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A tank initially contains 100 gal of pure water. Starting at \(t=0\), a brine containing \(4 \mathrm{lb}\) of salt per gallon flows into the tank at the rate of \(5 \mathrm{gal} / \mathrm{min}\). The mixture is kept uniform by stirring and the well-stirred mixture flows out at the slower rate of \(3 \mathrm{gal} / \mathrm{min}\). (a) How much salt is in the tank at the end of \(20 \mathrm{~min}\) ? (b) When is there \(50 \mathrm{lb}\) of salt in the tank?

Short Answer

Expert verified
(a) At the end of $20 \, \mathrm{min}$, there are approximately $15.78 \, \mathrm{lb}$ of salt in the tank. (b) There will be $50 \, \mathrm{lb}$ of salt in the tank at approximately $t = 38.99 \, \mathrm{min}$.

Step by step solution

01

Define the variables

Let t be the time in minutes, and let S(t) be the amount of salt in the tank at time t. The brine containing salt flows into the tank at a rate of 5 gal/min and with a concentration of 4 lb/gal. The mixture is kept uniform, and it flows out of the tank at a rate of 3 gal/min.
02

Write the differential equation for the salt amount

A differential equation can be derived by considering the balance of salt in the tank over time. The rate of salt entering the tank is 5 gal/min * 4 lb/gal = 20 lb/min. The rate of salt leaving the tank is (3 gal/min) * (S(t) lb / (100 + 2t) gal), where (100 + 2t) represents the total volume of liquid in the tank at time t. So, the equation can be written as: \[ \frac{dS}{dt} = 20 - \frac{3S}{100 + 2t} \]
03

Solve the differential equation

To solve the given differential equation, we can apply integrating factor method. Multiply the differential equation by the integrating factor (IF): \[ IF = e^{\int \frac{-3}{100 + 2t} dt} = e^{-\frac{3}{2} \ln (100 + 2t)} = (100 + 2t)^{-\frac{3}{2}} \] Now, the resulting equation will be exact: \[ (100 + 2t)^{-\frac{3}{2}} \frac{dS}{dt} + \frac{-3}{(100 + 2t)^{\frac{5}{2}}} S = 20(100 + 2t)^{-\frac{3}{2}} \] Now, integrate both sides: \[ \int (100 + 2t)^{-\frac{3}{2}} dS = \int 20(100 + 2t)^{-\frac{3}{2}} dt + C \] Solve for S(t): \[ S(t) = -\frac{20}{\sqrt{100 + 2t}} + \frac{C(100 + 2t)^{\frac{1}{2}}}{ \frac{1}{2}} \]
04

Determine the integration constant C

At t = 0, the tank contains no salt, i.e., S(0) = 0. Substituting these values into the equation, we can determine the value of C: \[ 0 = -\frac{20}{\sqrt{100}} + C \] Solving for C, we get: \[ C = \frac{20}{\sqrt{100}} = \frac{20}{10} = 2 \] Now we have the function for the amount of salt in the tank at any time t: \[ S(t) = -\frac{20}{\sqrt{100 + 2t}} + 2\sqrt{100 + 2t} \]
05

Find the amount of salt at the end of 20 minutes

To find the amount of salt at the end of 20 minutes (a), substitute t = 20 into the equation: \[ S(20) = -\frac{20}{\sqrt{100 + 2(20)}} + 2\sqrt{100 + 2(20)} = -\frac{20}{\sqrt{140}} + 2\sqrt{140} = 15.78 \, lb \] Thus, at the end of 20 minutes, there are approximately 15.78 lb of salt in the tank.
06

Find the time when there are 50 lb of salt in the tank

To find the time when there are 50 lb of salt in the tank (b), set S(t) = 50 and solve for t: \[ 50 = -\frac{20}{\sqrt{100 + 2t}} + 2\sqrt{100 + 2t} \] Solve for t using numerical methods (e.g., Newton-Raphson method) or using a computer software, or graphically: \[ t \approx 38.99 \, min \] Thus, there will be 50 lb of salt in the tank at approximately t = 38.99 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Factor Method
The Integrating Factor Method is a clever technique used to solve first-order linear differential equations. When faced with an equation like \( \frac{dS}{dt} + p(t)S = q(t) \), the task is to make it "exact" so it can be integrated directly.

To do this, we multiply the entire differential equation by an integrating factor, which is derived from the function \( p(t) \). For our specific problem, the equation was \( \frac{dS}{dt} = 20 - \frac{3S}{100 + 2t} \), and the integrating factor was
  • \( IF = e^{\int \frac{-3}{100 + 2t} dt} = (100 + 2t)^{-\frac{3}{2}} \).
By finding this multiplying factor, the equation becomes easier to integrate, as both sides align harmoniously. Post multiplication, you get an elegant equation that lets you calculate \( S(t) \) directly using integration.

It's crucial to remember that integrating factors come in handy for linear differential equations and are often instrumental in solving real-world problems like our salt concentration example.
Salt Concentration Problems
Salt concentration problems are a classic application of differential equations in real life. These involve tracking the amount of salt in a tank where there is a dynamic input and output.
  • Inflow: The substance flows into the tank with a known concentration and rate.
  • Outflow: The well-mixed liquid leaves the tank at maybe a different rate, changing the concentration over time.
For the exercise, we had **pure water** initially, with **brine flowing in** at **5 gal/min** carrying **4 lb of salt per gallon**. The overall volume changes due to input and output, specifically, inflow was faster than outflow causing a net change.

These problems are used often to teach rate-based dynamics through differential equations. By setting up the problem using rates of inflow and outflow, as well as initial conditions (like starting with pure water), we can develop a differential equation that describes how the substance’s concentration evolves over time.
Time-Dependent Solutions
Time-dependent solutions are key when dealing with dynamic systems in differential equations. They describe how a system changes over a specified time frame. Rather than a static solution, these evolve with every moment.
  • In our exercise, we sought to find the function \( S(t) \) which tells us the amount of salt at any given time \( t \).
  • This was done via the differential equation solution, which provided a precise function to predict salt levels not only at a specific point like 20 minutes but at any time.
The function we developed, \( S(t) = -\frac{20}{\sqrt{100 + 2t}} + 2\sqrt{100 + 2t} \), outlined how salt levels rise as brine continuously enters and leaves the system. By inputting specific times into this function, it's possible to answer key questions, such as the exact time for a specified salt amount or the amount at a predefined time.

Understanding these solutions is crucial in many scientific and engineering fields, from environmental science to chemical engineering.

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