Answer to Question #320390 in Calculus for koi

Question

It is estimated that x months from now the population of a certain town will be increasing at the rate of 2 + 6√𝑥 people per month. The current population is 5,000. What will be the population 9 months from now?

Answer 1

Let

$\frac{dp}{dx}=2+6\sqrt{x}$

$dp=(2+6\sqrt{x})dx$

Integrating both sides, we have that

$P(x)=2x+4x^{\frac{3}{2}}+k$

at $x=0, P=5000$

$5000=2(0)+4(0)^{\frac{3}{2}}+k$

$=>k=5000$

$P(x)=2x+4x^{\frac{3}{2}}+5000$

When $x=9$

$P=2(9)+4(9)^{\frac{3}{2}}+5000$

$P=18+108+5000=5126$

Hence, Population 9months from now is 5126