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Q43E

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Found in: Page 335

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

one method of slowing the growth of an insect population without using pesticides is to introduce into the population a number of sterile males that mate with fertile females but produce no offspring. if p represents the number of female insects in a population, s the number of sterile males introduced each generation, and r the population’s natural growth rate, then the female population is related to time t by$$t = \int {\frac{{p + s}}{{p\left( {(r - 1)p - s} \right)}}} dp$$Suppose an insect population with 10,000 females grows at a rate of r=0.10 and 900 sterile males are added. Evaluate the integral to the give an equation relating the female population to time.

Here p and t are variables. R is a constant characteristics of the particular insect population, and s is the constant introduction rate of sterile males per generation.

Separate the numerator into convenient parts to facilitate integration

Use initial condition values to evaluate constants of integration.

given relation between t and p is:$$t = \int {\frac{{p + s}}{{p\left( {(r - 1)p - s} \right)}}} dp$$

where r,s are characteristic constants.

The integration can be carried out by disintegration of the numerator.

Given initial condition at t=0is p=10,000

See the step by step solution

Step 1: simplifying integration by disintegration of the numerator

\begin{aligned}{l}t &= \int {\frac{{p + s}}{{p\left( {(r - 1)p - s} \right)}}} dp\\t &= \int {\frac{p}{{p\left( {(r - 1)p - s} \right)}}} dp + \int {\frac{s}{{p\left( {(r - 1)p - s} \right)}}} dp\\t &= \frac{1}{{(r - 1)}}\int {\frac{{(r - 1)}}{{\left( {(r - 1)p - s} \right)}}} dp + \int {\frac{{(r - 1)p - \left( {(r - 1)p - s} \right)}}{{p\left( {(r - 1)p - s} \right)}}} dp\\t &= \frac{1}{{(r - 1)}}\ln ((r - 1)p - s) - \ln p + c\end{aligned}

Step 2: introducing initial condition value

Given constant quantities $$r = 0.10,s = 900$$

Given initial condition $$t = 0,p = 10,000$$

Substituting the initial condition value in the above equation

\begin{aligned}{l}c &= \ln 10,000 - \frac{r}{{r - 1}}\ln \left( {(r - 1)10,000 - s} \right)\\t &= \ln \left( {\frac{{10,000}}{p}} \right) - \frac{r}{{r - 1}}\ln \left( {\frac{{(r - 1)p - s}}{{(r - 1)10,000 - s}}} \right)\\t &= \ln \left( {\frac{{10,000}}{p}} \right) - \frac{r}{{1 - r}}\ln \left( {\frac{{(1 - r)p + s}}{{(1 - r)10,000 + s}}} \right)\\t &= \ln \left( {\frac{{10,000}}{p}} \right) + \frac{1}{9}\ln \left( {\frac{{960 + 0.9p}}{{960 + 9000}}} \right)\\t &= \ln \left( {\frac{{10,000}}{p}} \right) + \frac{1}{9}\ln \left( {\frac{{100 + 0.1p}}{{1100}}} \right)\end{aligned}

Hence, the integration of the given equation is: $$t = \ln \left( {\frac{{10,000}}{p}} \right) + \frac{1}{9}\ln \left( {\frac{{100 + 0.1p}}{{1100}}} \right)$$.