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The rate at which the population of a city increases at any time is propotional to the population at that time. If there were $1,30,000$ people in the city in $1960 $ and $1,60,000$ in $1990$ what population may be anticipated in $2020$.$[\log_{e}(\large\frac{16}{13})=$$.2070;$$e^{.42}=1.52]$

$\begin{array}{1 1} 97600 \\ 197600 \\ 234600 \\ 234419 \end{array} $

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  • First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • It can be written as $\large\frac{g_1(y)}{g_2(y)}$$dy=\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
Let $x$ be the population of the city at time $t$ in years
$t=0$ in the year 1960,when the initial population $x_0=130,000$
Step 2:
Now $\large\frac{dx}{dt}\propto x$
$\therefore \large\frac{dx}{dt}$$=kx$
Step 3:
$\int \large\frac{dx}{x}$$=\int kdt+\log c$
$\log x=kt+\log c$
$\log \large\frac{x}{c}=$$kt$
$x=ce^{kt}$
Step 4:
When $t=0,x=1,30,000$
$\therefore c=1,30,000$
$\Rightarrow x=130000e^{kt}$
Step 5:
In 1990,t=30 years and $x=16,000$
$\therefore 1,60,000=1,30,000e^{30k}=\large\frac{16}{13}$
$30k\log e=\log \large\frac{16}{13}$
$\log_e\large\frac{16}{13}$$=0.2070$
$k=\large\frac{0.2070}{30}$
$k=0.0069$
Step 6:
In 2020,$t=60years$
$x=130000e^{60k}$
$\;\;=130000e^{0.0069\times 60}$
$\;\;\;=130000e^{0.42}$
$e^{0.42}=1.52$
$\;\;\;=130000\times 1.52$
$\;\;\;=197600$
The population in 2020 is anticipated to be 197600
answered Sep 8, 2013 by sreemathi.v
edited Mar 24, 2014 by thanvigandhi_1
 

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