Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
Let ‘x’ be the population at time ‘t’ years.
∴ `dx/dt prop x`
∴ `dx/dt = kx`, where k is the constant of proportionality.
∴ `dx/x = kdt`
Integrating on both sides, we get
`int dx/x = int kdt`
∴ logx = kt + c …(i)
When t = 0, x = 50,000
∴ log(50,000) = k(0) + c
∴ c = log(50,000)
∴ logx = kt + log(50,000) ...(ii) [From (i)]
When t = 25, x = 1,00,000, we have
log(1,00,000) = 25k + log(50000)
∴ log2 = 25k
∴ k = `1/25 log 2` …(iii)
When x = 4,00,000, we get
`log(4,00,000) = [1/25 log (2)]t + log(50,000)` ...[From (ii) and (iii)]
∴ `log[400000/50000] = t/25 log2`
∴ log8 = `t/25 log2`
∴ 3log2 = `t/25 log2`
∴ 3 = `t/25`
∴ t = 75 years.
Thus, the population will be 4,00,000 after 75 – 25 = 50 years from present date.
संबंधित प्रश्न
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
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)
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after `2 1/2` hours.
[Take `sqrt2 = 1.414`]
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.
A right circular cone has height 9 cm and radius of the base 5 cm. It is inverted and water is poured into it. If at any instant the water level rises at the rate of `(pi/"A")`cm/sec, where A is the area of the water surface A at that instant, show that the vessel will be full in 75 seconds.
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.
Choose the correct option from the given alternatives:
The decay rate of certain substances is directly proportional to the amount present at that instant. Initially there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is
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 normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.
Choose the correct alternative:
The solution of `("d"y)/("d"x) + x^2/y^2` = 0 is
The solution of `("d"y)/("d"x) + y` = 3 is ______
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution: Let x be the number of bacteria in the culture at time t.
Then the rate of increase of x is `"dx"/"dt"` which is proportional to x.
∴ `"dx"/"dt" ∝ "x"`
∴ `"dx"/"dt"` = kx, where k is a constant
∴ `square`
On integrating, we get
`int "dx"/"x" = "k" int "dt"`
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x0
∴ log x0 = k × 0 + c
∴ c = `square`
∴ log x = kt + log x0
∴ log x - log x0 = kt
∴ `log ("x"/"x"_0)`= kt ......(1)
Since the number doubles in 4 hours, i.e. when t = 4,
x = 2x0
∴ `log ((2"x"_0)/"x"_0)` = 4k
∴ k = `square`
∴ equation (1) becomes, `log ("x"/"x"_0) = "t"/4` log 2
When t = 12, we get
`log ("x"/"x"_0) = 12/4` log 2 = 3 log 2
∴ `log ("x"/"x"_0)` = log 23
∴ `"x"/"x"_0 = 8`
∴ x = `square`
∴ number of bacteria will be 8 times the original number in 12 hours.
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 = ec
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
∴ ek = `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 rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are 27 gm of certain substance and 3 h later it is found that 8 gm are left, then the amount left after one more hour is ______.
If r is the radius of spherical balloon at time t and the surface area of balloon changes at a constant rate K, then ______.
The population of a town increases at a rate proportional to the population at that time. If the population increases from 26,000 to 39,000 in 50 years, then the population in another 25 years will be ______ `(sqrt(3/2) = 1.225)`
Let the population of rabbits surviving at a time t be governed by the differential equation `(dp(t))/dt = 1/2p(t) - 200`. If p(0) = 100, then p(t) equals ______
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 ______
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, complete the following activity to find the number of times the bacteria are increased in 12 hours.
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution:
Let N be the number of bacteria present at time ‘t’.
Since the rate of increase of N is proportional to N, the differential equation can be written as –
`(dN)/dt αN`
∴ `(dN)/dt` = KN, where K is constant of proportionality
∴ `(dN)/N` = k . dt
∴ `int 1/N dN = K int 1 . dt`
∴ log N = `square` + C ...(1)
When t = 0, N = N0 where N0 is initial number of bacteria.
∴ log N0 = K × 0 + C
∴ C = log N0
Also when t = 4, N = 2N0
∴ log (2 N0) = K . 4 + `square` ...[From (1)]
∴ `log((2N_0)/N_0)` = 4K,
∴ log 2 = 4K
∴ K = `square` ...(2)
Now N = ? when t = 12
From (1) and (2)
log N = `1/4 log 2 . (12) + log N_0`
log N – log N0 = 3 log 2
∴ `log(N_0/N_0)` = `square`
∴ N = 8 N0
∴ Bacteria are increased 8 times in 12 hours.
Bacteria increase at the rate proportional to the number of bacteria present. If the original number N doubles in 3 hours, find in how many hours the number of bacteria will be 4N?
