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 The solution in the tank is kept well stirred and is withdrawn at the rate of \(r_{2}\) gallons per minute. A 200 -gallon tank is full of a solution containing 25 pounds of concentrate. Starting at time \(t=0\) , distilled water is admitted to the tank at a rate of 10 gallons per minute, and the well-stirred solution is withdrawn at the same rate. (a) Find the amount of concentrate \(Q\) in the solution as a function of \(t .\) (b) Find the time at which the amount of concentrate in the tank reaches 15 pounds. (c) Find the quantity of the concentrate in the solution as \(t \rightarrow \infty\) .

Short answer

The amount of concentrate in the solution as a function of time is given by Q(t) = 25*\(e^{-t/20}\). The time at which the amount of concentrate in the tank reaches 15 pounds is approximately -20*ln(0.6) minutes. The quantity of concentrate in the solution as time t tends to infinity is zero.

Step-by-step solution

Differential Equation Setup

The rate of change of weight of the concentrate in the tank is the difference between the rate at which the concentrate enters the tank and the rate at which it leaves. However, since distilled water is admitted to the tank, it does not bring in any concentrate. Therefore, the rate of entry of the concentrate is 0. The rate at which concentrate leaves the tank is proportional to the percentage of concentrate in the tank i.e., \(Q/200\) where Q(t) represents the amount of concentrate in the tank at time t. Hence we have the following differential equation: dQ/dt = -10*\(Q/200\) = -\(Q/20\).

Differential Equation Solution

Separating variables and integrating, we get \(\int dQ/Q = -\int (1/20) dt\), which yields ln|Q| = -t/20 + C. After applying the exponential function to both sides we get Q = \(e^{-t/20 + C}\) = \(e^{-t/20}\)*\(e^C\), where \(e^C\) becomes a constant we can denote by A for simplicity. Thus the amount Q of concentrate in the solution as a function of time t is Q(t) = A*\(e^{-t/20}\).

Determination of the Constant of Integration A

Now we use the initial condition provided in the problem: at time t = 0, Q = 25 pounds. Therefore: 25 = A*\(e^{0}\), from which we find A = 25. So, the function Q(t) is given by Q(t) = 25*\(e^{-t/20}\).

Finding the Time When the Amount of Concentrate is 15 Pounds

We set 15 = 25*\(e^{-t/20}\) and solve for t, yielding t = -20*ln(0.6).

Long-term Quantity of Concentrate as t -> infinity

We consider the limit as t approaches infinity of our quantity function. Since the base of the natural logarithm e is positive but less than one, and since t is in the exponent, the exponential term \(e^{-t/20}\) tends to zero as t tends to infinity. Therefore, the quantity Q tends to zero as t approaches infinity.

Key concepts

Rate of Change

The concept of 'Rate of Change' is fundamental in understanding differential equations, particularly in scenarios involving fluid dynamics in a tank. Here, the rate of change of the amount of concentrate, denoted as \( Q(t) \), describes how quickly this quantity changes over time.
In this tank problem, the key point is that the rate at which the concentrate leaves the tank is proportional to its concentration in the solution. If 10 gallons of water are admitted and withdrawn each minute, and each gallon is uniformly mixed, then the rate of change can be represented by the differential equation: \( \frac{dQ}{dt} = -\frac{Q}{20} \).
Understanding this equation boils down to recognizing that the amount leaving is based on the proportion left in the tank. This differentiation shows how even the absence of incoming concentrate results in the solution's concentration diminishing over time.

Initial Conditions

When solving differential equations, the 'Initial Conditions' are vital as they provide a necessary starting point to uniquely determine the solution. For this exercise, at time \( t = 0 \), the tank contains 25 pounds of concentrate.
This initial condition allows us to determine the constant in the solution expression for \( Q(t) \).

• At \( t = 0 \), the known amount of concentrate is represented as \( Q(0) = 25 \).
• This sets the context for solving the equation \( 25 = A \cdot e^0 \), leading to the determination of the constant \( A = 25 \).

This step is crucial for slope field analysis, allowing integration results to specify a particular trajectory in solution plots.

Exponential Decay

This problem beautifully illustrates 'Exponential Decay', a key pattern in mathematics and physics. An exponentially decaying function decreases quickly at first and then more slowly as time passes.
The function derived for the amount of concentrate is \( Q(t) = 25 \cdot e^{-t/20} \). The essence of exponential decay here is that the rate of loss of concentrate is proportional to its current amount.

• The base of the natural exponent \( e^{-t/20} \) describes how this decaying effect occurs at each step.
• Real-world applications often include radioactive decay and capacitor discharge.

Understanding exponential decay is crucial as it informs us why even without additional factors, the solution itself will asymptotically approach zero, signifying an almost complete absence of the original state—the concentrated solution.

Separation of Variables

'Separation of Variables' is a method used to solve simple differential equations, particularly useful in this problem. When equations are in a separable form, this technique involves manipulating the equation to isolate different types of terms on either side.
In the original equation \( \frac{dQ}{dt} = -\frac{Q}{20} \), both \( Q \) and \( t \) need manipulation. Here's what happens:

• Rearrange terms: \( \int \frac{1}{Q} \; dQ = -\int \frac{1}{20} \; dt \).
• Integration then provides: \( \ln|Q| = -\frac{t}{20} + C \), where \( C \) is the constant of integration.

The technique makes solving the differential equation straightforward by utilizing basic calculus integration, revealing the foundational link between interest in rates of change and their resulting functions.