Question

suppose that the per capita growth rate of a population is 3%; that is, if n(t) denotes the population size at time t, then $\frac{1}{n}\frac{dn}{dt}=0.03$. suppose also that the population size at time t = 3 is equal to 200. use a linear approximation to compute the population size at time t = 3.2. the population size at time t = 3.2 is $\square$.

Explanation:

Step1: Recall the linear approximation formula

The linear approximation formula is \(N(t)\approx N(t_0)+N^{\prime}(t_0)(t - t_0)\). We know that \(\frac{1}{N}\frac{dN}{dt}=0.03\), so \(N^{\prime}(t)=0.03N(t)\). Given \(t_0 = 3\) and \(N(3)=200\), then \(N^{\prime}(3)=0.03\times200 = 6\).

Step2: Apply the linear approximation formula

We want to find \(N(3.2)\) with \(t = 3.2\) and \(t_0=3\). Substitute into the formula \(N(3.2)\approx N(3)+N^{\prime}(3)(3.2 - 3)\).
Substitute \(N(3) = 200\), \(N^{\prime}(3)=6\) and \(t - t_0=3.2 - 3=0.2\) into the formula:
\(N(3.2)\approx200+6\times0.2\).
Calculate \(6\times0.2 = 1.2\), then \(N(3.2)\approx200 + 1.2\).

Answer:

\(201.2\)