The population of a town increases at a rate | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $p$ be the population at time $t$ years. Given that the rate of change of population is proportional to the population: $\frac{dp}{dt} = kp$ $

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(C) Let $p$ be the population at time $t$ years. Given that the rate of change of population is proportional to the population: $\frac{dp}{dt} = kp$ $\Rightarrow \frac{dp}{p} = k dt$ Integrating both sides,we get: $\log p = kt + c$ At $t = 0$,$p = 40,000$: $\log 40,000 = k(0) + c \Rightarrow c = \log 40,000$ So,$\log p = kt + \log 40,000 \Rightarrow \log \left(\frac{p}{40,000}ight) = kt$ At $t = 40$ years,$p = 80,000$: $\log \left(\frac{80,000}{40,000}ight) = 40k \Rightarrow \log 2 = 40k \Rightarrow k = \frac{\log 2}{40}$ We need to find the population after another $40$ years,i.e.,at $t = 80$ years: $\log \left(\frac{p}{40,000}ight) = \left(\frac{\log 2}{40}ight) 	imes 80$ $\log \left(\frac{p}{40,000}ight) = 2 \log 2 = \log 2^2 = \log 4$ $\frac{p}{40,000} = 4$ $p = 4 	imes 40,000 = 160,000$

The population of a town increases at a rate proportional to the population at that time. If the population increases from $40,000$ to $80,000$ in $40$ years,then the population in another $40$ years will be (in $,000$)

Similar Questions

$A$ body cools according to Newton's law of cooling from $100^{\circ} C$ to $60^{\circ} C$ in $15$ minutes. If the temperature of the surrounding is $20^{\circ} C$,then the temperature of the body after cooling down for one hour is (in $^{\circ} C$)

$A$ radioactive sample has a half-life period of $10 \text{ days}$. If a sample originally has a mass of $1000 \text{ mg}$,then the mass remaining after $50 \text{ days}$ is:

The area bounded by a curve,the $x$-axis,and the ordinate of some point $(x, y)$ of the curve is equal to the length of the corresponding arc of the curve from a fixed point. If the curve passes through the point $P(0, 1)$,then the equation of this curve is:

Water at $100^{\circ} C$ cools in $15 \text{ minutes}$ to $75^{\circ} C$ in a room temperature of $25^{\circ} C$. Then the temperature of water after $30 \text{ minutes}$ is

If the population grows at the rate of $8 \%$ per year,then the time taken for the population to be doubled is $\quad$ (given $\log 2=0.6912$ ) (in $years$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module