Let's solve the brine tank problem step by step. --- ## **Given:** - **Tank capacity:** 500 gal - **Initially:** Pure water (no salt) - **Brine in:** 2 lb/gal at 5 gal/min - **Well-mixed solution out:** (a) 5 gal/min, (b) 10 gal/min --- ## **Let:** - \( A(t) \): pounds of salt in tank at time \( t \) - \( V(t) \): volume of solution in tank at time \( t \) --- ### **(a) Outflow = Inflow = 5 gal/min** #### **Volume:** - \( V(t) = 500 \) gal (constant) #### **Salt In Rate:** - \( 2 \) lb/gal \(\times\) \( 5 \) gal/min \( = 10 \) lb/min #### **Salt Out Rate:** - Outflow: \( 5 \) gal/min - Salt concentration: \( \frac{A(t)}{500} \) lb/gal - Salt out rate: \( 5 \cdot \frac{A(t)}{500} = \frac{A(t)}{100} \) lb/min #### **Differential Equation:** \[ \frac{dA}{dt} = \text{(in)} - \text{(out)} = 10 - \frac{A}{100} \] #### **Solve the DE:** \[ \frac{dA}{dt} + \frac{A}{100} = 10 \] This is a linear ODE. Integrating factor is \( e^{t/100} \): \[ e^{t/100} \frac{dA}{dt} + \frac{A}{100} e^{t/100} = 10 e^{t/100} \] \[ \frac{d}{dt} [A(t) e^{t/100}] = 10 e^{t/100} \] Integrate both sides: \[ A(t) e^{t/100} = 1000 e^{t/100} + C \] \[ A(t) = 1000 + Ce^{-t/100} \] **Initial condition:** \( A(0) = 0 \) (pure water) \[ 0 = 1000 + C \implies C = -1000 \] \[ \boxed{A(t) = 1000 \left(1 - e^{-t/100}\right)} \] #### **As \( t \to \infty \):** \[ \boxed{A(\infty) = 1000 \text{ lb of salt}} \] --- ### **(b) Outflow = 10 gal/min** #### **Volume:** - \( V(t) = 500 - 5t \) gal - The tank empties when \( V(t) = 0 \implies t = 100 \) min #### **Salt In Rate:** - Still \( 10 \) lb/min #### **Salt Out Rate:** - Outflow: \( 10 \) gal/min - Salt concentration: \( \frac{A(t)}{V(t)} \) lb/gal - Salt out rate: \( 10 \cdot \frac{A(t)}{500-5t} \) lb/min #### **DE:** \[ \frac{dA}{dt} = 10 - 10\cdot \frac{A}{500-5t} \] Rewrite: Let’s use \( V(t) = 500-5t \): \[ \frac{dA}{dt} + \frac{10}{500-5t}A = 10 \] #### **Integrating Factor:** \[ \mu(t) = \exp\left( \int \frac{10}{500-5t} dt \right) \] Let \( u = 500-5t \implies du = -5 dt \implies dt = -du/5 \): \[ \int \frac{10}{u} \cdot (-du/5) = -2 \int \frac{1}{u} du = -2 \ln|u| = -2 \ln|500-5t| \] So the integrating factor is: \[ \mu(t) = (500-5t)^{-2} \] #### **Solve:** Multiply both sides by integrating factor: \[ (500-5t)^{-2} \frac{dA}{dt} + \frac{10}{500-5t} (500-5t)^{-2} A = 10 (500-5t)^{-2} \] \[ \frac{d}{dt} \left[ A(t) (500-5t)^{-2} \right] = 10 (500-5t)^{-2} \] Integrate both sides: \[ A(t)(500-5t)^{-2} = \int 10 (500-5t)^{-2} dt + C \] Let \( u = 500-5t \implies du = -5 dt \implies dt = -du/5 \): \[ \int 10 (500-5t)^{-2} dt = 10 \int u^{-2} \cdot (-du/5) = -2 \int u^{-2} du = 2u^{-1} \] So, \[ A(t)(500-5t)^{-2} = 2(500-5t)^{-1} + C \] \[ A(t) = 2(500-5t) + C(500-5t)^2 \] Initial condition: \( A(0) = 0 \): \[ 0 = 2 \cdot 500^{-1} + C \cdot 500^{-2} \] Wait, plug in at \( t=0 \): \[ A(0) = 2 \cdot 500 + C \cdot 500^2 = 0 \] \[ 0 = 1000 + 250000C \implies C = -\frac{1000}{250000} = -\frac{1}{250} \] So: \[ A(t) = 2(500-5t) - \frac{1}{250}(500-5t)^2 \] --- #### **When does the tank empty?** \( V(t) = 500-5t = 0 \implies t = 100 \) min #### **Concentration at that time:** \( A(100) = 2(0) - \frac{1}{250}(0)^2 = 0 \) lb (the salt also goes to zero as the last drops leave!) **But let's check the concentration just _before_ the tank empties:** Set \( t = 100 \) min, \( V(100) = 0 \), so the formula gives \( A(100) = 0 \). But what's the **concentration** as the tank empties? It's \( \frac{A(t)}{V(t)} \): Let \( t \to 100^- \): \[ A(t) = 2(500-5t) - \frac{1}{250}(500-5t)^2 \] \[ V(t) = 500-5t \] \[ \text{Concentration} = \frac{A(t)}{V(t)} = \frac{2V(t) - \frac{1}{250}V(t)^2}{V(t)} = 2 - \frac{1}{250}V(t) \] As \( V(t) \to 0 \) (i.e., as \( t \to 100 \)), \( \text{Concentration} \to 2 \) lb/gal. **So, the answer:** > **The concentration approaches 2 lb/gal** as the tank empties. --- ### **(c)** - **Initial volume:** 250 gal - **Inflow:** 5 gal/min of 2 lb/gal brine (so, in at 10 lb/min) - **Outflow:** 2.5 gal/min - **Net inflow:** 2.5 gal/min - **\( V(t) = 250 + 2.5 t \)** - **Tank is full when \( V(t) = 500 \implies 250 + 2.5t = 500 \implies t = 100 \) min** #### **Salt Out Rate:** - Outflow: 2.5 gal/min - Salt concentration: \( \frac{A(t)}{V(t)} \) lb/gal - Out rate: \( 2.5 \cdot \frac{A(t)}{V(t)} \) lb/min #### **DE:** \[ \frac{dA}{dt} = 10 - 2.5 \frac{A}{250 + 2.5t} \] #### **Linear ODE:** \[ \frac{dA}{dt} + \frac{2.5}{250+2.5t} A = 10 \] Integrating factor: \[ \mu(t) = \exp\left( \int \frac{2.5}{250+2.5t} dt \right) \] Let \( u = 250 + 2.5t \implies du = 2.5 dt \implies dt = du/2.5 \): \[ \int \frac{2.5}{u} dt = \int \frac{2.5}{u} \cdot \frac{du}{2.5} = \int \frac{1}{u} du = \ln u \] So integrating factor is \( \mu(t) = 250 + 2.5t \) Multiply both sides: \[ (250 + 2.5t) \frac{dA}{dt} + 2.5A = 10(250 + 2.5t) \] \[ \frac{d}{dt}[A(t)(250 + 2.5t)] = 10(250 + 2.5t) \] Integrate both sides: \[ A(t)(250 + 2.5t) = \int 10(250 + 2.5t) dt + C \] \[ = 10 \int (250 + 2.5t) dt + C \] \[ = 10(250t + 1.25t^2) + C \] \[ = 2500t + 12.5t^2 + C \] So, \[ A(t) = \frac{2500t + 12.5t^2 + C}{250 + 2.5t} \] Initial condition: \( t = 0, A(0) = 0 \): \[ 0 = \frac{0 + 0 + C}{250} \implies C = 0 \] So, \[ \boxed{A(t) = \frac{2500t + 12.5t^2}{250 + 2.5t}} \] Or, factor numerator: \[ A(t) = \frac{12.5 t (20 + t)}{250 + 2.5t} \] #### **Concentration when tank is full:** At \( t = 100 \), \( V(100) = 250 + 2.5 \times 100 = 500 \) gal \[ A(100) = \frac{12.5 \cdot 100 \cdot (20 + 100)}{500} = \frac{12.5 \cdot 100 \cdot 120}{500} = \frac{12.5 \cdot 12,000}{500} = \frac{150,000}{500} = 300 \text{ lb} \] Concentration: \[ \frac{A(100)}{V(100)} = \frac{300}{500} = 0.6 \text{ lb/gal} \] --- ## **Summary Table** | Case | \( A(t) \) | Salt after long time / at empty | Concentration at end | |------|------------|----------------------------------|----------------------| | (a) | \( 1000 (1-e^{-t/100}) \) | \( 1000 \) lb | \( 2 \) lb/gal (steady) | | (b) | \( 2(500-5t) - \frac{1}{250}(500-5t)^2 \) | 0 lb | \( 2 \) lb/gal (as tank empties) | | (c) | \( \frac{12.5t(20+t)}{250+2.5t} \) | 300 lb at \( t=100 \) | \( 0.6 \) lb/gal | --- ## **Final Answers** ### **(a)** \[ \boxed{ A(t) = 1000 \left(1 - e^{-t/100}\right) } \] After a long time: \[ \boxed{A(\infty) = 1000\ \text{lb}} \] --- ### **(b)** \[ \boxed{ A(t) = 2(500-5t) - \frac{1}{250}(500-5t)^2 } \] When the tank is empty (\( t = 100 \) min), the **concentration approaches**: \[ \boxed{2\ \text{lb/gal}} \] --- ### **(c)** \[ \boxed{ A(t) = \frac{12.5\, t \, (20 + t)}{250 + 2

(a) \(A(t) = 1000(1 - e^{-t/100})\), with \(A(\infty) = 1000\) lb; (b) \(A(t) = 2(500 - 5t) - \frac{(500 - 5t)^2}{250}\), concentration approaches 2 lb/gal as tank empties at \(t=100\) min; (c) \(A(t) = \frac{12.5\, t (20 + t)}{250 + 2.5 t}\), with 300 lb of salt at full tank (\(t=100\) min), concentration 0.6 lb/gal.