Question

the size p of a certain insect population at time t (in days) obeys the function ( p(t) = 700e^{0.02t} ).
(a) determine the number of insects at ( t = 0 ) days.
(b) what is the growth rate of the insect population?
(c) what is the population after 10 days?
(d) when will the insect population reach 910?
(e) when will the insect population double?

(c) what is the population after 10 days?
approximately 855 insects.
(do not round until the final answer. then round to the nearest whole number as needed.)
(d) when will the population reach 910 insects?
in approximately 13.1 days.
(do not round until the final answer. then round to the nearest tenth as needed.)
(e) when will the insect population double?
in about days.
(do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Set up the equation

The initial population is \(P(0) = 700\). When the population doubles, \(P(t)=2\times700 = 1400\). Substitute into the formula \(P(t)=700e^{0.02t}\), we get \(1400 = 700e^{0.02t}\).

Step2: Simplify the equation

Divide both sides of \(1400 = 700e^{0.02t}\) by \(700\), we have \(2=e^{0.02t}\).

Step3: Take the natural logarithm

Take the natural logarithm of both sides: \(\ln(2)=\ln(e^{0.02t})\).

Step4: Use the property of logarithm

Since \(\ln(e^{x}) = x\), then \(\ln(2)=0.02t\).

Step5: Solve for \(t\)

Solve for \(t\): \(t=\frac{\ln(2)}{0.02}\).

Step6: Calculate the value

We know that \(\ln(2)\approx0.6931\), so \(t=\frac{0.6931}{0.02}=34.655\approx34.7\).

Answer:

\(34.7\)