential in the expression.
$\int{dP}=\int{k\sqrt{t}dt}$
$P=\frac{2}{3}kt\sqrt{t}+c$
What remains now is to solve for the two values of $k$ and $t$. The question states that the intial population of bacteria is 500. This means that the Population at $t=0$ must be 500.
$500=\frac{2}{3}k0\sqrt{0}+c$
$500=c$
Solving for $k$ now, we follow the same steps. The question states that after 1 day, the population reached a size of 600. Thus,
$600=\frac{2}{3}k(1)\sqrt{1}+500$
$100=\frac{2}{3}k$
$k=150$
Predicting the population at day 7 is now possible, using the population function $P(t)$
$P(7)=\frac{2}{3}*(150)*7\sqrt{7}+500$
$P(7)=2352.02592$
$P(7)\approx2352$
