The growth of population is proportional to t | vedclass

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

Question

(B) Let $P_{0}$ be the initial population. Given that the rate of growth is proportional to the population: $\frac{dP}{dt} = \lambda P$. Integrating b

Vedclass · Accepted Answer

$50\left(\frac{\log 3}{\log 2}ight) 	ext{ yrs}$

Vedclass · Answer

(B) Let $P_{0}$ be the initial population. Given that the rate of growth is proportional to the population: $\frac{dP}{dt} = \lambda P$. Integrating both sides: $\int \frac{dP}{P} = \int \lambda dt \Rightarrow \ln P = \lambda t + C$. At $t = 0$,$P = P_{0}$,so $C = \ln P_{0}$. Thus,$\ln P = \lambda t + \ln P_{0} \Rightarrow \ln \left(\frac{P}{P_{0}}ight) = \lambda t$. Given that the population doubles in $50$ years: $\ln \left(\frac{2P_{0}}{P_{0}}ight) = 50\lambda \Rightarrow \ln 2 = 50\lambda \Rightarrow \lambda = \frac{\ln 2}{50}$. Now,we need to find $t$ when the population triples $(P = 3P_{0})$: $\ln \left(\frac{3P_{0}}{P_{0}}ight) = \lambda t \Rightarrow \ln 3 = \left(\frac{\ln 2}{50}ight) t$. Solving for $t$: $t = 50 \left(\frac{\ln 3}{\ln 2}ight) 	ext{ years}$.

The growth of population is proportional to the number present. If the population of a colony doubles in $50$ years,then the population will become triple in . . . . . . years.

Similar Questions

If $y = At^2 + \frac{B}{t}$ ($A, B$ are parameters) is the general solution of the differential equation $f(t) y''(t) + g(t) y'(t) + h(t) y = 0$,then $2 f(t) + t^2 h(t) =$

$A$ wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the open air loses half its moisture during the first hour,then the time $t$,in which $99 \%$ of the moisture will be lost,is

In a bank,the principal increases continuously at the rate of $6 \%$ per year. Then the time required to double $₹ 6000$ is (in years)

Let $\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} dt = \int_0^x y(t) dt, 0 \leq x \leq 3, y \geq 0$,$y(0)=0$. Then at $x=2, y^{\prime \prime}+y+1$ is equal to :

The population $P = P(t)$ at time $t$ of a certain species follows the differential equation $\frac{dP}{dt} = 0.5P - 450$. If $P(0) = 850$,then the time at which the population becomes zero is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module