Skip to main content
Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 9, Problem 9.PE.44c

In Exercises 43 and 44, let S represent the pounds of salt in a tank at time t minutes. Set up a differential equation representing the given information and the rate at which S changes. Then solve for S and answer the particular questions.


Pure water flows into a tank at the rate of 4 gal/min, and the well-stirred mixture flows out of the tank at the rate of 5 gal/min. The tank initially holds 200 gal of solution containing 50 pounds of salt.


c. When will the tank have exactly 5 pounds of salt and how many gallons of solution will be in the tank?

Verified step by step guidance
1
Identify the variables and given information: Let \(S(t)\) be the amount of salt (in pounds) in the tank at time \(t\) (in minutes). The inflow rate of pure water is 4 gal/min, and the outflow rate of the mixture is 5 gal/min. The initial volume of solution is 200 gallons, and initially, there are 50 pounds of salt.
Express the volume of solution in the tank at time \(t\): Since water flows in at 4 gal/min and flows out at 5 gal/min, the volume decreases by 1 gal/min. Therefore, the volume at time \(t\) is \(V(t) = 200 - t\) gallons.
Set up the differential equation for the rate of change of salt \(S(t)\): The rate of salt entering is zero because pure water contains no salt. The rate of salt leaving is proportional to the concentration of salt in the tank times the outflow rate. The concentration is \(\frac{S(t)}{V(t)}\), so the rate out is \(5 \times \frac{S(t)}{V(t)}\). Thus, the differential equation is: \[ \frac{dS}{dt} = 0 - 5 \times \frac{S(t)}{200 - t} = - \frac{5S}{200 - t} \]
Solve the differential equation: This is a separable equation. Rewrite it as: \[ \frac{dS}{S} = - \frac{5}{200 - t} dt \] Integrate both sides to find \(S(t)\), including the constant of integration determined by the initial condition \(S(0) = 50\).
Use the solution \(S(t)\) to find when the tank has exactly 5 pounds of salt by solving \(S(t) = 5\). Then, find the volume at that time using \(V(t) = 200 - t\) gallons.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
11m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Setting Up a Differential Equation for Mixing Problems

In mixing problems, the rate of change of the quantity of a substance (like salt) in a tank is modeled by a differential equation. This equation accounts for the rate at which the substance enters and leaves the tank, considering inflow and outflow rates and concentrations. Understanding how to express these rates mathematically is essential to formulating the correct differential equation.
Recommended video:
07:39
Classifying Differential Equations

Solving First-Order Linear Differential Equations

The differential equation derived from the mixing problem is typically a first-order linear ODE. Solving it involves finding an integrating factor or using separation of variables to obtain an explicit formula for the amount of substance over time. Mastery of these solution techniques allows one to predict the system's behavior at any given time.
Recommended video:
06:06
Solving Separable Differential Equations

Interpreting the Solution to Find Specific Values

After solving the differential equation, interpreting the solution involves substituting values to answer specific questions, such as when the salt quantity reaches a certain amount. Additionally, understanding how volume changes over time due to differing inflow and outflow rates is necessary to determine the solution volume at that time.
Recommended video:
05:12
Finding Global Extrema (Extreme Value Theorem)