The rate at which the population of a city in | vedclass

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

Question

(C) Let $P(t)$ be the population at time $t$. According to the problem,the rate of increase is proportional to the population: $\frac{dP}{dt} = kP$. I

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let $P(t)$ be the population at time $t$. According to the problem,the rate of increase is proportional to the population: $\frac{dP}{dt} = kP$. Integrating this,we get $\ln P = kt + C$,or $P(t) = P_0 e^{kt}$. At $t = 0$,$P(0) = 4$ lakhs,so $P_0 = 4$. At $t = 20$,$P(20) = 6$ lakhs. Thus,$6 = 4 e^{20k}$,which gives $e^{20k} = \frac{6}{4} = 1.5$. We need to find the population after another $20$ years,i.e.,at $t = 40$. $P(40) = P_0 e^{40k} = 4 (e^{20k})^2$. Substituting $e^{20k} = 1.5$,we get $P(40) = 4 	imes (1.5)^2 = 4 	imes 2.25 = 9$ lakhs.

The rate at which the population of a city increases varies as the population. In a period of $20$ years,the population increased from $4$ lakhs to $6$ lakhs. In another $20$ years,the population will be (in $lakhs$)

Similar Questions

The solution of the differential equation $\frac{dy}{dx} = \frac{a + bx}{c + dy}$ represents a family of circles centered at the origin if...

The family of curves in which the sub-tangent at any point to any curve is double the abscissa is given by

Let $f$ be a twice differentiable non-negative function such that $(f(x))^2 = 25 + \int_{0}^{x} ((f(t))^2 + (f'(t))^2) dt$. Then the mean of $f(\log_e(1)), f(\log_e(2)), \ldots, f(\log_e(625))$ is equal to:

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is $3$ units and after $3$ seconds it is $6$ units,find the radius of the balloon after $t$ seconds.

Bacteria increases at a rate proportional to the number of bacteria present. If the original number $N$ doubles in $4$ hours,then the number of bacteria will be $4N$ in (in $hours$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module