the size p of a certain insect population at time t (in days) obeys the function p(t)=700e^{0.05t}. (a)…

Explanation:

Step1: Substitute t = 0

Substitute $t = 0$ into $P(t)=700e^{0.05t}$. $P(0)=700e^{0.05\times0}$

Step2: Simplify the exponent

Since $0.05\times0 = 0$, and $e^0=1$, we have: $P(0)=700\times1 = 700$

Step3: Find the growth - rate

The general form of an exponential growth function is $P(t)=P_0e^{rt}$, where $r$ is the growth rate. In $P(t)=700e^{0.05t}$, the growth rate $r = 0.05$ or $5%$.

Step4: Find population at t = 10

Substitute $t = 10$ into $P(t)=700e^{0.05t}$. $P(10)=700e^{0.05\times10}=700e^{0.5}$ Using a calculator, $e^{0.5}\approx1.64872$, so $P(10)=700\times1.64872 = 1154.104\approx1154$.

Step5: Solve for t when P(t)=1050

Set $P(t)=1050$, so $1050 = 700e^{0.05t}$. First, divide both sides by 700: $\frac{1050}{700}=e^{0.05t}$, which simplifies to $1.5 = e^{0.05t}$. Then take the natural - logarithm of both sides: $\ln(1.5)=\ln(e^{0.05t})$. Since $\ln(e^{x})=x$, we have $\ln(1.5)=0.05t$. Solve for $t$: $t=\frac{\ln(1.5)}{0.05}$. Using a calculator, $\ln(1.5)\approx0.4055$, so $t=\frac{0.4055}{0.05}=8.11\approx8.1$ days.

Step6: Solve for t when population doubles

The initial population $P_0 = 700$. When the population doubles, $P(t)=1400$. Set $1400 = 700e^{0.05t}$. Divide both sides by 700: $2 = e^{0.05t}$. Take the natural - logarithm of both sides: $\ln(2)=\ln(e^{0.05t})$. Since $\ln(e^{x})=x$, we have $\ln(2)=0.05t$. Solve for $t$: $t=\frac{\ln(2)}{0.05}$. Using a calculator, $\ln(2)\approx0.6931$, so $t=\frac{0.6931}{0.05}=13.862\approx13.9$ days.

Answer:

(a) 700 insects (b) 5% (c) 1154 insects (d) 8.1 days (e) 13.9 days