What is Mixture Problems using Differential Equations? | Simple Explanation for Class 12

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What is the Concept of Mixture Problems using Differential Equations?

Grade Level:

Class 12

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Definition

What is it?

Mixture problems using differential equations help us understand how the amount of a substance changes in a container (like a tank) over time, especially when liquids are flowing in and out. It involves setting up an equation that describes the rate of change of the substance's quantity.

Simple Example

Quick Example

Imagine you have a big jug of water with some salt dissolved in it. If you continuously pour in fresh water and simultaneously drain out the salty mixture, the amount of salt in the jug will change. Differential equations help us calculate exactly how much salt is left after, say, 10 minutes.

Worked Example

Step-by-Step

Let's say a tank initially contains 100 liters of water with 10 kg of salt dissolved. Fresh water flows in at 5 liters/minute, and the well-mixed solution flows out at 5 liters/minute. How much salt is in the tank after 't' minutes?

Step 1: Define variables. Let A(t) be the amount of salt (in kg) at time t (in minutes).
---Step 2: Calculate the concentration of salt in the tank at time t. Since the volume is constant (100 liters in, 100 liters out), concentration C(t) = A(t) / 100 kg/liter.
---Step 3: Determine the rate of salt entering the tank. Fresh water has 0 kg/liter of salt. So, rate in = 0 kg/minute.
---Step 4: Determine the rate of salt leaving the tank. Rate out = (Concentration) x (Outflow rate) = (A(t)/100) kg/liter * 5 liters/minute = A(t)/20 kg/minute.
---Step 5: Set up the differential equation. The rate of change of salt is (Rate in) - (Rate out). So, dA/dt = 0 - A(t)/20 = -A(t)/20.
---Step 6: Solve the differential equation. This is a separable differential equation. dA/A = -1/20 dt. Integrating both sides gives ln|A| = -t/20 + C. So, A(t) = e^(-t/20 + C) = K * e^(-t/20).
---Step 7: Use the initial condition to find K. At t=0, A(0) = 10 kg. So, 10 = K * e^(0) => K = 10.
---Step 8: The final solution is A(t) = 10 * e^(-t/20).

Answer: The amount of salt in the tank after 't' minutes is 10 * e^(-t/20) kg.

Why It Matters

Understanding mixture problems is crucial in fields like Biotechnology for drug delivery systems, Environmental Science for tracking pollutants in rivers, and Chemical Engineering for reactor design. Engineers use these concepts to predict how chemicals mix or dilute, ensuring safety and efficiency in various processes.

Common Mistakes

MISTAKE: Assuming the volume of the mixture in the tank always stays constant. | CORRECTION: Always check if the inflow rate is equal to the outflow rate. If they are different, the volume of the mixture will change over time, and this change must be included in the concentration calculation.

MISTAKE: Forgetting to include the initial amount of the substance when solving the differential equation. | CORRECTION: After finding the general solution, use the given initial condition (amount of substance at time t=0) to find the specific constant of integration.

MISTAKE: Incorrectly calculating the rate of the substance leaving the tank. | CORRECTION: The rate of substance leaving is always (Concentration of substance in the tank) x (Outflow rate of the mixture). Make sure the concentration reflects the current amount in the tank.

Practice Questions

Try It Yourself

QUESTION: A 200-liter tank contains 50 kg of sugar dissolved in water. Water containing 0.5 kg of sugar per liter flows into the tank at 4 liters/minute. The well-mixed solution flows out at 4 liters/minute. Find the differential equation for the amount of sugar A(t) at time t. | ANSWER: dA/dt = 2 - A(t)/50

QUESTION: A tank contains 100 liters of brine with 10 kg of salt. Brine containing 0.2 kg/liter of salt flows in at 6 liters/minute, and the well-mixed solution flows out at 4 liters/minute. Set up the differential equation for the amount of salt A(t) in the tank. (Hint: Volume changes!) | ANSWER: The volume at time t is V(t) = 100 + 2t. dA/dt = 1.2 - (4A(t))/(100 + 2t)

QUESTION: A 500-liter tank is full of pure water. A solution containing 0.1 kg of chemical per liter enters at 5 liters/minute. The mixture leaves at 5 liters/minute. How much chemical is in the tank after 1 hour? | ANSWER: Approximately 47.58 kg (A(t) = 50 - 50 * e^(-t/100), where t is in minutes. For t=60, A(60) = 50 - 50 * e^(-0.6))

MCQ

Quick Quiz

In a mixture problem, if the inflow rate of a solution is 5 liters/minute and the outflow rate is 7 liters/minute, what happens to the total volume of the mixture in the tank?

It increases over time.

It decreases over time.

It remains constant.

It first increases then decreases.

The Correct Answer Is:

Since more liquid is flowing out (7 liters/minute) than flowing in (5 liters/minute), the total volume of the mixture in the tank will continuously decrease over time. Options A and C are incorrect because the rates are not equal or inflow is not greater. Option D is incorrect as there's no mechanism for it to increase first.

Real World Connection

In the Real World

Imagine a water treatment plant in your city. Engineers use these concepts to model how quickly chlorine added to water mixes and reduces bacteria, or how pollutants dilute in a river flowing past a factory. This ensures the water you drink is safe and clean. It's also used in pharmaceutical companies to design how drugs dissolve and spread in the body.

Key Vocabulary

Key Terms

DIFFERENTIAL EQUATION: An equation involving an unknown function and its derivatives, describing how a quantity changes. | MIXTURE PROBLEM: A type of problem where substances are mixed and flow in/out of a container. | RATE OF CHANGE: How quickly a quantity increases or decreases over time. | CONCENTRATION: The amount of a substance dissolved in a given volume of solution. | INITIAL CONDITION: The state of the system (e.g., amount of substance) at the beginning of the process (time t=0).

What's Next

What to Learn Next

Next, you can explore solving different types of differential equations, like linear first-order equations, which are very common in these problems. This will give you more tools to tackle complex real-world scenarios in engineering and science.