Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
According to the given condition,
`(dV)/dt ∝ 1/(t+1)^2`
∴ `(dV)/dt = (-k)/(t+1)^2` …[Negative sign indicates disintegration]
∴ `dV = (-kdt)/(t+1)^2`
Integrating on both sides, we get
`int dV = - k int dt/(t+1)^2`
∴ `V = k/(t+1) + c`
when t = 0, V = 8,00,000
∴ `8,00,000 = k/((0+1)) +c`
∴ 8,00,000 = k + c …(i)
when t = 1, V = 7,00,000
∴ `7,00,000 = k /((1 +1)) + c`
∴ `7,00,000 = k/ 2 + c` …(ii)
From (i) – (ii), we get
`1,00,000 = k /2`
∴ k = 2,00,000 …(iii)
Substituting (iii) in (i), we get
c = 6,00,000 …(iv)
when t = 6, we get
`V = k/((6+ 1)) + c`
=`(2,00,000 )/7 + 6,00,000`
= 6,28,571.4286
≈6,28,571
∴ Value of the machine after 6 years is ₹ 6,28,571.
APPEARS IN
संबंधित प्रश्न
The rate of disintegration of a radioactive element at any time t is proportional to its mass at that time. Find the time during which the original mass of 1.5 gm will disintegrate into its mass of 0.5 gm.
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.
Find the population of a city at any time t, given that the 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 30,000 to 40,000.
A body cools according to Newton’s law from 100° C to 60° C in 20 minutes. The temperature of the surrounding being 20° C. How long will it take to cool down to 30° C?
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.
Radium decomposes at the rate proportional to the amount present at any time. If p percent of the amount disappears in one year, what percent of the amount of radium will be left after 2 years?
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.
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)`
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?
Choose the correct alternative:
The integrating factor of `("d"y)/("d"x) + y` = e–x is
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 ______
Integrating factor of `("d"y)/("d"x) + y/x` = x3 – 3 is ______
The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and present population is 1 lac., when will the city have population 4,00,000?
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 = 100000
∴ log 100000 = k × 0 + c
∴ 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
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 = 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
If the lengths of the transverse axis and the latus rectum of a hyperbola are 6 and `8/3` respectively, then the equation of the hyperbola 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 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 ______.
The length of the perimeter of a sector of a circle is 24 cm, the maximum area of the sector is ______.
The rate of growth of bacteria is proportional to the number present. If initially, there are 1000 bacteria and the number doubles in 1 hour, the number of bacteria after `21/2` hours will be ______. `(sqrt(2) = 1.414)`
If `(dy)/(dx)` = y + 3 > 0 and y = (0) = 2, then y (in 2) is equal to ______.
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?
