Question

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Compare this concentration with theoretical limiting concentration if the tank had infinite capacity.

Short answer

Based on the given information, the amount of salt in the tank at any time can be represented by the equation S(t) = (100t + (1/2)t^2 + 20000) / (200 + t). When the tank is about to overflow at t = 100 minutes, the amount of salt present in the tank is approximately 116.67 pounds, and the concentration of salt at this point is approximately 0.2333 pounds per gallon.

Step-by-step solution

Determine the rate of change of salt in the tank

To find the rate of change of salt, consider the incoming and outgoing rates: Incoming rate: 3 gallons/min * 1 lb of salt/gallon = 3 lb/min Outgoing rate: 2 gallons/min * (S(t) / (200 + t)) lb/gallon (since there are S(t) pounds of salt in (200 + t) gallons of water) Hence, the rate of change of salt in the tank (dS/dt) is dS/dt = 3 - 2(S(t) / (200 + t)).

Create a differential equation for the amount of salt

The rate of change of salt (dS/dt) is given by the following differential equation: dS/dt + (2/(200 + t))S(t) = 3

Solve the differential equation

To solve this linear first-order differential equation, we will use an integrating factor. The integrating factor (IF) can be found as: IF = e^(∫(2/(200 + t))dt) Calculating the integral, we get: IF = e^(ln(200 + t)) = 200 + t Now, we will multiply both sides of the differential equation by the integrating factor: (200 + t)dS/dt + 2S(t) = 3(200 + t) Now, the left part of the equation is the derivative of the product of the function S(t) and the integrating factor (IF): d(S(t)*(200+t))/dt = 3(200 + t) To find S(t), we need to integrate both sides of the equation: ∫(d(S(t)*(200+t))/dt)dt = ∫(3(200 + t))dt After integrating, we get: S(t) * (200 + t) = 100t + (1/2)t^2 + C Solve the equation for S(t): S(t) = (100t + (1/2)t^2 + C) / (200 + t) Use the initial condition at t=0, S(0)=100 to find C: 100 = (0 + 0 + C) / (200 + 0) C = 20000 So, the amount of salt in the tank at any time is: S(t) = (100t + (1/2)t^2 + 20000) / (200 + t)

Find the amount of salt when the tank is about to overflow

When the tank is about to overflow, there will be 500 gallons of water in it. Since water is entering at a rate of 3 gallons/min and there were initially 200 gallons, this will occur at time t = (500-200) / 3 = 100 minutes. The amount of salt in the tank at this time is: S(100) = (100*100 + (1/2)*100^2 + 20000) / (200 + 100) S(100) = (10000 + 5000 + 20000) / 300 S(100) = 35000 / 300 S(100) = 116.67 lb

Find the concentration of salt when the tank is about to overflow

The concentration of salt can be found by dividing the amount of salt in the tank by the total amount of water in the tank: Concentration = S(100) / 500 Concentration = 116.67 / 500 Concentration = 0.2333 lb/gallon The concentration of salt when the tank is about to overflow is 0.2333 pounds per gallon.

Key concepts

Rate of Change of Salt

Understanding the rate of change of salt in a solution is crucial for tackling problems involving mixtures in a tank. In our exercise, the rate of change is determined by the difference in the amount of salt entering and leaving the tank.

The incoming salt rate is straightforward: the tank receives salt from the water entering at a constant rate, which is simply the product of the incoming water's flow rate and the concentration of salt in that water. In this case, it's 3 gallons per minute each with 1 pound of salt, yielding a constant 3 pounds of salt per minute being added to the tank.

On the other hand, the outgoing rate is variable as it depends on the amount of salt currently in the tank, which changes over time. The salt leaves the tank with the outgoing water, which has its salt concentration calculated as the amount of salt divided by the volume of water at any given time. The calculation of the differential equation that models this situation takes into account these dynamics, leading to a more complex understanding of the changes in salt concentration over time.

Linear First-Order Differential Equation

The situation described in the problem translates into a linear first-order differential equation. Such equations have the general form \(\frac{dS}{dt} + P(t)S = Q(t)\), where \(S\) represents the unknown function we want to find (in this case, the amount of salt in the tank), and \(P(t)\) and \(Q(t)\) are known functions of time.

In our scenario, the function \(P(t)\) corresponds to the outflow rate of the solution divided by the current volume, while \(Q(t)\) is the constant inflow of salt. In practice, solving these types of equations involves finding an expression for \(S\) in terms of \(t\), that satisfies the initial conditions of the problem. The integration of mathematical techniques, such as an integrating factor, provides the solution that models the physical situation accurately.

Integrating Factor

An integrating factor is a mathematical tool used to find the solutions to certain types of differential equations, specifically linear first-order ones. The idea is to multiply every term of the differential equation by an integrating factor, which is a function of \(t\), so that the left-hand side of the equation becomes the derivative of a product.

This integrating factor is usually denoted as \(e^{\int P(t)dt}\), where \(P(t)\) is the coefficient of \(S\) in the differential equation. Once the integrating factor is applied, the once-daunting differential equation simplifies to a form where both sides can be integrated directly, allowing us to solve for \(S\) more easily. The integrating factor is not just a mathematical trick; it's the keystone that aligns the equation into a solvable format.

Initial Condition

The initial condition in a differential equation problem provides the starting point for solving the equation. It's the value of the unknown function (in this case, the salt concentration) at a specific time (usually \(t=0\)).

For the given problem, we are told initially there are 100 pounds of salt in the tank. This initial state plays a significant role in finding the particular solution to our differential equation. When we solve for the constant of integration \(C\), the initial condition is used to pin down the exact solution that pertains to our real-world scenario. Without this piece of information, we could only determine a general solution, which wouldn't provide the specific answers we're looking for.