1. **State the problem:** A tank initially contains 100 gallons of fresh water. Saltwater with concentration $\frac{1}{2}$ lb/gal flows in at 2 gal/min, and the mixture flows out at the same rate. After 10 minutes, the inflow changes to fresh water (0 lb/gal salt) at 2 gal/min, with outflow continuing at 2 gal/min. We need to find the amount of salt in the tank after the additional 10 minutes. 2. **Set up variables and equations:** Let $Q(t)$ be the amount of salt (in pounds) in the tank at time $t$ minutes. - For $0 \leq t \leq 10$, saltwater flows in. - For $10 < t \leq 20$, fresh water flows in. The volume remains constant at 100 gallons because inflow and outflow rates are equal. 3. **Formulate the differential equation for $0 \leq t \leq 10$: ** Salt inflow rate = concentration $\times$ inflow rate = $\frac{1}{2} \times 2 = 1$ lb/min. Salt outflow rate = concentration in tank $\times$ outflow rate = $\frac{Q}{100} \times 2 = \frac{2Q}{100} = \frac{Q}{50}$ lb/min. Differential equation: $$\frac{dQ}{dt} = \text{inflow} - \text{outflow} = 1 - \frac{Q}{50}$$ Initial condition: $Q(0) = 0$ (fresh water initially). 4. **Solve the differential equation for $0 \leq t \leq 10$: ** Rewrite: $$\frac{dQ}{dt} + \frac{1}{50}Q = 1$$ Integrating factor: $$\mu(t) = e^{\int \frac{1}{50} dt} = e^{\frac{t}{50}}$$ Multiply both sides: $$e^{\frac{t}{50}} \frac{dQ}{dt} + \frac{1}{50} e^{\frac{t}{50}} Q = e^{\frac{t}{50}}$$ Left side is derivative: $$\frac{d}{dt} \left(Q e^{\frac{t}{50}}\right) = e^{\frac{t}{50}}$$ Integrate both sides: $$Q e^{\frac{t}{50}} = \int e^{\frac{t}{50}} dt + C = 50 e^{\frac{t}{50}} + C$$ Divide both sides by $e^{\frac{t}{50}}$: $$Q = 50 + C e^{-\frac{t}{50}}$$ Apply initial condition $Q(0) = 0$: $$0 = 50 + C \Rightarrow C = -50$$ So, $$Q(t) = 50 - 50 e^{-\frac{t}{50}}$$ Calculate $Q(10)$: $$Q(10) = 50 - 50 e^{-\frac{10}{50}} = 50 - 50 e^{-0.2}$$ 5. **For $10 < t \leq 20$, fresh water flows in:** Now inflow salt concentration is 0, so: $$\frac{dQ}{dt} = 0 - \frac{Q}{50} = -\frac{Q}{50}$$ Initial condition at $t=10$ is $Q(10)$ from above. 6. **Solve for $10 < t \leq 20$: ** $$\frac{dQ}{dt} = -\frac{Q}{50}$$ Separate variables: $$\frac{dQ}{Q} = -\frac{1}{50} dt$$ Integrate: $$\ln|Q| = -\frac{t}{50} + C$$ Exponentiate: $$Q = K e^{-\frac{t}{50}}$$ Apply initial condition at $t=10$: $$Q(10) = K e^{-\frac{10}{50}} = 50 - 50 e^{-0.2}$$ Solve for $K$: $$K = \frac{50 - 50 e^{-0.2}}{e^{-0.2}} = 50 e^{0.2} - 50$$ So, $$Q(t) = \left(50 e^{0.2} - 50\right) e^{-\frac{t}{50}}$$ Calculate $Q(20)$: $$Q(20) = \left(50 e^{0.2} - 50\right) e^{-\frac{20}{50}} = \left(50 e^{0.2} - 50\right) e^{-0.4}$$ Simplify: $$Q(20) = 50 e^{0.2 - 0.4} - 50 e^{-0.4} = 50 e^{-0.2} - 50 e^{-0.4}$$ 7. **Final answer:** The amount of salt in the tank after the additional 10 minutes is $$\boxed{50 e^{-0.2} - 50 e^{-0.4} \text{ pounds}}$$ This is approximately: $$50 (e^{-0.2} - e^{-0.4}) \approx 50 (0.8187 - 0.6703) = 50 \times 0.1484 = 7.42 \text{ pounds}$$