Problem 81 - College Algebra with Corequisite Support

Question

For the following exercises, express a rational function that describes the situation. A large mixing tank currently contains 300 gallons of water, into which 8 pounds of sugar have been mixed. A tap will open, pouring 20 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 2 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after \(t\) minutes.

Step-by-Step Solution

Answer

The concentration function is \( C(t) = \frac{8 + 2t}{300 + 20t} \).

Step 1: Understand initial conditions

The tank initially has 300 gallons of water and 8 pounds of sugar. The initial concentration of sugar is \( \frac{8}{300} \) pounds per gallon.

Step 2: Recognize changes over time

Water is added at 20 gallons per minute, and sugar is added at 2 pounds per minute. Thus, after \( t \) minutes, the volume of water is \( 300 + 20t \) gallons, and the total mass of sugar is \( 8 + 2t \) pounds.

Step 3: Formulate the rational function

The concentration of sugar as a function of time, \( C(t) \), is given by the ratio of the total sugar to the total water. This is expressed by the rational function:\[C(t) = \frac{8 + 2t}{300 + 20t}\]This function represents the concentration in pounds per gallon at any time \( t \).

Step 4: Verify the function's behavior

Check the function at \( t = 0 \): \( C(0) = \frac{8 + 2 \times 0}{300 + 20 \times 0} = \frac{8}{300} \), confirming it matches the initial concentration. As \( t \to \infty \), \( C(t) \to 0.1 \), matching the continuous rate inputs at equilibrium.

Key Concepts

Concentration Calculation

The concept of concentration is essential when we want to describe how much solute (in this case, sugar) is present relative to the solvent (water). Here, concentration is measured in pounds per gallon, which gives us an idea of how 'sweet' the mixture in the tank is. Initially, we start with 8 pounds of sugar spread throughout 300 gallons of water. To find this initial concentration, we calculate \( \frac{8}{300} \) pounds per gallon.

Now, as time progresses, both water and sugar are being added simultaneously. After \( t \) minutes, we've added \( 2t \) additional pounds of sugar and \( 20t \) additional gallons of water. Thus, the new concentration becomes \( \frac{8 + 2t}{300 + 20t} \) which continuously measures how the mixture's sweetness changes over time. This calculation gives us a clear snapshot of sugar concentration at any time \( t \).

• Initial concentration: \( \frac{8}{300} \) pounds per gallon
• Sugar added per minute: 2 pounds
• Water added per minute: 20 gallons
• Concentration equation: \( C(t) = \frac{8 + 2t}{300 + 20t} \)

Understanding this helps us see the balance of increasing sugar and water, maintaining a specific concentration level.

Rate of Change

The term 'rate of change' refers to how a quantity shifts over time—in our scenario, it's how quickly sugar and water quantities change within the tank. Here, the problem specifies two rates:

• Sugar is added at a rate of 2 pounds per minute.
• Water is added at a rate of 20 gallons per minute.

This steady addition means that the concentration's rate of change isn't static but adjusts as both sugar and water inflate proportionally. The formula \( C(t) = \frac{8 + 2t}{300 + 20t} \) captures this dynamic change—it's a rational function that adjusts for both added elements.

By focusing on these rates, we understand how quickly or slowly the mixing process impacts the mixture's concentration. The rate at which both components are added ultimately defines the curve or decrease of concentration over time; it describes a gradual decline or approach towards an equilibrium, bridging the function towards a stable concentration.

Function of Time

When we describe a 'function of time,' we're talking about how variables like concentration change as time, \( t \), progresses. In this exercise, both the sugar amount and water volume are functions that rely directly on the passage of time.

The rational function \( C(t) = \frac{8 + 2t}{300 + 20t} \) effectively demonstrates how concentration is related to time. This is because:

• For each passing minute, \( t \), both the numerator \( 8 + 2t \) and the denominator \( 300 + 20t \) grow steadily.
• This aligned growth suggests that initially, the concentration changes significantly, responding to the new inputs.
• As \( t \to \infty \), the concentration approaches \( 0.1 \), indicating a balance won't change much as time continues.

Understanding concentration as a function of time allows us to predict future scenarios and comprehend how systems evolve—especially important in real-world applications, like chemical mixing or water treatment processes. The judgment of how smoothly it approaches stability informs us about the efficiency and effect of the mixing system described.