Respuesta :
The relation between population and time is:
[tex]t = \int { \frac{P+S}{P[(r - 1)P - S]} } \, dP [/tex]
Since r = 1.2 and S = 400, we get:
[tex]t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]} } \, dP [/tex]
= [tex] \int { \frac{P+400}{P(0.2P - 400)} } \, dP [/tex]
This is a rational function that we can write as:
[tex]\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} + \frac{B}{(0.2P - 400)} [/tex]
In order to evaluate A and B, let's calculate the LCD:
[tex]\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P - 400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)} [/tex]
which brings to:
P + 400 = 0.2AP - 400A + BP
= P(0.2A + B) - 400A
Since, the left side must be equal to the right side, we get the following system of equations:
[tex] \left \{ {{0.2A + B = 1} \atop {-400A = 400}} \right. [/tex]
Which gives:
[tex] \left \{ {{B=1.2} \atop {A=-1}} \right. [/tex]
Therefore, our integral can be written as:
[tex]t = \int { \frac{P+400}{P(0.2P - 400)} } \, dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP [/tex]
= [tex]- \int { \frac{1}{P} \, dP + 1.2\int { \frac{1}{0.2P-400} } \, dP[/tex]
= [tex]- \int { \frac{1}{P} \, dP + 6\int { \frac{0.2}{0.2P-400} } \, dP[/tex]
= - ln |P| + 6 ln |0.2P - 400| + C
We now have to evaluate the constant C. In order to do so, we consider that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5)
Therefore, the equation relating female population with time is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)
[tex]t = \int { \frac{P+S}{P[(r - 1)P - S]} } \, dP [/tex]
Since r = 1.2 and S = 400, we get:
[tex]t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]} } \, dP [/tex]
= [tex] \int { \frac{P+400}{P(0.2P - 400)} } \, dP [/tex]
This is a rational function that we can write as:
[tex]\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} + \frac{B}{(0.2P - 400)} [/tex]
In order to evaluate A and B, let's calculate the LCD:
[tex]\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P - 400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)} [/tex]
which brings to:
P + 400 = 0.2AP - 400A + BP
= P(0.2A + B) - 400A
Since, the left side must be equal to the right side, we get the following system of equations:
[tex] \left \{ {{0.2A + B = 1} \atop {-400A = 400}} \right. [/tex]
Which gives:
[tex] \left \{ {{B=1.2} \atop {A=-1}} \right. [/tex]
Therefore, our integral can be written as:
[tex]t = \int { \frac{P+400}{P(0.2P - 400)} } \, dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP [/tex]
= [tex]- \int { \frac{1}{P} \, dP + 1.2\int { \frac{1}{0.2P-400} } \, dP[/tex]
= [tex]- \int { \frac{1}{P} \, dP + 6\int { \frac{0.2}{0.2P-400} } \, dP[/tex]
= - ln |P| + 6 ln |0.2P - 400| + C
We now have to evaluate the constant C. In order to do so, we consider that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5)
Therefore, the equation relating female population with time is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)
The equation relating females population with time is [tex]\boxed{t = - \ln \left| P \right| + 6\ln \left| {0.2P - 400} \right| - 12\ln \left( 5 \right) + 4\ln \left( {10} \right) - 36\ln \left( 2 \right)}.[/tex]
Further explanation:
Explanation:
The female population is related to time and it is given as follows,
[tex]t = \int {\dfrac{{p + s}}{{p\left[ {\left( {r - 1} \right)p - s} \right]}}dp}[/tex]
The sterile males are 400 and the growth rate is 1.2.
[tex]\begin{aligned}t&= \int {\frac{{p + 400}}{{p\left[ {\left( {1.2 - 1} \right)p - 400} \right]}}dp}\\&= \int {\frac{{p + 400}}{{p\left[ {0.2p - 400} \right]}}dp}\\\end{aligned}[/tex]
The expression [tex]\dfrac{{p + 400}}{{p\left[ {0.2p - 400} \right]}}[/tex] can also be expressed as follows,
[tex]\dfrac{{p + 400}}{{p\left[ {0.2p - 400} \right]}} = \dfrac{A}{p} + \dfrac{B}{{\left( {0.2p - 400} \right)}}[/tex]
Find the least common denominator.
[tex]\dfrac{{p + 400}}{{p\left( {0.2p - 400} \right)}} = \dfrac{{A\left( {0.2p - 400} \right)}}{{p\left( {0.2p - 400} \right)}} + \dfrac{{Bp}}{{p\left( {0.2p - 400} \right)}}[/tex]
[tex]\begin{aligned}p + 400 &= 0.2Ap - 400A + Bp\\p + 400 &= p\left( {0.2A + B} \right) - 400A\\\end{aligned}[/tex]
Equate left and right side of the expression to obtain the value of A and B.
[tex]\begin{aligned}0.2A + B &= 1\\- 400A &= 400\\\end{aligned}[/tex]
The value of A can be obtained as follows,
[tex]\begin{aligned}A&= \dfrac{{400}}{{ - 400}}\\A& =- 1\\\end{aligned}[/tex]
The value of B is 1.2.
Integral can be expressed as follows,
[tex]\begin{aligned}t &= \int {\frac{{p + 400}}{{p\left( {0.2p - 400} \right)}}dp} \\ &= \int {\left( {\frac{{ - 1}}{p} + \frac{{1.2}}{{\left( {0.2p - 400} \right)}}} \right)dp} \\&= - \int {\frac{1}{p}dp} + 6 \times \int {\frac{{0.2}}{{0.2p - 400}}dp}\\&= - \ln \left| p \right| + 6\ln \left| {0.2p - 400} \right| + C\\\end{aligned}[/tex]
At initial time the population is 10000.
[tex]\begin{aligned}- \ln \left| {10000} \right| + 6\ln \left| {0.2 \times 10000 - 400} \right| + C &= 0\\\ln \left| {10000} \right| - 6\ln \left| {1600} \right| &= C\\4\ln 10 - 36\ln \left( 2 \right) - 12\ln \left( 5 \right) &= C\\\end{aligned}[/tex]
Substitute [tex]4\ln 10 - 36\ln \left( 2 \right) - 12\ln \left( 5 \right)[/tex] for [tex]C.[/tex]
[tex]t = - \ln \left| p \right| + 6\ln \left| {0.2p - 400} \right| + 4\ln 10 - 36\ln \left( 2 \right) - 12\ln \left( 5 \right)[/tex]
The equation relating females population with time is [tex]\boxed{t = - \ln \left| P \right| + 6\ln \left| {0.2P - 400} \right| - 12\ln \left( 5 \right) + 4\ln \left( {10} \right) - 36\ln \left( 2 \right)}.[/tex]
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Exponential function
Keywords: exponential growth, growth rate model, pesticides, slowing, insect population, sterile males, mate, fertile, females, explicitly, female grows, integral, female insects, male insects, capita rate, chosen mate, generation.
