Let's solve **Problem 1** step-by-step. ## **Problem 1 Restated** A well stirred tank originally contains 100 gal of fresh water. Then brine containing 1/4 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave the tank at the same rate. What is the amount of salt, in lbs, in the tank at any instant? --- ### **Step 1: Define variables** Let: - \( n(t) \): amount of salt (in lbs) in the tank at time \( t \) (in minutes) - \( v(t) \): volume of fluid in the tank at time \( t \) (in gallons) We are told: - Initial volume: \( v(0) = 100 \) gal - Initial salt: \( n(0) = 0 \) lbs (fresh water) - Brine in: \( r_{\text{in}} = 2 \) gal/min, \( c_{\text{in}} = 0.25 \) lb/gal - Outflow: \( r_{\text{out}} = 2 \) gal/min Since the inflow and outflow rates are equal, the volume remains constant: \( v(t) = 100 \) gal for all \( t \). --- ### **Step 2: Write the differential equation** Using the formula from the worksheet: \[ \frac{dn}{dt} = r_{\text{in}} c_{\text{in}} - r_{\text{out}} \frac{n}{v} \] Plugging in values: \[ \frac{dn}{dt} = 2 \cdot 0.25 - 2 \cdot \frac{n}{100} \] \[ \frac{dn}{dt} = 0.5 - \frac{2n}{100} \] \[ \frac{dn}{dt} = 0.5 - 0.02n \] --- ### **Step 3: Solve the ODE** This is a linear ODE: \[ \frac{dn}{dt} + 0.02n = 0.5 \] The integrating factor is: \[ \mu(t) = e^{\int 0.02 dt} = e^{0.02t} \] Multiply both sides by integrating factor: \[ e^{0.02t}\frac{dn}{dt} + 0.02 e^{0.02t} n = 0.5 e^{0.02t} \] \[ \frac{d}{dt}\left(e^{0.02t} n \right) = 0.5 e^{0.02t} \] Integrate both sides with respect to \( t \): \[ e^{0.02t} n = \int 0.5 e^{0.02t} dt \] \[ e^{0.02t} n = 0.5 \int e^{0.02t} dt \] \[ e^{0.02t} n = 0.5 \cdot \frac{1}{0.02} e^{0.02t} + C \] \[ e^{0.02t} n = 25 e^{0.02t} + C \] \[ n(t) = 25 + C e^{-0.02t} \] --- ### **Step 4: Apply Initial Condition** At \( t = 0 \), \( n(0) = 0 \): \[ 0 = 25 + C \implies C = -25 \] So the solution is: \[ \boxed{n(t) = 25 \left(1 - e^{-0.02t}\right)} \] --- ### **Step 5: Interpretation** - At any time \( t \), the amount of salt in the tank (in lbs) is \( n(t) = 25(1 - e^{-0.02t}) \). - As \( t \to \infty \), \( n(t) \to 25 \) lbs (the tank approaches equilibrium with the incoming brine). --- **Final Answer:** > **The amount of salt in the tank at any instant \( t \) is** > \[ > \boxed{n(t) = 25\left(1 - e^{-0.02t}\right)\ \text{lbs}} > \] Let me know if you'd like the step-by-step for another problem!

Certainly! Below are detailed step-by-step solutions for all four problems based on the modeling approach described. --- ## **Problem 1** **Restatement:** A tank initially contains 100 gallons of fresh water. Brine with 1/4 lb/gal salt enters at 2 gal/min, and the mixture leaves at 2 gal/min. Find the amount of salt \( n(t) \) in the tank at any time. ### **Solution:** ### Step 1: Define variables - \( v(t) = 100 \) gallons (constant) - \( n(t) \): salt in lbs at time \( t \) - \( c_{\text{in}} = 0.25 \) lb/gal - Inflow rate \( r_{in} = 2 \) gal/min - Outflow rate \( r_{out} = 2 \) gal/min ### Step 2: Write the differential equation \[ \frac{dn}{dt} = r_{in} \cdot c_{\text{in}} - r_{out} \cdot \frac{n}{v} \] \[ \frac{dn}{dt} = 2 \times 0.25 - 2 \times \frac{n}{100} = 0.5 - 0.02 n \] ### Step 3: Solve the linear ODE \[ \frac{dn}{dt} + 0.02 n = 0.5 \] Integrating factor: \[ \mu(t) = e^{0.02 t} \] Multiply through: \[ \frac{d}{dt} \left( e^{0.02 t} n \right) = 0.5 e^{0.02 t} \] Integrate: \[ e^{0.02 t} n = \frac{0.5}{0.02} e^{0.02 t} + C = 25 e^{0.02 t} + C \] Solve for \( n(t) \): \[ n(t) = 25 + Ce^{-0.02 t} \] ### Step 4: Apply initial condition At \( t=0 \), \( n(0) = 0 \): \[ 0 = 25 + C \Rightarrow C = -25 \] ### **Final solution:** \[ \boxed{ n(t) = 25 \left( 1 - e^{-0.02 t} \right) \text{ lbs} } \] --- ## **Problem 2** **Restatement:** A 500-gallon tank initially contains 200 gallons of water with 100 lbs of salt. Brine with 1 lb/gal salt enters at 3 gal/min; outflow is 2 gal/min. Find the salt concentration when the tank overflows and 10 minutes later. ### **Solution:** ### Step 1: Volume over time \[ v(t) = 200 + (3 - 2)t = 200 + t \] The tank overflows when \( v(t) = 500 \): \[ 200 + t = 500 \Rightarrow t_{overflow} = 300 \text{ min} \] ### Step 2: Differential equation Inflow: \[ r_{in} = 3 \text{ gal/min}, \quad c_{in} = 1 \text{ lb/gal} \] Outflow: \[ r_{out} = 2 \text{ gal/min} \] Salt in: \[ \frac{dn}{dt} = r_{in} c_{in} - r_{out} \frac{n}{v(t)} = 3 \times 1 - 2 \frac{n}{v(t)} = 3 - \frac{2 n}{v(t)} \] ### Step 3: Recognize the variable volume The ODE: \[ \frac{dn}{dt} + \frac{2}{v(t)} n = 3 \] ### Step 4: Solve the ODE - Integrating factor: \[ \mu(t) = e^{\int \frac{2}{v(t)} dt} = e^{2 \int \frac{1}{200 + t} dt} = e^{2 \ln(200 + t)} = (200 + t)^2 \] - Multiply through: \[ (200 + t)^2 \frac{dn}{dt} + 2 (200 + t) n = 3 (200 + t)^2 \] which simplifies to: \[ \frac{d}{dt} \left[ (200 + t)^2 n \right] = 3 (200 + t)^2 \] - Integrate: \[ (200 + t)^2 n = \int 3 (200 + t)^2 dt + C \] \[ = 3 \times \frac{(200 + t)^3}{3} + C = (200 + t)^3 + C \] - Solve for \( n(t) \): \[ n(t) = \frac{(200 + t)^3 + C}{(200 + t)^2} = (200 + t) + \frac{C}{(200 + t)^2} \] ### Step 5: Initial condition At \( t=0 \), \( n(0) = 100 \): \[ 100 = 200 + \frac{C}{(200)^2} \Rightarrow \frac{C}{40000} = 100 - 200 = -100 \] \[ C = -100 \times 40000 = -4,000,000 \] ### **Part 1: Salt concentration at overflow (\( t=300 \))** \[ n(300) = (200 + 300) + \frac{-4,000,000}{(200 + 300)^2} = 500 - \frac{4,000,000}{500^2} \] \[ = 500 - \frac{4,000,000}{250,000} = 500 - 16 = 484 \text{ lbs} \] Concentration: \[ \text{Concentration} = \frac{n(300)}{v(300)} = \frac{484}{500} = 0.968 \text{ lbs/gal} \approx 0.97 \] --- ### **Part 2: Salt amount 10 minutes later (\( t=10 \))** \[ n(10) = 210 + \frac{-4,000,000}{210^2} = 210 - \frac{4,000,000}{44100} \approx 210 - 90.7 = 119.3 \text{ lbs} \] Volume: \[ v(10) = 200 + 10 = 210 \text{ gallons} \] Concentration: \[ \frac{119.3}{210} \approx 0.568 \text{ lbs/gal} \] --- ## **Problem 3** **Restatement:** A 1500-gallon tank initially contains 1000 gallons of fresh water. Two inflows: - Pipe 1: 1/2 lb/gal salt, 4 gal/min - Pipe 2: 1/3 lb/gal salt, 6 gal/min outflow at 5 gal/min. Determine \( n(t) \) prior to overflow. ### **Solution:** ### Step 1: Volume over time \[ v(t) = 1000 + (4 + 6 - 5)t = 1000 + 5 t \] Tank overflows when \( v(t) = 1500 \): \[ 1000 + 5 t = 1500 \Rightarrow t=100 \text{ min} \] ### Step 2: Differential equation Total inflow: \[ r_{in} = 4 + 6 = 10\ \text{gal/min} \] - Salt from pipe 1: \( 4 \times \frac{1}{2} = 2 \) lbs/min - Salt from pipe 2: \( 6 \times \frac{1}{3} = 2 \) lbs/min Total salt inflow: \[ r_{\text{salt in}} = 2 + 2 = 4 \text{ lbs/min} \] Salt out: \[ r_{\text{salt out}} = r_{out} \times \frac{n}{v(t)} = 5 \times \frac{n}{v(t)} \] ODE: \[ \frac{dn}{dt} = 4 - 5 \frac{n}{v(t)} \] \[ v(t) = 1000 + 5 t \] ### Step 3: Rewrite ODE \[ \frac{dn}{dt} + \frac{5}{v(t)} n = 4 \] \[ \frac{dn}{dt} + \frac{5}{1000 + 5 t} n = 4 \] ### Step 4: Find integrating factor \[ \mu(t) = e^{\int \frac{5}{1000 + 5 t} dt} \] \[ = e^{\int \frac{5}{5 (200 + t)} dt} = e^{\int \frac{1}{200 + t} dt} = e^{\ln(200 + t)} = 200 + t \] ### Step 5: Multiply through \[ (200 + t) \frac{dn}{dt} + n = 4 (200 + t) \] \[ \frac{d}{dt} [ (200 + t) n ] = 4 (200 + t) \] ### Step 6: Integrate both sides \[ (200 + t) n = \int 4(200 + t) dt + C \] \[ = 4 \times \frac{(200 + t)^2}{2} + C = 2 (200 + t)^2 + C \] ### Step 7: Solve for \( n(t) \) \[ n(t) = \frac{2 (200 + t)^2 + C}{200 + t} = 2(200 + t) + \frac{C}{200 + t} \] ### Step 8: Initial condition At \( t=0 \): \[ n(0) = 0 = 2 \times 200 + \frac{C}{200} \Rightarrow 0 = 400 + \frac{C}{200} \] \[ C = -400 \times 200 = -80,000 \] ### **Final expression:** \[ \boxed{ n(t) = 2 (200 + t) - \frac{80,000}{200 + t} } \] At \( t=100 \) (just before overflow): \[ n(100) = 2 \times 300 - \frac{80,000}{300} = 600 - 266.67 \approx 333.33 \text{ lbs} \] --- ## **Problem 4** **Restatement:** A 500-gallon tank contains 4% alcohol (by volume). Beer with 6% alcohol is pumped in at 5 gal/min, mixture flows out at same rate. Find the percentage after 1 hour and how long to reach 6%. ### **Solution:** ### Step 1: Variables - \( v(t) = 500 \) gallons (constant) - \( n(t) \): lbs of alcohol at time \( t \) - Initial alcohol: \[ n(0) = 0.04 \times 500 = 20 \text{ lbs} \] - Inflow concentration \( c_{\text{in}} = 0.06 \times 500 = 30 \text{ lbs/gal} \)? No, since the in-flow is 5 gal/min of 6% alcohol: \[ c_{\text{in}} = 0.06 \text{ (fraction)} \] - Inflow rate: \( r_{in} = 5 \) gal/min - Outflow rate: \( r_{out} = 5 \) gal/min ### Step 2: Differential equation \[ \frac{dn}{dt} = r_{in} \times c_{\text{in}} - r_{out} \times \frac{n}{v} = 5 \times 0.06 \times 500 / 500 - 5 \times \frac{n}{500} \] But more straightforwardly: \[ \frac{dn}{dt} = 5 \times 0.06 \times 500 / 500 - 5 \times \frac{n}{500} \] Since the concentration of incoming beer is 6% (0.06), and the volume remains constant at 500 gallons: Inflow salt: \[ r_{in} \times c_{\text{in}} = 5 \times 0.06 \times 500 = 5 \times 30 = 150 \text{ lbs/min} \] Outflow: \[ r_{out} \times \frac{n}{v} = 5 \times \frac{n}{500} = \frac{n}{100} \] ### **Step 3: Write ODE** \[ \frac{dn}{dt} = 150 - \frac{n}{100} \] ### Step 4: Solve the ODE \[ \frac{dn}{dt} + \frac{1}{100} n = 150 \] Integrating factor: \[ \mu(t) = e^{t/100} \] Multiply through: \[ \frac{d}{dt} \left( e^{t/100} n \right) = 150 e^{t/100} \] Integrate: \[ e^{t/100} n = 150 \times 100 e^{t

Apologies for the interruption. Continuing from where I left off: --- ## **Problem 4 Continued** ### Step 4: Solve the ODE We have: \[ \frac{dn}{dt} + \frac{1}{100} n = 150 \] The integrating factor: \[ \mu(t) = e^{\int \frac{1}{100} dt} = e^{t/100} \] Multiply both sides by the integrating factor: \[ e^{t/100} \frac{dn}{dt} + \frac{1}{100} e^{t/100} n = 150 e^{t/100} \] \[ \frac{d}{dt} \left( e^{t/100} n \right) = 150 e^{t/100} \] Integrate both sides: \[ e^{t/100} n = \int 150 e^{t/100} dt + C \] Calculate the integral: \[ \int 150 e^{t/100} dt = 150 \times 100 e^{t/100} + C' = 15,000 e^{t/100} + C' \] So: \[ e^{t/100} n = 15,000 e^{t/100} + C \] \[ n(t) = 15,000 + C e^{-t/100} \] ### Step 5: Use initial condition At \( t=0 \), \( n(0) = 20 \) lbs: \[ 20 = 15,000 + C e^{0} \Rightarrow C = 20 - 15,000 = -14,980 \] ### **Final expression for \( n(t) \):** \[ n(t) = 15,000 - 14,980 e^{-t/100} \] --- ### **Part A: Percentage of alcohol after 1 hour** At \( t=60 \) minutes: \[ n(60) = 15,000 - 14,980 e^{-60/100} = 15,000 - 14,980 e^{-0.6} \] Calculate \( e^{-0.6} \): \[ e^{-0.6} \approx 0.5488 \] \[ n(60) \approx 15,000 - 14,980 \times 0.5488 \approx 15,000 - 8,219 \approx 6,781 \text{ lbs} \] Since the total volume is 500 gallons, the percentage of alcohol: \[ \frac{n(60)}{500} \times 100\% \approx \frac{6,781}{500} \times 100\% \approx 13.56\% \] --- ### **Part B: Time to reach 6% alcohol** 6% of 500 gallons: \[ 0.06 \times 500 = 30 \text{ lbs} \] Set \( n(t) = 30 \) and solve for \( t \): \[ 30 = 15,000 - 14,980 e^{-t/100} \] \[ 14,980 e^{-t/100} = 15,000 - 30 = 14,970 \] \[ e^{-t/100} = \frac{14,970}{14,980} \approx 0.99933 \] \[ -t/100 = \ln(0.99933) \approx -0.00067 \] \[ t/100 \approx 0.00067 \] \[ t \approx 0.067 \text{ minutes} \approx 4.0 \text{ seconds} \] **Interpretation:** The tank reaches 6% alcohol very quickly—almost immediately after starting the process. --- ## **Summary of Final Results** ### **Problem 1:** \[ \boxed{ n(t) = 25 \left(1 - e^{-0.02 t}\right) \text{ lbs} } \] ### **Problem 2:** - Salt when the tank overflows (\( t=300 \) min): \[ n \approx 484 \text{ lbs} \] - Salt 10 min later: \[ n \approx 119.3 \text{ lbs} \] ### **Problem 3:** \[ \boxed{ n(t) = 2 (200 + t) - \frac{80,000}{200 + t} } \] At just before overflow (\( t=100 \) min): \[ n \approx 333.33 \text{ lbs} \] ### **Problem 4:** \[ n(t) = 15,000 - 14,980 e^{-t/100} \] - After 1 hour (\( t=60 \) min): \[ \text{Alcohol \%} \approx 13.56\% \] - To reach 6\% (\( n=30 \) lbs): \[ t \approx 4 \text{ seconds} \] --- If you'd like, I can prepare a summarized version or assist with specific parts further!