Question

suppose ( p(t) ) represents the population of a certain mosquito colony, where ( t ) is measured in days. the current population of the colony is known to be 597 mosquitos; that is, ( p(0)=597 ). if ( p^{prime}(0)=139 ) mosquitos per day, estimate the size of the population in 10 days.

Explanation:

Step1: Recall the formula for population growth

If the population growth is modeled by \(P(t)\), and we assume an exponential - like growth model (since we are given \(P(0)\) and \(P^{\prime}(0)\)). The general form of a function for population growth is \(P(t)=P(0)e^{kt}\), and the derivative \(P^{\prime}(t)=kP(0)e^{kt}\). When \(t = 0\), \(P^{\prime}(0)=kP(0)\). We know \(P(0)=597\) and \(P^{\prime}(0)=139\). So, \(k=\frac{P^{\prime}(0)}{P(0)}=\frac{139}{597}\approx0.233\).

Step2: Use the population growth formula for \(t = 10\)

The population formula is \(P(t)=P(0)e^{kt}\). Substitute \(P(0) = 597\), \(k=\frac{139}{597}\), and \(t = 10\) into the formula.
\(P(10)=597e^{\frac{139}{597}\times10}\)
First, calculate \(\frac{139\times10}{597}=\frac{1390}{597}\approx2.328\)
Then, \(e^{2.328}\approx10.26\) (using a calculator, since \(y = e^{x}\), and for \(x = 2.328\), \(e^{2.328}=\sum_{n = 0}^{\infty}\frac{(2.328)^{n}}{n!}\approx10.26\))
\(P(10)=597\times10.26\)
\(P(10)=597\times(10 + 0.26)=597\times10+597\times0.26=5970+155.22 = 6125.22\)

Answer:

\(6125\)