From the question it is given that,
The present population is 1,00,000
Then, the population of a city doubled in the past 25 years,
So, let us assume P be the surface area of balloon,
(\( \dfrac{dP}{dt}\)) ∝ P
Then,
\( \dfrac{dP}{dt}\)= λP
\( \dfrac{dP}{dt}\)= λ dt
Integrating on both side we get,
∫\( \dfrac{dP}{dt}\)= λ ∫dt
Log P = λt + c … [equation (i)]
From question, P = Po t when t = 0,
log (Po) = 0 + c
c = log (Po)
Then, equation (i) becomes,
log (P) = λt + log (Po)
log \( \dfrac{P}{P_o}\) = λt … [equation (ii)]
And also form question, given P = 2Po when t = 25
log\( \dfrac{2P_o}{2P_o}\) = 25λ
log 2 = 25λ
By cross multiplication we get,
λ =\(\dfrac{ log2}{25}\)
So, now equation (ii) becomes,
log \( \dfrac{P}{P_o}\) = (\( \dfrac{ log2}{25}\))t
let us assume that t1 be the time to become population 5,00,000 from 1,00,000,
Then, \( log \dfrac{5,00,000}{1,00,000}\) = (\( \dfrac{ log2}{25}\))t1
By cross multiplication we get,
t1 = \(\dfrac{ 25 log 5}{log 2}\)
= 25\(\dfrac{ (1.609)}{(0.6931)}\)
= 58
Therefore, the required time is 58 years.
Answered by Aaryan | 1 year agoFind the particular solution satisfying the given condition \( (\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y\)
If the interest is compounded continuously at 6% per annum, how much worth Rs. 1000 will be after 10 years? How long will it take to double Rs. 1000?
In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
A population grows at the rate of 5% per year. If x=x(t) denotes the number of individuals in the population after t years, then the rate of change of x is equal to 5% of x. Form the desired differential equation . Find the time period
The surface area of a spherical balloon,being inflated changes at a rate proportional to time the if initially its radius is 1 units and after 3 seconds it is 2 units, find the radius after t seconds.