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Essential Calculus: Early Transcendentals
Found in: Page 335
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

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 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: \(t = \ln \left( {\frac{{10,000}}{p}} \right) + \frac{1}{9}\ln \left( {\frac{{100 + 0.1p}}{{1100}}} \right)\).

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