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Question
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.
Solution
Let r be the radius of the water surface and h be the height of the water at time t.
∴ area of the water surface A = πr2 sq cm.
Since the height of the right circular cone is 9 cm and radius of the base is 5 cm.
`"r"/"h" = 5/9` ∴ r = `5/9"h"`
∴ area of water surface, i.e. A = `pi(5/9 "h")^2`
∴ A = `(25 pi"h"^2)/81` .....(1)
The water level, i.e. the rate of change of h is `"dh"/"dt"` rises at the rate of `(pi/"A")` cm/sec.
∴ `"dh"/"dt" = pi/"A" = (pi xx 81)/(25 pi "h"^2)` ....[By (1)]
∴ `"dh"/"dt" = 81/(25 "h"^2)`
∴ `"h"^2 "dh" = 81/25 "dt"`
On integrating, we get
`int "h"^2 "dh" = 81/25 int "dt" + "c"`
∴ `"h"^3/3 = 81/25 * "t + c"`
Initially, i.e. when t = 0, h = 0
∴ 0 = 0 + c ∴ c = 0
∴ `"h"^3/3 = 81/25 "t"`
When the vessel will be full, h = 9
∴ `(9)^3/3 = 81/25 xx "t"`
∴ t = `(81 xx 9 xx 25)/(3 xx 81) = 75`
Hence, the vessel will be full in 75 seconds.
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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`
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∴ 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?
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 1,00,000, when will the city have population 4,00,000?
Let ‘p’ be the population at time ‘t’ years.
∴ `("dp")/"dt" prop "p"`
∴ Differential equation can be written as `("dp")/"dt" = "kp"`
where k is constant of proportionality.
∴ `("dp")/"p" = "k.dt"`
On integrating we get
`square` = kt + c ...(i)
(i) Where t = 0, p = 1,00,000
∴ from (i)
log 1,00,000 = k(0) + c
∴ c = `square`
∴ log `("p"/(1,00,000)) = "kt"` ...(ii)
(ii) When t = 25, p = 2,00,000
as population doubles in 25 years
∴ from (ii) log2 = 25k
∴ k = `square`
∴ log`("p"/(1,00,000)) = (1/25log2).t`
(iii) ∴ when p = 4,00,000
`log ((4,00,000)/(1,00,000)) = (1/25log2).t`
∴ `log 4 = (1/25 log2).t`
∴ t = `square ` years
