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Question
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.
Solution
Let P(x, y) be a point on the curve y = f(x).
Then slope of the normal to the curve is `- 1/("dy"/"dx")`
∴ equation of the normal is
`("Y - y") = - 1/("dy"/"dx") ("X - x")`
∴ `("Y - y") "dy"/"dx" = - ("X - x")`
∴ `("Y - y") "dy"/"dx" + ("X - x") = 0`
Since, this normal passes through (2, 0), we get
`(0 - "y") "dy"/"dx" + (2 - "x") = 0`
∴ `- "y" "dy"/"dx" = "x - 2"`
∴ `- "y" "dy" = ("x - 2")"dx"`
Integrating both sides, we get
`int - "y" "dy" = int ("x - 2")"dx"`
∴ `- "y"^2/2 = "x"^2/2 - 2"x" + "c"_1`
∴ `"x"^2/2 + "y"^2/2 - 2"x" + "c"_1 = 0`
∴ x2 + y2 - 4x + 2c1 = 0
∴ x2 + y2 = 4x - 2c1
∴ x2 + y2 = 4x + c, where c = - 2c1
This is the general equation of the curve.
Since, the required curve passed through the point (2, 3), we get
22 + 32 = 4(2) + c
∴ c = 5
∴ equation of the required curve is
x2 + y2 = 4x + 5.
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Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.
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∴ `"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 = 100000
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∴ c = `square`
∴ log p = kt + log 100000
∴ log p – log 100000 = kt
∴ `log ("P"/100000)` = kt ......(i)
Since the number doubled in 25 years, i.e., when t = 25, p = 200000
∴ `log (200000/100000)` = 25k
∴ k = `square`
∴ equation (i) becomes, `log("p"/100000) = square`
When p = 400000, then find t.
∴ `log(400000/100000) = "t"/25 log 2`
∴ `log 4 = "t"/25 log 2`
∴ t = `25 (log 4)/(log 2)`
∴ t = `square` years
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∴ `"dp"/"dt" prop "p"`
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∴ `("d"x)/"dt" ∝ x`
∴ `("d"x)/"dt"` = kx, where k is a constant
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On integrating, we get
`int ("d"x)/x = "k" int "dt"`
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∴ x = aekt where a = ec
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∴ 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
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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`
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?
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
