give answer in 2 step with explanation at the end of each step and final answer at the end:(1 point)
The flow system shown in the figure is activated at time ¢ = 0. Let Q;(t) denote the amount of sol

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:(1 point)
The flow system shown in the figure is activated at time ¢ = 0. Let Q;(t) denote the amount of solute present eo if on J
in the ith tank at time ¢. Assume that all the flow rates are a constant 10 L/min. It follows that the volume of
solution in each tank remains constant; assume this volume to be 1000 L. The concentration of solute in the
inflow to Tank 1 (from a source other than Tank 2) is 0.3 kg/L, and the concentration of solute in the inflow to
Tank 2 (from a source other than Tank 1) is 0 kg/L. Assume each tank is mixed perfectly.
b. If Q1(0) = 10 kg and Q,(0) = 20 kg, find the amount of solute in each tank after ¢ minutes.
(
QM =" kg
Qn =| | kg

VIPSolutions · Accepted Answer

Let's solve the problem step by step.

### **Step 1: Set up the system of differential equations**  
Let $ Q_1(t) $ and $ Q_2(t) $ be the amount of solute in tanks 1 and 2 at time $ t $.

- **Tank 1:**
  - Inflow: $ 10 $ L/min × $ 0.3 $ kg/L = $ 3 $ kg/min (from outside)
  - Outflow to Tank 2: $ 10 $ L/min, concentration in Tank 1 = $ Q_1/1000 $ kg/L
  - Outflow rate of solute: $ 10 \times Q_1/1000 = Q_1/100 $ kg/min

So,
  $$
  \frac{dQ_1}{dt} = 3 - \frac{Q_1}{100}
  $$

- **Tank 2:**
  - Inflow from Tank 1: $ 10 $ L/min, concentration = $ Q_1/1000 $ kg/L ⇒ $ 10 \times Q_1/1000 = Q_1/100 $ kg/min
  - Inflow from outside: 0 kg/min (since concentration is 0)
  - Outflow: $ 10 $ L/min, concentration in Tank 2 = $ Q_2/1000 $ kg/L ⇒ $ 10 \times Q_2/1000 = Q_2/100 $ kg/min

So,
  $$
  \frac{dQ_2}{dt} = \frac{Q_1}{100} - \frac{Q_2}{100}
  $$

**Explanation for Step 1:**  
We set up the system of differential equations based on the inflow and outflow rates and solute concentrations in each tank.

---

### **Step 2: Solve the system with initial conditions $ Q_1(0) = 10 $, $ Q_2(0) = 20 $**

#### **Solve for $ Q_1(t) $:**

$$
\frac{dQ_1}{dt} + \frac{1}{100}Q_1 = 3
$$
This is a first-order linear ODE. The integrating factor is $ e^{t/100} $:

$$
e^{t/100} \frac{dQ_1}{dt} + \frac{1}{100} e^{t/100} Q_1 = 3e^{t/100}
$$
$$
\frac{d}{dt}\left( e^{t/100} Q_1 \right) = 3e^{t/100}
$$
Integrate both sides:
$$
e^{t/100} Q_1 = 300 e^{t/100} + C
$$
$$
Q_1(t) = 300 + C e^{-t/100}
$$
Initial condition $ Q_1(0) = 10 $:
$$
10 = 300 + C \implies C = -290
$$
$$
Q_1(t) = 300 - 290 e^{-t/100}
$$

**Explanation for Step 2a:**  
We solved the first ODE for $ Q_1(t) $ using integrating factors and initial conditions.

---

#### **Solve for $ Q_2(t) $:**

$$
\frac{dQ_2}{dt} + \frac{1}{100}Q_2 = \frac{Q_1}{100}
$$
This is a first-order linear ODE with a nonhomogeneous term. Substitute $ Q_1(t) $:

$$
\frac{dQ_2}{dt} + \frac{1}{100}Q_2 = \frac{1}{100} (300 - 290 e^{-t/100})
$$
$$
\frac{dQ_2}{dt} + \frac{1}{100}Q_2 = 3 - 2.9 e^{-t/100}
$$
Integrating factor: $ e^{t/100} $

$$
e^{t/100} \frac{dQ_2}{dt} + \frac{1}{100} e^{t/100} Q_2 = 3e^{t/100} - 2.9 e^{t/100} e^{-t/100}
$$
$$
\frac{d}{dt}(e^{t/100} Q_2) = 3e^{t/100} - 2.9
$$
Now integrate both sides:
$$
e^{t/100} Q_2 = 300 e^{t/100} - 2.9 t + C
$$
$$
Q_2(t) = 300 - 2.9 t e^{-t/100} + C e^{-t/100}
$$
But integrating $ -2.9 $, you get $ -2.9 t $, so:
$$
Q_2(t) = 300 - 2.9 t + C e^{-t/100}
$$
Use initial condition $ Q_2(0) = 20 $:
$$
20 = 300 - 0 + C \implies C = -280
$$
Therefore,
$$
Q_2(t) = 300 - 2.9 t - 280 e^{-t/100}
$$

**Explanation for Step 2b:**  
We solved the second ODE for $ Q_2(t) $, substituted $ Q_1(t) $, used integrating factor, and applied the initial condition.

---

### **Final Answers:**

$$
\boxed{
Q_1(t) = 300 - 290 e^{-t/100} \text{ kg}
}
$$
$$
\boxed{
Q_2(t) = 300 - 2.9 t - 280 e^{-t/100} \text{ kg}
}
$$

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