Samacheer Kalvi Class 12 Maths Solutions Chapter 10 Ordinary Differential Equations Exercise 10.8

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?
Answer: Let \( x \) be the number of bacteria at time \( t \) hours.
Given that the rate of increase in bacteria is proportional to the number present, we can write:
\( \frac { dx }{ dt } = kx \)
Now, we rearrange the equation to separate variables:
\( \frac { dx }{ x } = k \, dt \)
Next, we integrate both sides:
\[ \int \frac { dx }{ x } = \int k \, dt \]
\( \log x = kt + C \)
We can write \( C \) as \( \log C_1 \), so:
\( \log x = kt + \log C_1 \)
\( \log x - \log C_1 = kt \)
\( \log \left( \frac { x }{ C_1 } \right) = kt \)
This means:
\( x = C_1 e^{kt} \)
Let \( x_0 \) be the number of bacteria at time \( t = 0 \). Substituting these values:
\( x_0 = C_1 e^{k(0)} \)
\( x_0 = C_1 e^0 \)
\( x_0 = C_1 \)
So, the equation becomes:
\( x = x_0 e^{kt} \)
We are given that the number of bacteria triples in 5 hours. So, when \( t = 5 \) hours, \( x = 3x_0 \).
\( 3x_0 = x_0 e^{k(5)} \)
\( 3 = e^{5k} \)
We need to find the number of bacteria after 10 hours. So, when \( t = 10 \):
\( x = x_0 e^{k(10)} \)
We can write \( e^{k(10)} \) as \( (e^{5k})^2 \).
\( x = x_0 (e^{5k})^2 \)
Since we found \( e^{5k} = 3 \), we substitute this value:
\( x = x_0 (3)^2 \)
\( x = x_0 \times 9 \)
\( x = 9x_0 \)
Therefore, after 10 hours, the number of bacteria will be 9 times the original number.
In simple words: The bacteria grow in a way that their growth rate depends on how many there already are. Because they tripled in 5 hours, after another 5 hours (total 10 hours), they will have multiplied by 3 again, making the total number 9 times the start.

🎯 Exam Tip: Always clearly define your variables, especially initial conditions (like \( x_0 \) for initial quantity) and the time unit used in the problem.

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.
Answer: Let \( P \) denote the population of the city at time \( t \).
Given that the rate of increase of population is proportional to the population at that instant, we can write:
\( \frac { dP }{ dt } = kP \)
To solve this differential equation, we separate the variables:
\( \frac { dP }{ P } = k \, dt \)
Next, we integrate both sides:
\[ \int \frac { dP }{ P } = \int k \, dt \]
\( \log P = kt + \log C \)
This can be written as:
\( \log P - \log C = kt \)
\( \log \left( \frac { P }{ C } \right) = kt \)
Exponentiating both sides, we get:
\( \frac { P }{ C } = e^{kt} \)
So, the population equation is:
\( P = Ce^{kt} \quad \ldots (1) \)
Now, we use the initial condition: when \( t = 0 \), the population \( P = 3,00,000 \).
Substitute these values into equation (1):
\( 3,00,000 = Ce^{k(0)} \)
\( 3,00,000 = Ce^0 \)
Since \( e^0 = 1 \):
\( 3,00,000 = C \)
Substitute the value of \( C \) back into equation (1):
\( P = 3,00,000 e^{kt} \quad \ldots (2) \)
We are also given that in a period of 40 years, the population increased from 3,00,000 to 4,00,000. So, when \( t = 40 \) years, \( P = 4,00,000 \).
Substitute these values into equation (2):
\( 4,00,000 = 3,00,000 e^{40k} \)
Divide both sides by 3,00,000:
\( \frac { 4,00,000 }{ 3,00,000 } = e^{40k} \)
\( \frac { 4 }{ 3 } = e^{40k} \)
Now, we take the natural logarithm of both sides to find \( k \):
\( \log \left( \frac { 4 }{ 3 } \right) = 40k \)
\( k = \frac { 1 }{ 40 } \log \left( \frac { 4 }{ 3 } \right) \)
This can also be written as \( k = \log \left( \frac{4}{3} \right)^{\frac{1}{40}} \).
Finally, substitute the expression for \( k \) back into equation (2) to find the population \( P \) at any time \( t \):
\( P = 3,00,000 e^{\left( \frac{1}{40} \log \left( \frac{4}{3} \right) \right) t} \)
\( P = 3,00,000 e^{\log \left( \frac{4}{3} \right)^{\frac{t}{40}}} \)
Using the property \( e^{\log A} = A \):
\( P = 3,00,000 \left( \frac{4}{3} \right)^{\frac{t}{40}} \)
This equation gives the population of the city at any time \( t \).
In simple words: When something grows proportionally to its current size, like a city's population, we use a special math formula. We used the starting population and how much it grew over 40 years to find the growth rate. Then we made a general formula to know the population at any future time.

🎯 Exam Tip: Remember that "rate of increase is proportional to the number present" always leads to an exponential growth model \( P = Ce^{kt} \).

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.
Answer: First, let's visualize the circuit described:

The given equation for electromotive force (EMF) in an RL circuit is:
\( E = Ri + L \frac { di }{ dt } \)
We need to find the current \( i \) at time \( t \) when \( E = 0 \).
Substitute \( E = 0 \) into the equation:
\( 0 = Ri + L \frac { di }{ dt } \)
Rearrange the equation to isolate the derivative term:
\( L \frac { di }{ dt } = -Ri \)
Now, separate the variables \( i \) and \( t \):
\( \frac { di }{ i } = -\frac { R }{ L } \, dt \)
Integrate both sides of the equation:
\[ \int \frac { di }{ i } = \int -\frac { R }{ L } \, dt \]
\( \log i = -\frac { R }{ L } t + \log C \)
Here, \( \log C \) is the integration constant.
Rearrange the logarithmic terms:
\( \log i - \log C = -\frac { R }{ L } t \)
\( \log \left( \frac { i }{ C } \right) = -\frac { R }{ L } t \)
Exponentiate both sides:
\( \frac { i }{ C } = e^{-\frac { R }{ L } t} \)
Finally, solve for the current \( i \):
\( i = C e^{-\frac { R }{ L } t} \)
This equation shows how the current decays over time in an RL circuit when the external EMF is removed (i.e., \( E=0 \)). The constant \( C \) depends on the initial current in the circuit.
In simple words: We used a special formula for how current changes in a circuit with a resistor and an inductor when the power is turned off. By doing some math steps, we found that the current will slowly get smaller over time, following an exponential decay pattern.

🎯 Exam Tip: Recognizing that \( E = Ri + L \frac { di }{ dt } \) is a first-order linear differential equation is key. When \( E=0 \), it becomes a homogeneous equation, leading to an exponential decay solution.

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.
Answer: Let \( v \) be the velocity of the motorboat at time \( t \).
The problem states that the engine is shut off when the motorboat is moving at 10 m/s. This means the initial velocity is \( v_0 = 10 \) m/s at \( t = 0 \).
Retardation is the negative of acceleration. We are given that the retardation is equal to the velocity at that time. So, if acceleration is \( \frac{dv}{dt} \), then retardation is \( -\frac{dv}{dt} \).
Thus, we have:
\( -\frac { dv }{ dt } = v \)
This can be rewritten as:
\( \frac { dv }{ dt } = -v \)
To solve this differential equation, we separate the variables:
\( \frac { dv }{ v } = -dt \)
Now, integrate both sides:
\[ \int \frac { dv }{ v } = \int -dt \]
\( \log v = -t + \log C \)
Rearrange the logarithmic terms:
\( \log v - \log C = -t \)
\( \log \left( \frac { v }{ C } \right) = -t \)
Exponentiate both sides to solve for \( v \):
\( \frac { v }{ C } = e^{-t} \)
So, the velocity equation is:
\( v = Ce^{-t} \quad \ldots (1) \)
Now, we use the initial condition: when \( t = 0 \), \( v = 10 \) m/s.
Substitute these values into equation (1):
\( 10 = Ce^{-0} \)
\( 10 = Ce^0 \)
Since \( e^0 = 1 \):
\( 10 = C \)
Substitute the value of \( C \) back into equation (1):
\( v = 10e^{-t} \)
We need to find the velocity after 2 seconds. So, substitute \( t = 2 \) into the equation:
\( v = 10e^{-2} \)
This can also be written as:
\( v = \frac {10}{e^2} \)
Therefore, the velocity of the motorboat after 2 seconds is \( \frac{10}{e^2} \) m/s.
In simple words: The boat slows down at a rate equal to its current speed. Starting at 10 m/s, after 2 seconds, its speed will be 10 divided by \( e \) squared. This is because when something changes at a rate proportional to its own value, it usually involves the number \( e \).

🎯 Exam Tip: Remember that "retardation" means negative acceleration. If the retardation is proportional to velocity, it leads to an exponential decay model for velocity, similar to radioactive decay.

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?
Answer: Let \( P \) be the principal amount (money) in the bank account at time \( t \).
The initial deposit is Rs. 10,000.
The interest rate is 5% per annum, compounded continuously.
For continuous compounding, the rate of change of the principal amount is proportional to the principal itself:
\( \frac { dP }{ dt } = P \left( \frac { 5 }{ 100 } \right) \)
\( \frac { dP }{ dt } = 0.05P \)
To solve this differential equation, we separate the variables:
\( \frac { dP }{ P } = 0.05 \, dt \)
Now, integrate both sides:
\[ \int \frac { dP }{ P } = \int 0.05 \, dt \]
\( \log P = 0.05t + \log C \)
Rearrange the logarithmic terms:
\( \log P - \log C = 0.05t \)
\( \log \left( \frac { P }{ C } \right) = 0.05t \)
Exponentiate both sides:
\( \frac { P }{ C } = e^{0.05t} \)
So, the principal amount at any time \( t \) is:
\( P = Ce^{0.05t} \quad \ldots (1) \)
Now, we use the initial condition: when \( t = 0 \) (initial deposit time), \( P = 10,000 \).
Substitute these values into equation (1):
\( 10,000 = Ce^{0.05(0)} \)
\( 10,000 = Ce^0 \)
Since \( e^0 = 1 \):
\( 10,000 = C \)
Substitute the value of \( C \) back into equation (1):
\( P = 10,000 e^{0.05t} \quad \ldots (2) \)
We need to find the amount of money in the account 18 months later. First, convert 18 months to years:
\( t = 18 \text{ months} = \frac{18}{12} \text{ years} = 1.5 \text{ years} = \frac{3}{2} \text{ years} \)
Now, substitute \( t = 1.5 \) into equation (2):
\( P = 10,000 e^{0.05 (1.5)} \)
\( P = 10,000 e^{0.075} \)
Therefore, the amount of money in the bank account 18 months later will be \( \text{Rs. } 10,000 e^{0.075} \).
In simple words: When money grows with "continuous compounding," it means it's always earning interest, even on the smallest amount of time. We used a special formula to calculate that after 18 months, the initial Rs. 10,000 will become Rs. 10,000 multiplied by \( e \) raised to the power of 0.075.

🎯 Exam Tip: Continuous compounding problems always use the formula \( P(t) = P_0 e^{rt} \), where \( P_0 \) is the initial principal, \( r \) is the annual interest rate (as a decimal), and \( t \) is the time in years.

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?
Answer: Let \( N \) be the number of radioactive nuclei present at time \( t \), and \( N_0 \) be the initial number of radioactive nuclei at \( t = 0 \).
The rate of decay of radioactive nuclei is proportional to the number of nuclei present. Since it's decay, we use a negative constant:
\( \frac { dN }{ dt } = -kN \)
Here, \( k \) is a positive constant (decay constant).
To solve this differential equation, we separate the variables:
\( \frac { dN }{ N } = -k \, dt \)
Integrate both sides:
\[ \int \frac { dN }{ N } = \int -k \, dt \]
\( \log N = -kt + \log C \)
Rearrange the logarithmic terms:
\( \log N - \log C = -kt \)
\( \log \left( \frac { N }{ C } \right) = -kt \)
Exponentiate both sides:
\( \frac { N }{ C } = e^{-kt} \)
So, the number of nuclei at any time \( t \) is:
\( N = Ce^{-kt} \quad \ldots (1) \)
Now, use the initial condition: when \( t = 0 \), \( N = N_0 \).
Substitute these values into equation (1):
\( N_0 = Ce^{-k(0)} \)
\( N_0 = Ce^0 \)
Since \( e^0 = 1 \):
\( N_0 = C \)
Substitute the value of \( C \) back into equation (1):
\( N = N_0 e^{-kt} \quad \ldots (2) \)
We are given that in 100 years, 10% of the original nuclei have disintegrated. This means 90% of the original nuclei remain.
So, when \( t = 100 \) years, \( N = 90\% \text{ of } N_0 = \frac{90}{100} N_0 = \frac{9}{10} N_0 \).
Substitute these values into equation (2):
\( \frac{9}{10} N_0 = N_0 e^{-k(100)} \)
Divide by \( N_0 \):
\( \frac{9}{10} = e^{-100k} \)
To find \( k \), take the natural logarithm of both sides:
\( \log \left( \frac{9}{10} \right) = -100k \)
\( k = -\frac{1}{100} \log \left( \frac{9}{10} \right) \)
Using the logarithm property \( \log(A/B) = -\log(B/A) \):
\( k = \frac{1}{100} \log \left( \frac{10}{9} \right) \)
Now, substitute the value of \( k \) back into equation (2):
\( N = N_0 e^{-\left( \frac{1}{100} \log \left( \frac{10}{9} \right) \right) t} \)
\( N = N_0 e^{\log \left( \frac{10}{9} \right)^{-\frac{t}{100}}} \)
\( N = N_0 e^{\log \left( \frac{9}{10} \right)^{\frac{t}{100}}} \)
Using the property \( e^{\log A} = A \):
\( \frac{N}{N_0} = \left( \frac{9}{10} \right)^{\frac{t}{100}} \quad \ldots (3) \)
We need to find the percentage of original nuclei remaining after 1000 years. So, substitute \( t = 1000 \) into equation (3):
\( \frac{N}{N_0} = \left( \frac{9}{10} \right)^{\frac{1000}{100}} \)
\( \frac{N}{N_0} = \left( \frac{9}{10} \right)^{10} \)
To express this as a percentage, multiply by 100:
Percentage remaining \( = \left( \frac{9}{10} \right)^{10} \times 100 \)
This can also be written as:
Percentage remaining \( = \frac{9^{10}}{10^{10}} \times 100 = \frac{9^{10}}{10^8} \% \)
In simple words: Radioactive decay means the number of atoms goes down over time. We found a formula for this decay. Since 10% decayed in 100 years (meaning 90% was left), we used this to figure out how much would be left after a much longer time, 1000 years. The answer is a very small percentage because it decays by a fixed fraction over equal time periods.

🎯 Exam Tip: For decay problems, the decay constant \( k \) is usually positive, so the term in the exponent should be \( -kt \). Ensure you correctly calculate the remaining fraction (100% - disintegrated percentage).

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 5 = 1.6094]
Answer: Let \( T \) be the temperature of the water at time \( t \), and \( T_0 \) be the constant room temperature.
Given room temperature \( T_0 = 25^\circ \text{C} \).
According to Newton's Law of Cooling, the rate of change of temperature is proportional to the difference between the object's temperature and the ambient temperature:
\( \frac { dT }{ dt } = -k(T - T_0) \)
Substitute \( T_0 = 25^\circ \text{C} \):
\( \frac { dT }{ dt } = -k(T - 25) \)
Separate the variables \( T \) and \( t \):
\( \frac { dT }{ T-25 } = -k \, dt \)
Integrate both sides:
\[ \int \frac { dT }{ T-25 } = \int -k \, dt \]
\( \log (T - 25) = -kt + C_1 \)
Now, let's find the constant \( C_1 \) using the initial condition: when \( t = 0 \), \( T = 100^\circ \text{C} \).
\( \log (100 - 25) = -k(0) + C_1 \)
\( \log 75 = C_1 \)
Substitute \( C_1 \) back into the equation:
\( \log (T - 25) = -kt + \log 75 \)
Rearrange the terms:
\( \log (T - 25) - \log 75 = -kt \)
\( \log \left( \frac { T-25 }{ 75 } \right) = -kt \quad \ldots (2) \)
Next, we use the condition that the water cools to 80°C in 10 minutes. So, when \( t = 10 \), \( T = 80^\circ \text{C} \).
Substitute these values into equation (2):
\( \log \left( \frac { 80-25 }{ 75 } \right) = -k(10) \)
\( \log \left( \frac { 55 }{ 75 } \right) = -10k \)
\( k = -\frac{1}{10} \log \left( \frac{55}{75} \right) \)
Using the logarithm property \( \log(A/B) = -\log(B/A) \):
\( k = \frac{1}{10} \log \left( \frac{75}{55} \right) \)
Simplify the fraction \( \frac{75}{55} = \frac{15}{11} \):
\( k = \frac{1}{10} \log \left( \frac{15}{11} \right) \)
Now, substitute the value of \( k \) back into equation (2):
\( \log \left( \frac { T-25 }{ 75 } \right) = -\left( \frac{1}{10} \log \left( \frac{15}{11} \right) \right) t \quad \ldots (3) \)

**(i) The temperature of the water after 20 minutes:**
Substitute \( t = 20 \) into equation (3):
\( \log \left( \frac { T-25 }{ 75 } \right) = -\left( \frac{1}{10} \log \left( \frac{15}{11} \right) \right) (20) \)
\( \log \left( \frac { T-25 }{ 75 } \right) = -2 \log \left( \frac{15}{11} \right) \)
Using logarithm properties \( -n \log A = n \log(1/A) = \log(1/A)^n \):
\( \log \left( \frac { T-25 }{ 75 } \right) = 2 \log \left( \frac{11}{15} \right) \)
\( \log \left( \frac { T-25 }{ 75 } \right) = \log \left( \frac{11}{15} \right)^2 \)
Exponentiate both sides:
\( \frac { T-25 }{ 75 } = \left( \frac{11}{15} \right)^2 \)
\( T - 25 = 75 \times \left( \frac{11}{15} \right)^2 \)
\( T = 75 \times \left( \frac{121}{225} \right) + 25 \)
\( T = \frac{121}{3} + 25 \)
\( T = 40.333 \ldots + 25 \)
\( T \approx 65.33^\circ \text{C} \)
Therefore, the temperature of the water after 20 minutes is approximately \( 65.33^\circ \text{C} \).

**(ii) The time when the temperature is 40° C:**
Substitute \( T = 40 \) into equation (3):
\( \log \left( \frac { 40-25 }{ 75 } \right) = -\left( \frac{1}{10} \log \left( \frac{15}{11} \right) \right) t \)
\( \log \left( \frac { 15 }{ 75 } \right) = -\frac{t}{10} \log \left( \frac{15}{11} \right) \)
\( \log \left( \frac { 1 }{ 5 } \right) = -\frac{t}{10} \log \left( \frac{15}{11} \right) \)
We know \( \log(1/5) = -\log 5 \). So:
\( -\log 5 = -\frac{t}{10} \log \left( \frac{15}{11} \right) \)
\( \log 5 = \frac{t}{10} \log \left( \frac{15}{11} \right) \)
\( t = 10 \frac{\log 5}{\log \left( \frac{15}{11} \right)} \)
Using the given values: \( \log \left( \frac{11}{15} \right) = -0.3101 \implies \log \left( \frac{15}{11} \right) = 0.3101 \); \( \log 5 = 1.6094 \).
\( t = 10 \frac{1.6094}{0.3101} \)
\( t = 10 \times 5.1899 \ldots \)
\( t \approx 51.90 \text{ minutes} \)
Therefore, the water will reach \( 40^\circ \text{C} \) after approximately 51.90 minutes.
In simple words: This problem uses Newton's Law of Cooling, which describes how hot things get cooler in a cooler environment. We first used the given information to find the rate at which the water cools down. Then, we used this rate to calculate its temperature after 20 minutes and how long it would take to reach a specific lower temperature.

🎯 Exam Tip: Remember to convert time units if necessary (e.g., minutes to hours or vice-versa) for consistency. Pay close attention to the sign of the constant \( k \) in Newton's Law of Cooling; it should be positive as written \( \frac{dT}{dt} = -k(T-T_0) \).

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?
Answer: Let \( T \) be the temperature of the coffee at time \( t \) (in minutes from 10:00 AM), and \( T_0 \) be the constant kitchen temperature.
Given kitchen temperature \( T_0 = 70^\circ \text{F} \).
According to Newton's Law of Cooling:
\( \frac { dT }{ dt } = k(T - T_0) \)
Substitute \( T_0 = 70^\circ \text{F} \):
\( \frac { dT }{ dt } = k(T - 70) \)
Separate the variables:
\( \frac { dT }{ T-70 } = k \, dt \)
Integrate both sides:
\[ \int \frac { dT }{ T-70 } = \int k \, dt \]
\( \log (T - 70) = kt + \log C \)
Rearrange the terms:
\( \log (T - 70) - \log C = kt \)
\( \log \left( \frac { T-70 }{ C } \right) = kt \)
Exponentiate both sides:
\( \frac { T-70 }{ C } = e^{kt} \)
So, the temperature equation is:
\( T = Ce^{kt} + 70 \quad \ldots (1) \)
Now, use the initial condition: at 10:00 AM, \( t = 0 \), and \( T = 180^\circ \text{F} \).
Substitute these values into equation (1):
\( 180 = Ce^{k(0)} + 70 \)
\( 180 = Ce^0 + 70 \)
\( 180 - 70 = C \)
\( C = 110 \)
Substitute the value of \( C \) back into equation (1):
\( T = 110e^{kt} + 70 \quad \ldots (2) \)
Next, we use the condition that 10 minutes later, the temperature was 160°F. So, when \( t = 10 \), \( T = 160^\circ \text{F} \).
Substitute these values into equation (2):
\( 160 = 110e^{k(10)} + 70 \)
\( 160 - 70 = 110e^{10k} \)
\( 90 = 110e^{10k} \)
\( \frac{90}{110} = e^{10k} \)
\( \frac{9}{11} = e^{10k} \)
From this, we can find \( e^k \):
\( e^k = \left( \frac{9}{11} \right)^{1/10} \quad \ldots (3) \)

**(i) What was the temperature of the coffee at 10.15 AM?**
This means \( t = 15 \) minutes (since 10:00 AM is \( t=0 \)).
Substitute \( t = 15 \) into equation (2):
\( T = 110e^{k(15)} + 70 \)
We can rewrite \( e^{k(15)} \) as \( (e^k)^{15} \). Substitute \( e^k \) from equation (3):
\( T = 110 \left( \left( \frac{9}{11} \right)^{1/10} \right)^{15} + 70 \)
\( T = 110 \left( \frac{9}{11} \right)^{15/10} + 70 \)
\( T = 110 \left( \frac{9}{11} \right)^{3/2} + 70 \)
\( T = 110 \times \frac{9 \sqrt{9}}{11 \sqrt{11}} + 70 \)
\( T = 110 \times \frac{9 \times 3}{11 \times \sqrt{11}} + 70 \)
\( T = 10 \times \frac{27}{\sqrt{11}} + 70 \)
\( T = \frac{270}{\sqrt{11}} + 70 \)
Using \( \sqrt{11} \approx 3.3166 \):
\( T \approx \frac{270}{3.3166} + 70 \)
\( T \approx 81.41 + 70 \)
\( T \approx 151.41^\circ \text{F} \)
Therefore, the temperature of the coffee at 10.15 AM was approximately \( 151.41^\circ \text{F} \).

**(ii) Between what times should she have drunk the coffee?**
The woman likes to drink coffee when its temperature is between 130°F and 140°F. We need to find the times \( t \) when the coffee is at these temperatures.

**(a) When \( T = 130^\circ \text{F} \):**
Substitute \( T = 130 \) into equation (2):
\( 130 = 110e^{kt} + 70 \)
\( 130 - 70 = 110e^{kt} \)
\( 60 = 110e^{kt} \)
\( \frac{60}{110} = e^{kt} \)
\( \frac{6}{11} = e^{kt} \)
Substitute \( e^k = \left( \frac{9}{11} \right)^{1/10} \) from equation (3):
\( \frac{6}{11} = \left( \left( \frac{9}{11} \right)^{1/10} \right)^t \)
\( \frac{6}{11} = \left( \frac{9}{11} \right)^{t/10} \)
Take the natural logarithm of both sides:
\( \log \left( \frac{6}{11} \right) = \frac{t}{10} \log \left( \frac{9}{11} \right) \)
\( t = 10 \frac{\log \left( \frac{6}{11} \right)}{\log \left( \frac{9}{11} \right)} \)
Using a calculator for logarithms (or values from a table, if provided, though not for these specific fractions in the question description):
\( \log(6/11) \approx \log(0.5454) \approx -0.606 \)
\( \log(9/11) \approx \log(0.8181) \approx -0.201 \)
\( t = 10 \frac{-0.606}{-0.201} \)
\( t \approx 10 \times 3.015 \)
\( t \approx 30.15 \text{ minutes} \)

**(b) When \( T = 140^\circ \text{F} \):**
Substitute \( T = 140 \) into equation (2):
\( 140 = 110e^{kt} + 70 \)
\( 140 - 70 = 110e^{kt} \)
\( 70 = 110e^{kt} \)
\( \frac{70}{110} = e^{kt} \)
\( \frac{7}{11} = e^{kt} \)
Substitute \( e^k = \left( \frac{9}{11} \right)^{1/10} \) from equation (3):
\( \frac{7}{11} = \left( \frac{9}{11} \right)^{t/10} \)
Take the natural logarithm of both sides:
\( \log \left( \frac{7}{11} \right) = \frac{t}{10} \log \left( \frac{9}{11} \right) \)
\( t = 10 \frac{\log \left( \frac{7}{11} \right)}{\log \left( \frac{9}{11} \right)} \)
Using a calculator for logarithms:
\( \log(7/11) \approx \log(0.6363) \approx -0.452 \)
\( \log(9/11) \approx \log(0.8181) \approx -0.201 \)
\( t = 10 \frac{-0.452}{-0.201} \)
\( t \approx 10 \times 2.248 \)
\( t \approx 22.48 \text{ minutes} \)
So, the coffee temperature is 140°F at approximately 22.48 minutes past 10:00 AM (around 10:22 AM), and it is 130°F at approximately 30.15 minutes past 10:00 AM (around 10:30 AM).
Therefore, the woman should drink the coffee between approximately 10:22 AM and 10:30 AM.
In simple words: This problem is about how hot coffee cools down. We first found a math rule for how its temperature drops over time, considering the kitchen's constant temperature. Then, we used this rule to find the coffee's temperature at 10:15 AM. Finally, we calculated the time range when the coffee would be at her preferred drinking temperature, which is between 10:22 AM and 10:30 AM.

🎯 Exam Tip: When using Newton's Law of Cooling, ensure you correctly identify the initial temperature, ambient temperature, and all given time-temperature pairs. Calculations involving fractional exponents often require logarithms or using a calculator for accuracy.

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.
Answer: Let \( T \) be the water temperature at time \( t \), and \( T_m \) be the constant temperature of the kitchen. According to Newton's Law of Cooling, the rate of temperature change is proportional to the difference between the object's temperature and the surrounding temperature. The differential equation is given by: \( \frac{dT}{dt} = k(T - T_m) \) Integrating this, we get: \( \int \frac{dT}{T - T_m} = \int k dt \) \( \log(T - T_m) = kt + \log C \) \( T - T_m = Ce^{kt} \) ... (1) We are given the initial conditions: 1. When \( t = 0 \), \( T = 100^\circ C \). Substituting into (1): \( 100 - T_m = Ce^{k(0)} \) \( 100 - T_m = C \) So, \( C = 100 - T_m \). Substituting C back into (1) gives: \( T - T_m = (100 - T_m)e^{kt} \) ... (2) 2. When \( t = 5 \) minutes, \( T = 80^\circ C \). Substituting into (2): \( 80 - T_m = (100 - T_m)e^{5k} \) \( e^{5k} = \frac{80 - T_m}{100 - T_m} \) ... (3) 3. When \( t = 10 \) minutes, \( T = 65^\circ C \). Substituting into (2): \( 65 - T_m = (100 - T_m)e^{10k} \) We know that \( e^{10k} = (e^{5k})^2 \). So, we can substitute (3) into this equation: \( \frac{65 - T_m}{100 - T_m} = \left(\frac{80 - T_m}{100 - T_m}\right)^2 \) \( \frac{65 - T_m}{100 - T_m} = \frac{(80 - T_m)^2}{(100 - T_m)^2} \) Multiply both sides by \( (100 - T_m)^2 \): \( (65 - T_m)(100 - T_m) = (80 - T_m)^2 \) Expand both sides: \( 6500 - 65T_m - 100T_m + T_m^2 = 6400 - 160T_m + T_m^2 \) Combine like terms: \( 6500 - 165T_m + T_m^2 = 6400 - 160T_m + T_m^2 \) Subtract \( T_m^2 \) from both sides: \( 6500 - 165T_m = 6400 - 160T_m \) Rearrange the terms to solve for \( T_m \): \( 6500 - 6400 = 165T_m - 160T_m \) \( 100 = 5T_m \) \( T_m = \frac{100}{5} \) \( T_m = 20 \) The temperature of the kitchen is \( 20^\circ C \). This problem shows how temperatures approach the surrounding temperature over time, a common application of exponential decay.In simple words: We used how fast the water cooled down over two different times to figure out the constant temperature of the kitchen. The kitchen temperature, which is \( 20^\circ C \), is the final temperature the water will reach if left there forever.

🎯 Exam Tip: When solving problems involving Newton's Law of Cooling, clearly define your variables and constants. Pay close attention to the initial conditions and time intervals, as these are critical for setting up and solving the differential equation correctly.

Question 10. A tank initially contains 50 litres of pure water. Starting at time t = 0 a brine 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.
Answer: Let \( x \) be the amount of salt (in grams) in the tank at time \( t \) (in minutes). The volume of water in the tank remains constant at 50 litres because the inflow and outflow rates are the same (3 litres per minute). The rate of change of salt in the tank is given by: \( \frac{dx}{dt} = \text{rate of salt inflow} - \text{rate of salt outflow} \) ... (1) From the problem's context (and implicitly, the solution provides the missing value), the brine contains 2 grams of dissolved salt per litre. Rate of salt inflow: Concentration of salt in inflow = 2 grams/litre Volume inflow rate = 3 litres/minute Salt inflow rate = \( (2 \text{ g/L}) \times (3 \text{ L/min}) = 6 \text{ g/min} \) Rate of salt outflow: The concentration of salt in the tank at time \( t \) is \( \frac{x}{50} \) grams/litre (amount of salt divided by total volume). Volume outflow rate = 3 litres/minute Salt outflow rate = \( (\frac{x}{50} \text{ g/L}) \times (3 \text{ L/min}) = \frac{3x}{50} \text{ g/min} \) Substitute these rates into equation (1): \( \frac{dx}{dt} = 6 - \frac{3x}{50} \) To solve this differential equation, we first simplify the right side: \( \frac{dx}{dt} = \frac{300 - 3x}{50} = \frac{3(100 - x)}{50} \) Separate the variables \( x \) and \( t \): \( \frac{dx}{100 - x} = \frac{3}{50} dt \) To integrate easily, we can write \( \frac{dx}{x - 100} = -\frac{3}{50} dt \). Integrate both sides: \( \int \frac{dx}{x - 100} = \int -\frac{3}{50} dt \) \( \log|x - 100| = -\frac{3}{50}t + K_1 \) (where \( K_1 \) is the integration constant) We can write \( \log(x - 100) = -\frac{3}{50}t + \log C \) (assuming \( x < 100 \), which is true as salt accumulates towards a maximum, so \( x-100 \) is negative and \( \log|x-100| = \log(100-x) \), but the source uses \( x-100 \), absorbing the negative sign into C, which is common). \( \log\left(\frac{x - 100}{C}\right) = -\frac{3}{50}t \) \( x - 100 = Ce^{-\frac{3}{50}t} \) Now, apply the initial condition: When \( t = 0 \), the tank contains pure water, so \( x = 0 \) grams of salt. \( 0 - 100 = Ce^{-\frac{3}{50}(0)} \) \( -100 = Ce^0 \) \( -100 = C \) Substitute the value of \( C \) back into the equation for \( x \): \( x - 100 = -100e^{-\frac{3}{50}t} \) \( x = 100 - 100e^{-\frac{3}{50}t} \) \( x = 100(1 - e^{-\frac{3}{50}t}) \) This equation describes the amount of salt in the tank at any time \( t > 0 \). Over time, as \( t \) approaches infinity, the exponential term \( e^{-\frac{3}{50}t} \) approaches 0, and the amount of salt \( x \) approaches 100 grams. This makes sense because the incoming brine has a concentration such that 50 litres would contain 100 grams of salt.In simple words: The tank starts with no salt. Salt water flows in, and mixed water flows out. We found a formula that tells us how much salt is in the tank at any moment. As time passes, the amount of salt in the tank gets closer to 100 grams.

🎯 Exam Tip: For mixing problems, remember to calculate the rate of change of the substance (salt in this case) by subtracting the outflow rate from the inflow rate. Always consider the initial conditions to find the constant of integration, which helps specify the unique solution for the problem.