Problem 22 - Calculus Early Transcendentals

Question

A tank initially contains 200 gal of pure water. Then at time \(t=0\) brine containing 5 lb of salt per gallon of brine is allowed to enter the tank at a rate of 20 gal/min and the mixed solution is drained from the tank at the same rate. (a) How much salt is in the tank at an arbitrary time \(t ?\) (b) How much salt is in the tank after 30 min?

Step-by-Step Solution

Answer

(a) \( y(t) = 1000 - 1000e^{-t/10} \); (b) approximately 950.21 lb.

Step 1: Understand the problem setup

We have a tank with 200 gallons of pure water, which means initially there is no salt in the tank. Brine with 5 lb/gal salt concentration enters at 20 gal/min, and the mixture leaves the tank at the same rate, maintaining a constant volume.

Step 2: Define the Variables

Let \( y(t) \) be the amount of salt (in pounds) in the tank at time \( t \) minutes. Initially \( y(0) = 0 \). The inflow of salt is \( 5 \text{ lb/gal} \times 20 \text{ gal/min} = 100 \text{ lb/min} \).

Step 3: Determine the outflow rate

Since the solution (water + salt) is being drained at 20 gal/min and there are \(y(t)\) pounds of salt in the 200 gallons, the outflow rate of salt is \(\frac{20y(t)}{200} = \frac{y(t)}{10} \text{ lb/min}\).

Step 4: Set up the differential equation

The rate of change of salt in the tank is the inflow rate of salt minus the outflow rate. Therefore, \( \frac{dy}{dt} = 100 - \frac{y}{10} \).

Step 5: Solve the differential equation

The equation \( \frac{dy}{dt} = 100 - \frac{y}{10} \) can be rewritten and solved as a linear differential equation: \[\frac{dy}{dt} + \frac{y}{10} = 100.\]The integrating factor is \( e^{t/10} \), so multiplying through, we have:\[e^{t/10} \frac{dy}{dt} + \frac{y}{10}e^{t/10} = 100e^{t/10}.\]This simplifies to:\[\frac{d}{dt}(y \, e^{t/10}) = 100e^{t/10}.\]Integrate both sides:\[y \, e^{t/10} = 1000e^{t/10} + C.\]

Step 6: Apply initial conditions

Since \( y(0) = 0 \), substituting gives: \[0 = 1000 + C \rightarrow C = -1000\]Thus, \[ y \, e^{t/10} = 1000e^{t/10} - 1000.\]Dividing through by \( e^{t/10} \),\[ y(t) = 1000 - 1000e^{-t/10}.\]

Step 7: Calculate salt at arbitrary time \( t \)

The expression \( y(t) = 1000 - 1000e^{-t/10} \) gives the amount of salt at any time \( t \).

Step 8: Calculate salt in the tank after 30 minutes

Substitute \( t = 30 \) into the expression: \[y(30) = 1000 - 1000e^{-3}.\]Calculate \( 1000e^{-3} \) to find the exact amount of salt after 30 minutes.

Key Concepts

Rate of Change

The concept of "rate of change" is fundamental when working with differential equations. It describes how much a quantity, such as the amount of salt in a tank, changes over a period of time. In the context of the provided exercise, the rate at which salt is changing in the tank is determined by both its inflow and outflow rates.

Here’s a breakdown of how these rates function:

• **Inflow Rate:** Salt is entering the tank at 5 pounds per gallon, with a flow of 20 gallons per minute. Thus, the inflow rate is calculated as: \(5 \times 20 = 100 \text{ lb/min}\).
• **Outflow Rate:** The salt mixture is draining out at the same rate (20 gallons per minute). Given that there are \(y(t)\) pounds of salt in the mixture, the outflow rate becomes: \( \frac{y(t)}{10} \text{ lb/min}\).

The differential equation \( \frac{dy}{dt} = 100 - \frac{y}{10} \) defines the rate of change of salt within the tank, emphasizing that it is the inflow minus the outflow. Understanding this rate is crucial for solving when and how the amount of salt changes over time.

Integrating Factor

When faced with a linear first-order differential equation, like \( \frac{dy}{dt} + \frac{y}{10} = 100 \), employing the integrating factor method can significantly simplify finding the solution.

The integrating factor, derived from the standard form of the differential equation, is \( e^{t/10} \). Follow these steps to apply it:

• **Multiply Through by Integrating Factor:** Multiply every term in the equation by \( e^{t/10} \) to transform the left-hand side into an exact derivative. This results in: \( e^{t/10} \frac{dy}{dt} + \frac{y}{10}e^{t/10} = 100e^{t/10} \).
• **Simplify to Obtain an Exact Derivative:** The equation simplifies to: \( \frac{d}{dt}(y \, e^{t/10}) = 100e^{t/10} \). This form makes integration straightforward.
• **Integrate Both Sides:** Integrate to find \( y \, e^{t/10} = 1000e^{t/10} + C \), where \( C \) is a constant of integration.

Using the integrating factor helps convert the differential equation into a form that's much easier to integrate and solve, providing a direct path to the solution.

Initial Conditions

Initial conditions in differential equations are values of the variables at a specific point in time, which provide a starting point for the solution. They are crucial for determining the particular solution to a differential equation. In the tank problem, the initial condition given is that there is no salt in the tank at time \( t = 0 \).

This translates to the condition \( y(0) = 0 \). By applying this specific condition to the solution derived from the equation, you can solve for the constant \( C \) involved in the integration process:

• Substitute the initial condition into the integrated form: \( 0 = 1000 + C \).
• Solve for \( C \): Here, you find \( C = -1000 \).

Applying the initial condition allows us to refine our general solution \( y \, e^{t/10} = 1000e^{t/10} - 1000 \) to a particular solution that fits the specific situation of the tank initially containing no salt. This ensures the solution accurately models the real-world scenario described in the problem.