English

The population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousands to 60 thousands in 40 years ? - Mathematics and Statistics

Advertisements
Advertisements

Question

The population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousands to 60 thousands in 40 years, what will be the population in another 20 years?

(Given: `sqrt(3/2)= 1.2247)`

Sum

Solution

Step 1: Use the Exponential Growth Model

P(t) = P0​ekt

where:

  • P = Population at time t
  • P0 = Initial population
  • k = Growth constant
  • t = Time in years

Step 2: Find k Using Given Data

We are given:

  • P0 = 40,000 
  • P(40) = 60,000
  • t = 40 years

60,000 = 40,000e40k

`(60,000)/(40,000) = e^(40k)`

1.5 = e40k

Taking the natural logarithm:

ln (1.5) = 40k

`k = ln(1.5)/40`

Using ln⁡ (1.5) = 0.4055

`k = 0.4055/40 = 0.01014`

Step 3: Find Population After 60 Years

Now, we need to find P(60)

P(60) = 40,000e60k

P(60) = 40,000e60×0.01014

P(60) = 40,000e0.6084

P(60) = 40,000 × 1.837

P(60) ≈ 73,480

shaalaa.com
Application of Differential Equations
  Is there an error in this question or solution?
Chapter 8: Differential Equation and Applications - Exercise 8.6 [Page 170]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) [English] 12 Standard HSC Maharashtra State Board
Chapter 8 Differential Equation and Applications
Exercise 8.6 | Q 2 | Page 170

RELATED QUESTIONS

If the population of a country doubles in 60 years, in how many years will it be triple (treble) under the assumption that the rate of increase is proportional to the number of inhabitants?
(Given log 2 = 0.6912, log 3 = 1.0986)


The rate of decay of certain substances is directly proportional to the amount present at that instant. Initially, there is 25 gm of certain substance and two hours later it is found that 9 gm are left. Find the amount left after one more hour.


Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.


The rate of growth of the population of a city at any time t is proportional to the size of the population. For a certain city, it is found that the constant of proportionality is 0.04. Find the population of the city after 25 years, if the initial population is 10,000. [Take e = 2.7182]


Choose the correct option from the given alternatives:

If the surrounding air is kept at 20° C and a body cools from 80° C to 70° C in 5 minutes, the temperature of the body after 15 minutes will be


The rate of depreciation `(dV)/ dt` of a machine is inversely proportional to the square of t + 1, where V is the value of the machine t years after it was purchased. The initial value of the machine was ₹ 8,00,000 and its value decreased ₹1,00,000 in the first year. Find its value after 6 years.


The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lac, when will the city have population 4,00,000?


If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years? `("Given" sqrt(3/2) = 1.2247)`


Choose the correct alternative:

Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in


Choose the correct alternative:

The integrating factor of `("d"y)/("d"x) + y` = e–x is


Choose the correct alternative:

The integrating factor of `("d"^2y)/("d"x^2) - y` = ex, is e–x, then its solution is


Choose the correct alternative:

The solution of `dy/dx` = 1 is ______.


Find the population of city at any time t given that rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 30000 to 40000.

Solution: Let p be the population at time t.

Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.

∴ `"dp"/"dt" prop "p"`

∴ `"dp"/"dt"` = kp, where k is a constant.

∴ `"dp"/"p"` = k dt

On integrating, we get

`int "dp"/"p" = "k" int "dt"`

∴ log p = kt + c

Initially, i.e. when t = 0, let p = 30000

∴ log 30000 = k × 0 + c       

∴ c = `square`

∴ log p = kt + log 30000

∴ log p - log 30000 = kt

∴ `log("p"/30000)` = kt          .....(1)     

when t = 40, p = 40000

∴ `log (40000/30000) = 40"k"`

∴ k = `square`

∴ equation (1) becomes, `log ("p"/30000)` = `square`

∴ `log ("p"/30000) = "t"/40 log (4/3)`

∴ p = `square`


Bacteria increases at the rate proportional to the number of bacteria present. If the original number N doubles in 4 hours, find in how many hours the number of bacteria will be 16N.

Solution: Let x be the number of bacteria in the culture at time t.

Then the rate of increase of x is `("d"x)/"dt"` which is proportional to x.

∴ `("d"x)/"dt" ∝ x`

∴ `("d"x)/"dt"` = kx, where k is a constant

∴ `("d"x)/x` = kdt

On integrating, we get

`int ("d"x)/x = "k" int "dt"`

∴ log x = kt + c    .....(1)

∴ x = aekt where a = e

Initially, i.e.,when t = 0, let x = N

∴ N = aek(0)

∴ a = `square`

∴ a = N, x = Nekt    ......(2)

When t = 4, x = 2N

From equation (2), 2N = Ne4k

∴ e4k = 2

∴  e= `square`

Now we have to find out t, when x = 16N

From equation (2),

16N = Nekt 

∴ 16 = ekt 

∴ `"t"/4 = square` hours

Hence, number of bacteria will be 16N in `square` hours


The population of city doubles in 80 years, in how many years will it be triple when the rate of increase is proportional to the number of inhabitants. `("Given" log3/log2 = 1.5894)`

Solution: Let p be the population at time t.

Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.

∴ `"dp"/"dt"  ∝  "p"`

∴ `"dp"/"dt"` = kp, where k is a constant

∴ `"dp"/"p"` = kdt

On integrating, we get

`int "dp"/"p" = "k" int "dt"`

∴ log p = kt + c

Initially, i.e., when t = 0, let p = N

∴ log N = k × 0 + c

∴ c = `square`

When t = 80, p = 2N

∴ log 2N = 80k + log N

∴ log 2N – log N = 80k

∴ `log ((2"N")/"N")` = 80k

∴ log (2) = 80k

∴ k = `square`

∴ p = 3N, then t = ?

∴ log p = `log2/80  "t" + log "N"`

∴ log 3N – log N = `square`

∴ t = `square` = `square` years


The rate of increase of bacteria in a certain culture is proportional to the number present. If it doubles in 7 hours, then in 35 hours its number would be ______.


If a curve y = f(x) passes through the point (1, - 1) and satisfies the differential equation, y (1 + xy) dx = x dy, then `f(-1/2)` is equal to ______ 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×