Question

Mixture In Exercises \(35-38\) , consider a tank that at time \(t=0\) contains \(v_{0}\) gallons of a solution of which, by weight, \(q_{0}\) pounds is soluble concentrate. Another solution containing \(q_{1}\) pounds of the concentrate per gallon is running into the tank at the rate of \(r_{1}\) gallons per minute. The solution in the tank is kept well stirred and is withdrawn at the rate of \(r_{2}\) gallons per minute. Let \(Q\) be the amount of concentrate in the solution at any time t. Show that $$ \frac{d Q}{d t}+\frac{r_{2} Q}{v_{0}+\left(r_{1}-r_{2}\right) t}=q_{1} r_{1} $$

Short answer

By determining the rate of change of the amount of concentrate \(Q\) with respect to time \(t\) and accounting for the rates of concentrate flowing in and out, you derive the provided differential equation.

Step-by-step solution

Define the Variables

The variables in this problem are defined as follows: \(Q\) is the amount of concentrate in the solution at any time \(t\), \(v_{0}\) is the initial volume of the solution in the tank, \(q_{0}\) is the initial amount of concentrate in the solution by weight, \(q_{1}\) is the amount of concentrate per unit volume in the incoming solution, \(r_{1}\) is the rate at which the incoming solution is added, and \(r_{2}\) is the rate at which the solution is withdrawn from the tank.

Determine the Rate of Change of Concentrate

The rate of change of the amount of concentrate \(Q\) in the solution with respect to time \(t\) is the difference between the rate of concentrate flowing in and the rate of concentrate flowing out. The rate of concentrate flowing in is given by the product of the incoming rate of the solution and the concentration of the incoming solution, which is \(r_{1}q_{1}\). The rate of the concentrate flowing out is the total concentrate divided by the total volume times the outgoing rate (because the solution is well stirred), or \(\frac{Q}{v_{0}+(r_{1}-r_{2})t}r_{2}\)

Formulate the Differential Equation

Subtracting the outgoing rate from the incoming rate yields the differential equation representing the rate of change of concentrate in the tank with respect to time. This gives us \(\frac{d Q}{d t} + \frac{r_{2} Q}{v_{0}+\left(r_{1}-r_{2}\right) t} = q_{1} r_{1}\)

Key concepts

Rate of Change

The rate of change in a mathematical context refers to how quickly a quantity changes concerning another variable, often time. In real-world scenarios like mixture problems, this concept becomes vital for modeling situations where concentrations adjust over time due to external inputs and outputs.

Imagine filling a bathtub with water; the rate of change of the water level in the bathtub depends on the faucet flow rate (input) and the drain rate (output). Similarly, in mixture problems, we are concerned with the rate at which the concentration of a solution changes. This involves calculating the difference between the rate at which the substance is added and the rate at which it is removed.

In differential calculus, this is often represented as a derivative, written as \( \frac{dQ}{dt} \), where \( Q \) is the quantity of concentrate and \( t \) represents time. The solution to the problem mentioned above involves establishing a relationship that determines the instant rate at which concentrate changes within the tank. It's crucial in such problems to consider the system's dynamics: substances entering and leaving, which, when combined, define the overall rate of change.

Concentration of Solutions

Concentration of solutions is a measure of the amount of a given substance (solute) present in a mixture relative to the volume of solvent. It is another key factor in mixture problems involving differential equations. Concentration can be expressed in many units, such as moles per liter for chemical solutions or, as in our exercise, pounds of concentrate per gallon of solution.

In the context of the exercise provided, the concentration of the solution in the tank changes with time due to the inflow and outflow of the solution. The incoming solution has a known concentration \( q_{1} \) and is added at a rate \( r_{1} \) gallons per minute. However, as the solution is well mixed and some of it is withdrawn at a rate \( r_{2} \) gallons per minute, the concentration within the tank is not constant.

To tackle such mixture problems, a clear understanding of how these various factors play into the concentration equation is essential. One must account for the dilution effect that occurs as more solvent enters the system and the concentration effect from the continued addition of the solute.

Differential Calculus

Differential calculus is a branch of mathematics that deals with the study of rates at which quantities change. It's the tool we use in calculus to find the rate of change of one variable relative to another. It involves functions, derivatives, and the application of the derivative known as the differential.

In the context of the mixture problem, differential calculus is employed to formulate and solve the differential equation that describes the rate of change of the concentrate in the tank over time. A differential equation is a mathematical equation that relates a function with its derivatives. In our exercise, the presence of \( \frac{dQ}{dt} \) shows us we are dealing with a differential equation, where \( Q \) is a function that represents the quantity of concentrate over time \( t \).

To solve mixture problems using differential calculus, one typically sets up a relationship using the initial conditions and the rates of change, both for input and output. The goal is to find the function \( Q(t) \) that satisfies the differential equation. In doing so, we can predict the concentration of the solution at any given time \( t \) and manage the mixture process accordingly.