Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.8 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.8

Question 1.

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours, find how many bacteria will be present after 10 hours?

Solution:

Let x denote the number of bacteria at time t hours.

Given = \(\frac { dx }{ dt }\) = kx hence \(\frac { dx }{ x }\) = kdt

∴ x = C e^{kt}

Suppose x = x_{0} at time t = 0

x_{0} = C e^{k(0)} = C e° = C

∴ C = x_{0}

Hence x = x_{0} e^{kt}

At time 5, x = 3x_{0}

(∵ Number triple in 5 hrs)

∴ Hence 3x_{0} = x_{0} e^{5k}

∴ e^{5k} = 3

when t = 10, x = x_{0} e^{10k} = x_{0} (e^{5k})²

= x_{0} 3² = 9x_{0}

∴ After 10 hours, the number of bacteria as 9 times the original number of bacteria.

Question 2.

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 3,00,000 to 4,00,000.

Solution:

Let P denote the population of a city

Given \(\frac { dP }{ dt }\) = kP

This equation can be written as

\(\frac { dP }{ P }\) = kdt

Taking integration on both sides, we get

∫ \(\frac { dP }{ P }\) = k∫ dt

log P = kt + log c

log P – log c = kt

log (\(\frac { P }{ c }\)) = kt

\(\frac { P }{ c }\) = e^{kt}

P = ce^{kt} …….. (1)

Initial condition:

Given when t = 0 ; P = 3,00,000

Equation (1) becomes,

3.0. 000 = ce^{k(0) }= ce°

3,00,000 = c [∵ e° = 1]

∴ (1) ⇒ P = 3,00,000 e^{kt} ………. (2)

Again when t = 40; P = 4,00,000

Equation (2) becomes,

4,00,000 = 3,00,000 e^{40k}

\(\frac { 4,00,000 }{ 3,00,000 }\) = e^{40k}

\(\frac { 4 }{ 3 }\) = e^{40k}

Taking log,

log(\(\frac { 4 }{ 3 }\)) = 40k

k = \(\frac { 1 }{ 40 }\) log \(\frac { 4 }{ 3 }\)

k = log(\(\frac { 4 }{ 3 }\))^{\(\frac { 1 }{ 40 }\)} [∵ n log m = log mⁿ]

Substituting k values in equation (2), we get

P = 3,00,000 e^{log(\(\frac { 4 }{ 3 }\))\(\frac { 1 }{ 40 }\)}

P = 3,00,000 (\(\frac { 4 }{ 3 }\))^{\(\frac { 1 }{ 40 }\)} [∵ e^{log a x} = a^{x}]

Question 3.

The equation of electromotive force for an electric circuit containing resistance and self-inductance is E = Ri + L\(\frac { di }{ dt }\), where E is the electromotive force is given to the circuit, R the resistance and L, the coefficient of induction. Find the current i at time t when E = 0.

Solution:

This is a linear differential equation.

Integrating factor I.F = e^{∫ \(\frac { R }{ L }\)dt} = e^{\(\frac { R }{ L }\)t}

Its solution is given by

Question 4.

The engine of a motorboat moving at 10 m/s is shut off. Given that the retardation at any subsequent time (after shutting off the engine) equal to the velocity at that time. Find the velocity after 2 seconds of switching off the engine.

Solution:

Let v be the velocity

Given the engine of a motorboat moving 10 m/s.

After that the engine is shut off then the acceleration is negative.

So it be \(\frac { -dv }{ dt }\)

i.e., \(\frac { dv}{ dt}\) = -v

The equation can be written as taking integration on both sides, we get

\(\frac { dv}{ dt}\) = -dt

∫ \(\frac { dv}{ v }\) = ∫-dt

log v = -t + log c

log v – log c = -t

log (\(\frac { v }{ c }\)) = -t

\(\frac {v}{c}\) = e^{-t}

v = ce^{-t} ……….. (1)

Initial condition

Given that when t = 0, v = 10 m/sec i

substituting in equation (1), we get !

10 = ce^{-0}

10 = ce°

10 = c

∴ c = 10

(1) ⇒ v = 10e^{-t}

when t = 2 find v

v = 10 e^{-2}

v = \(\frac {10}{e^2}\)

The velocity after 2 seconds is \(\frac {10}{e^2}\)

i.e., v = \(\frac {10}{e^2}\)

Question 5.

Suppose a person deposits 10,000 Indian rupees in a bank account at the rate of 5% per annum compounded continuously. How much money will be in his bank account 18 months later?

Solution:

Let P be the principal amount

Given Rate of interest = 5% per annum.

∴ \(\frac {dp}{dt}\) = p(\(\frac {5}{100}\)) = 0.05P

The equation can be written as,

\(\frac {dP}{P}\) = 0.05 dt

Taking Integration on both sides, we get

∫\(\frac {dP}{P}\) = 0.051 ∫dt

log P = 0.05 t + log c

log P – log C = 0.05t

log (\(\frac {P}{C}\)) = 0.05 t

\(\frac {P}{C}\) = e^{0.05t}

P = C e^{0.05t} ………. (1)

Initial condition:

Given when t = 0; P = 10,000

Substituting these values in equation (1), we get

P = C e^{0.05t}

10,000 = C e^{0.05 (0)}

10,000 = C e°

C = 10,000

∴ Substituting the C value in equation (1), we get

P = 10,000 e^{0.05t} ……. (2)

When t= 18 months = 1\(\frac {1}{2}\)yr =3/2 years, we get

(2) ⇒ P = 10,000 e^{0.05 (3/2)}

P = 10,000 e^{0.075}

The amount in a bank account be

P = 10,000 e^{0.075}

Question 6.

Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. In a certain sample, 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. What percentage of the original radioactive nuclei will remain after 1000 years?

Solution:

Let N be the sample radioactive nuclei at any time t & N_{0} be the radioactive nuclei at the initial time.

Given \(\frac {dN}{dt}\) = -kN

Where k > 0 is a constant

The equation can be written as

\(\frac {dN}{N}\) = -kdt

Taking integration on both sides, we get

∫\(\frac {dN}{N}\) = ∫-kdt

log N = -kt + log C

log N – log C = -kt

log(\(\frac {N}{C}\)) = -kt

\(\frac {N}{C}\) = e^{-kt}

N = Ce^{-kt} ……… (1)

Initial condition:

when t = 0 we have N = N_{0}

N_{0} = Ce^{-k(0)}

N_{0} = Ce^{0}

N_{0} = C

Substituting C value in equation (1), we get ;

N = N_{0} e^{-kt} ……… (2)

Given: In a certain sample 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years be when t = p_{0}

Equation (2) becomes

The percentage of the original radioactive nuclei of remain after 1000 years is

\(\frac {N}{N_0}\) × 100 = (\(\frac {9}{10}\))^{10} × 100 = \(\frac {9^{10}}{10^{10}}\) × 10²

= \(\frac {9^{10}}{10^{8}}\) %

Question 7.

Water at temperature 100°C cools in 10 minutes to 80° C at a room temperature of 25°C. Find

(i) The temperature of the water after 20 minutes

(ii) The time when the temperature is 40° C

[log, \(\frac {11}{15}\) = -0.3101; log_{e} 5 = 1.6094]

Solution:

We apply “Newton’s law of cooling” while states that the rate of decrease of the temperature of a body is proportional to the difference between the temperature of the body and that of the medium.

\(\frac {dT}{dt}\) = -k(T – T_{0})

Where T is the temperature of the body at time t & T_{0} the constant temperature of the medium.

Thus \(\frac {dT}{dt}\) = -k(T – 25) or

\(\frac {dT}{T-25}\) = -kdt

On Integrating equation (1), we get

∫\(\frac {dT}{T-25}\) = ∫-kdt

log (T – 25) = -kt + c_{1}

Now T = 100, t = 0

log (100 – 25) = -k (0) + c_{1}

∴ log 75 = c_{1}

Substituting c_{1} value in equation (1), we get

log (T – 25) = -kt + log 75

∵ log m – log n = log m/n

log \(\frac {T-25}{75}\) = kt ……. (2)

Also T = 80 when t = 10 minutes

Substituting k value in equation (2), we get

T = 75 (0.5378) + 25

T = 40.33 + 25

T = 65.33°C

Question 8.

At 10.00 A.M. a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby kitchen counter to cool. At this instant, the temperature of the coffee was 180° F and 10 minutes later it was 160° F. Assume that the constant temperature of the kitchen was 70° F.

(i) What was the temperature of the coffee at 10.15 AM?

The woman likes to drink coffee when its temperature is between 130° F and 140° F. Between what times should she have drunk the coffee?

Solution:

Let T be the temperature of a coffee at time t.

T_{k} be the temperature of the kitchen

By Newton’s law of cooling

Initial condition:

when t = 0; T = 180°F

180 = ce^{k(0)} + 70

180 = ce° + 70

180 – 70 = c

∴ c = 110°

Substituting c value in equation (1), we get

T = ce^{+kt} + 70

T = 100 e^{kt} + 70 …….. (2)

Second condition:

when t = 10, T = 160

(2) ⇒ 160 = 110 e^{10k} + 70

160 – 70= 110 e^{10k}

T = 81.33 + 70

T = 151.3 F

∴ The temperature of the coffee at 10.15 A.M is 151.3° F

(ii) Woman’s like to drink a coffee between 130°F and 140°F.

(a) when T = 130°F

(2) ⇒ T = 110 e^{kt} + 70

130 – 70 = 110 e^{kt}

\(\frac { 60 }{ 110 }\)

\(\frac { 6 }{ 11 }\) = e^{kt}

b) When T = 140°

(2) ⇒ T = 110e^{kt} + 70

160 – 70 = 110 e^{kt}

t = 22.6 min

She drinks coffee between 10.22 & 10.30 approximately.

Question 9.

A pot of boiling water at 100° C is removed from a stove at time t = 0 and left to cool in the kitchen. After 5 minutes, the water temperature has decreased to 80° C and another 5 minutes later it has dropped to 65° C. Determine the temperature of the kitchen.

Solution:

Let T be the temperature of the boiling water.

T_{m} is the temperature of the kitchen.

By Newton’s law of cooling, we get

\(\frac { dT }{ dt }\) = k (T – T_{m})

The equation can be written as

Taking Integration on both sides, we get

(a – b)² = a² + b² – 2ab

a = 80 b = T_{m}

6400 + T\(_{m}^{2}\) – 160 T_{m} = 6500 – 100 T_{m} – 65 T_{m} + T\(_{m}^{2}\)

6400 – 6500 = 160 T_{m} – 165 T_{m}

– 100 = – 5T_{m}

T_{m} = \(\frac { 100 }{ 5 }\)

T_{m} = 20°C

Hence the temperature of the kitchen be 20° C

Question 10.

A tank initially contains 50 litres of pure water. Starting at time t = 0 a brine containing 2 grams of dissolved salt per litre flows into the tank at the rate of 3 litres per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time t > 0.

Solution:

Let x be the amount of salt in the tank at time t.

∴ \(\frac { dx }{ dt }\) = inflow rate – outflow rate ……… (1)

Given 2 grams of dissolved salt per litre flows into the tank at the rate of 3 litres per minute.

(i.e) inflow rate contain = 6 gram salt

[∵ for one litre = 2 gram salt]

[for 3 litre = 6 gram salt]

tank contain 50 litres of water

∴ out flow of rate of salt = \(\frac { 3x }{ 50 }\)

substitute inflow rate and outflow rate in equation (1), we get