Question

A large tank initially contains 200 gal of brine in which 15 lb of salt is dissolved. Starting at \(t=0\), brine containing 4 lb of salt per gallon flows into the tank at the rate of \(3.5 \mathrm{gal} / \mathrm{min} .\) The mixture is kept uniform by stirring and the well-stirred mixture leaves the tank at the rate of \(4 \mathrm{gal} / \mathrm{min}\). (a) How much salt is in the tank at the end of one hour? (b) How much salt is in the tank when the tank contains only 50 gal of brine?

Short answer

(a) At the end of one hour, the tank will contain approximately \(18.98\) lb of salt. (b) When the tank contains 50 gallons of brine, it will have approximately \(32.50\) lb of salt.

Step-by-step solution

Define the Variables

Let \(s(t)\) be the amount of salt in the tank at any time \(t\). The inflow rate of brine is given as \(3.5\) gal/min, and the outflow rate is given as \(4\) gal/min. The inflow rate of salt is given as \(3.5 \times 4\) lb/min. Let \(V(t)\) be the volume of brine at any time \(t\), so we have \(V(t) = 200 - 0.5t\). Since the brine is well-stirred, the salt concentration leaving the tank will be \(\frac{s(t)}{V(t)}\) lb/gal. Step 2: Formulate the Differential Equation

Formulate the Differential Equation

To find the rate of change of salt in the tank at any time \(t\), we can write a differential equation. The net rate of change of salt will be the difference between the inflow rate and outflow rate of salt: \(\frac{ds}{dt} = 3.5\times 4 - 4 \times \frac{s(t)}{V(t)}\) Step 3: Solve the Differential Equation

Solve the Differential Equation

We can now solve the differential equation to find the function \(s(t)\) that describes the salt quantity in the tank at any time \(t\): \(\frac{ds}{dt} + \frac{2s(t)}{200 - 0.5t} = 14\) This differential equation is linear and first-order. To solve it, we can find the integrating factor, which will be: \(e^{\int\frac{2}{200-0.5t} dt} = e^{\int\frac{4}{400 - t} dt} = e^{-4\ln(400 - t)} = \frac{1}{(400 -t)^4}\) Now, multiply the differential equation by this integrating factor: \(\frac{1}{(400 -t)^4}\frac{ds}{dt} + \frac{2s(t)}{(400 - 0.5t)(400 - t)^4} = \frac{14}{(400 -t)^4}\) Then, we integrate both sides with respect to \(t\): \(\int\frac{1}{(400 -t)^4}\frac{ds}{dt} dt = \int\frac{14}{(400 -t)^4} dt\) Using substitution, we find that the integral on the left side is simply \(\frac{-s}{(400-t)^3}\). The integral on the right side will be: \(\int\frac{14}{(400 -t)^4} dt = \frac{7}{(400-t)^3}\) By integrating both sides, we will get the desired equation for \(s(t)\). The result will be: \(\frac{-s(t)}{(400-t)^3} = \frac{7}{(400-t)^3} + C\) Step 4: Find the Integration Constant

Find the Integration Constant

We can determine the integration constant \(C\) using the initial condition given that the tank initially contains 200 gallons of brine and 15 lb of salt, so \(s(0) = 15\): \(\frac{-15}{(400-0)^3} = \frac{7}{(400-0)^3} + C\) \(C = \frac{8}{(400)^3}\) Step 5: Calculate Amount of Salt After One Hour

Calculate Amount of Salt After One Hour

We have the equation for the amount of salt in the tank at any time \(t\): \(s(t) = (400 - t)^3\left(\frac{-7}{(400 - t)^3} + \frac{8}{400^3}\right)\) After one hour, the time will be \(t = 60\) min. Plugging this value into our equation, we can find the amount of salt in the tank: \(s(60) = (400 - 60)^3 \left(\frac{-7}{(400 - 60)^3} + \frac{8}{400^3}\right)\) \(s(60) = 340^3 \left(\frac{-7}{340^3} + \frac{8}{400^3}\right)\) \(s(60) \approx 18.98\) lb Therefore, at the end of one hour, the tank will contain approximately 18.98 lb of salt. Step 6: Calculate Amount of Salt When Tank Contains 50 Gallons

Calculate Amount of Salt When Tank Contains 50 Gallons

We need to find the time \(t\) when the tank contains 50 gallons of brine: \(50 = 200 - 0.5t\) \(t = 300\) min Now, we can use the same equation for \(s(t)\) to find the amount of salt in the tank when it contains 50 gallons of brine: \(s(300) = (400 - 300)^3\left(\frac{-7}{(400 - 300)^3} + \frac{8}{400^3}\right)\) \(s(300) = 100^3\left(\frac{-7}{100^3} + \frac{8}{400^3}\right)\) \(s(300) \approx 32.50\) lb When the tank contains 50 gallons of brine, it will have approximately 32.50 lb of salt.

Key concepts

Mathematical Modeling

Mathematical modeling refers to the process of translating a real-world scenario into mathematical terms to analyze and predict behavior. For example, consider a tank that contains a saltwater solution, also known as brine. The dynamics of the salt concentration in the tank over time can be described using mathematical equations, which often take the form of ordinary differential equations (ODEs).

In the given exercise, we build a model that represents the rate of change of the amount of salt in the tank. This involves defining relevant variables, such as the amount of salt at time t, represented by s(t), and volume of brine V(t). We then create an equation that captures the balance between the salt coming into and going out of the tank system, which is a typical setup in mathematical modeling of dynamic systems.

Integration Factor

An integration factor is a powerful mathematical tool used to solve linear first-order differential equations. It is essentially a function used to multiply through an ordinary differential equation to facilitate the integration process. The beauty of the integration factor lies in its ability to turn a non-exact differential equation into an exact one, which can then be solved using straightforward integration.

In our tank example, the integrating factor is calculated as e to the power of the integral of the function multiplying s(t) in the ODE. It simplifies the process of finding the solution by transforming the ODE into an integrable form. By integrating the multiplied factors properly, one can find solutions to otherwise cumbersome differential equations in a more systematic and manageable way.

Linear Differential Equation

A linear differential equation is an equation that involves an unknown function and its derivatives, and has the property that the unknown function and its derivatives appear to the power of one and are not multiplied by each other.

For instance, in the context of the brine tank problem, we have a linear equation because the rate at which salt is added to or removed from the tank is directly proportional to the amount of brine flowing in or out. The key feature of a linear differential equation, especially a first-order one, is that it has a standard solution method involving an integrating factor as previously discussed. The structure of these equations also ensures that superposition applies, meaning that the net rate is just the sum of the individual rates.

First-Order Differential Equations

First-order differential equations contain the first derivative of the unknown function but no higher derivatives. These equations are widely used in various fields to model rates of change. The equation for the rate of change of salt in our tank is a first-order differential equation because it involves the first derivative of s(t) with respect to time, t, and no higher-order derivatives.

First-order equations can usually be solved using methods such as separation of variables, integrating factors, or substitution. Such equations are not only common in academic exercises but are also essential in real-life applications such as population growth, radioactive decay, and, as in this problem, fluid dynamics and chemical concentrations. Understanding how to solve these equations is crucial for students as they provide foundational skills for higher-level mathematical problems and real-world applications.