Let's break down and solve the problem step by step. ### **Given:** - Initial volume: \( V(0) = 500 \) gallons (pure water, so initial salt = 0) - Brine inflow: 2 lb/gal at 5 gal/min → Salt in per minute = \( 2 \times 5 = 10 \) lb/min - Outflow: 10 gal/min (solution is well-mixed) - Volume as a function of time: \( V(t) = 500 - 5t \) (since in: 5 gal/min, out: 10 gal/min, net: -5 gal/min) - We want: 1. \( A(t) \): amount of salt in the tank at time \( t \) 2. The concentration of salt in the tank when it is empty (\( t = 100 \) min) --- ## **Step 1: Differential Equation** Let \( A(t) \) be the amount of salt (in pounds) at time \( t \) (in minutes). ### **Inflow of Salt:** - \( 2 \) lb/gal × \( 5 \) gal/min = \( 10 \) lb/min ### **Outflow of Salt:** - The concentration in the tank at time \( t \) is \( \frac{A(t)}{V(t)} \) lb/gal - Outflow rate is \( 10 \) gal/min - So, outflow of salt = \( \frac{A(t)}{V(t)} \times 10 \) lb/min ### **Differential Equation:** \[ \frac{dA}{dt} = \text{(inflow)} - \text{(outflow)} = 10 - \frac{10A(t)}{V(t)} \] \[ \frac{dA}{dt} = 10 - \frac{10A(t)}{500 - 5t} \] \[ \frac{dA}{dt} + \frac{10}{500 - 5t}A = 10 \] --- ## **Step 2: Solve the Differential Equation** This is a linear first-order ODE. ### **Integrating Factor Method:** The integrating factor \( \mu(t) \) is \[ \mu(t) = \exp\left( \int \frac{10}{500 - 5t} dt \right) \] Let’s compute the integral: \[ \int \frac{10}{500 - 5t} dt = -2 \ln|500 - 5t| \] So, \[ \mu(t) = e^{-2 \ln(500 - 5t)} = (500 - 5t)^{-2} \] Multiply both sides of the ODE by \( (500 - 5t)^{-2} \): \[ (500 - 5t)^{-2} \frac{dA}{dt} + \frac{10}{500 - 5t}(500 - 5t)^{-2}A = 10(500 - 5t)^{-2} \] But notice: \[ \frac{d}{dt} \left( (500 - 5t)^{-2} A \right) = (500 - 5t)^{-2} \frac{dA}{dt} + A \frac{d}{dt}\left( (500 - 5t)^{-2} \right) \] \[ = (500 - 5t)^{-2} \frac{dA}{dt} + A \left( -2 (500 - 5t)^{-3} \cdot (-5) \right ) = (500 - 5t)^{-2} \frac{dA}{dt} + \frac{10A}{(500 - 5t)^{3}} \] Wait, that's not quite matching the above, so let's stick to integrating factor method (it's the standard approach). Recall: \[ \frac{dA}{dt} + P(t)A = Q(t) \] with integrating factor \( \mu(t) = e^{\int P(t) dt} \) Here, \( P(t) = \frac{10}{500 - 5t} \) So, \[ \int P(t) dt = \int \frac{10}{500 - 5t} dt = -2 \ln |500 - 5t| \] \[ \mu(t) = e^{-2 \ln(500 - 5t)} = (500 - 5t)^{-2} \] Multiply both sides by integrating factor: \[ (500 - 5t)^{-2} \frac{dA}{dt} + (500 - 5t)^{-2} \frac{10}{500-5t} A = 10 (500 - 5t)^{-2} \] But \[ (500 - 5t)^{-2} \frac{10}{500-5t} A = 10 (500 - 5t)^{-3} A \] So, \[ (500 - 5t)^{-2} \frac{dA}{dt} + 10 (500 - 5t)^{-3} A = 10 (500 - 5t)^{-2} \] Left side is the derivative: \[ \frac{d}{dt} \left[ (500 - 5t)^{-2} A \right ] = 10 (500 - 5t)^{-2} \] Integrate both sides with respect to \( t \): \[ \int \frac{d}{dt} \left[ (500 - 5t)^{-2} A \right ] dt = \int 10 (500 - 5t)^{-2} dt \] \[ (500 - 5t)^{-2} A = \int 10 (500 - 5t)^{-2} dt + C \] Let’s compute the integral: Let \( u = 500 - 5t \), \( du = -5 dt \), \( dt = -du/5 \) \[ \int 10 (500 - 5t)^{-2} dt = 10 \int u^{-2} \left( -\frac{1}{5} du \right ) = -2 \int u^{-2} du = -2( -u^{-1} ) = 2 u^{-1} = \frac{2}{500 - 5t} \] So, \[ (500 - 5t)^{-2} A = \frac{2}{500 - 5t} + C \] \[ A(t) = (500 - 5t)^2 \left[ \frac{2}{500 - 5t} + C \right ] = 2(500 - 5t) + C (500 - 5t)^2 \] --- ### **Step 3: Apply Initial Condition** At \( t = 0 \), \( A(0) = 0 \): \[ A(0) = 2(500) + C (500)^2 = 0 \] \[ 1000 + 250000 C = 0 \] \[ C = -\frac{1000}{250000} = -\frac{1}{250} \] --- ### **Step 4: Final Solution** \[ A(t) = 2(500 - 5t) - \frac{1}{250}(500 - 5t)^2 \] Or, expand if you like: \[ A(t) = 1000 - 10t - \frac{1}{250}(500 - 5t)^2 \] --- ### **Step 5: When is the tank empty?** Tank is empty when \( V(t) = 500 - 5t = 0 \implies t = 100 \) min. --- ### **Step 6: Final Concentration** At \( t = 100 \): \[ A(100) = 2(500 - 5*100) - \frac{1}{250}(500 - 5*100)^2 = 2(0) - \frac{1}{250} (0)^2 = 0 \] Volume at \( t=100 \) is 0, so the solution is empty. The concentration immediately prior to being empty is: Let’s look at \( t \to 100^- \): - As \( t \to 100 \), \( V(t) \to 0 \), \( A(t) \to 0 \), so the concentration approaches 0/0, which is indeterminate. - Let’s compute the limit: \[ \text{Concentration as } t \to 100^-: \quad \lim_{t \to 100^-} \frac{A(t)}{V(t)} \] Let’s use L'Hospital's Rule. Let \( x = V(t) = 500 - 5t \), so as \( t \to 100, x \to 0 \): \[ A(t) = 2x - \frac{1}{250}x^2 \] \[ \frac{A(t)}{V(t)} = \frac{2x - \frac{1}{250}x^2}{x} = 2 - \frac{1}{250}x \] As \( x \to 0 \), this approaches 2. So, **the final concentration just before emptying is 2 lb/gal**. --- ## **Summary** ### **Amount of salt in the tank at time \( t \):** \[ \boxed{ A(t) = 2(500 - 5t) - \frac{1}{250}(500 - 5t)^2 } \] for \( 0 \leq t \leq 100 \). ### **Final concentration as the tank empties:** \[ \boxed{ \text{The concentration approaches } 2 \text{ lb/gal as } t \to 100^- } \]