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Chapter 1: Problem 4

Flushing chlorine: City water, which is slightly chlorinated, is being used to flush a tank of heavily chlorinated water. The concentration \(C=C(t)\) of chlorine in the tank \(t\) hours after flushing begins is given by \(C=0.1+2.78 e^{-0.37 t}\) milligrams per gallon a. What is the initial concentration of chlorine in the tank? b. Express the concentration of chlorine in the tank after 3 hours using functional notation, and then calculate its value.

Short Answer

Expert verified
The initial concentration is 2.88 mg/gal. After 3 hours, the concentration is approximately 1.014 mg/gal.

Step by step solution

01

Determine Initial Chlorine Concentration

To find the initial concentration of chlorine in the tank, evaluate the concentration function at time \( t = 0 \). This is represented by \( C(0) = 0.1 + 2.78 e^{-0.37 \cdot 0} \). Simplify the expression by recognizing that \( e^0 = 1 \), resulting in \( C(0) = 0.1 + 2.78 \cdot 1 = 2.88 \) mg/gal.
02

Express Concentration After 3 Hours

Next, express the concentration as \( C(3) \), where 3 represents the number of hours after flushing begins. Substitute 3 for \( t \) in the equation \( C(t) = 0.1 + 2.78 e^{-0.37 t} \). This gives \( C(3) = 0.1 + 2.78 e^{-0.37 \cdot 3} \).
03

Calculate Concentration After 3 Hours

Now, calculate the value of the expression \( C(3) = 0.1 + 2.78 e^{-1.11} \). Using a calculator, first find \( e^{-1.11} \) which approximately equals 0.329, then multiply by 2.78 to get 0.914. Adding 0.1 gives approximately \( C(3) \approx 1.014 \) mg/gal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chlorination
Chlorination is the process of adding chlorine to water as a method of water purification. Chlorine is a powerful disinfectant, effective at killing bacteria and other harmful microorganisms. In our example, the city water has a slight chlorine concentration, and it's used to flush a tank containing heavily chlorinated water. This helps lower the concentration of chlorine in the tank over time. The initial high levels of chlorine in the tank need to be reduced, as too much chlorine can be harmful to both humans and the environment. The action of flushing鈥攚here fresh, gradually chlorinated city water is introduced鈥攆acilitates a process of exponential decay of chlorine concentration in the tank. Understanding chlorination is crucial for managing safe levels of chlorine in drinking water supplies and preventing health hazards from excessive chlorine exposure.
Concentration Function
In this scenario, the concentration function models how the level of chlorine decreases over time in the tank. The function is given by \[ C(t) = 0.1 + 2.78 e^{-0.37t} \] where \(C(t)\) represents the concentration of chlorine in milligrams per gallon at any time \(t\) in hours.- **0.1**: This represents the base concentration level in the city water. It indicates the residual concentration that remains after infinite time.- **2.78**: This is the initial additional concentration from the heavily chlorinated state of the tank at \(t=0\). The high initial concentration decays over time.- **\(e^{-0.37t}\)**: This term demonstrates exponential decay. The rate of \(0.37\) indicates how quickly the chlorine concentration decreases within the tank.This function is an example of an exponential decay model frequently used in scenarios where a substance reduces in quantity at a rate proportional to its current amount. These can include processes like radioactive decay, cooling, or, as in this context, chemical concentration change.
Initial Value Problem
An initial value problem in mathematics involves finding a function or a quantity that models a physical or real-life scenario, given an initial condition. Here, we have a differential equation scenario transforming into a problem of an exponential decay. The initial value problem is presented through identifying the concentration of chlorine at time \(t = 0\), referred to as the initial condition. The initial concentration of chlorine, calculated earlier as \(C(0) = 2.88\) mg/gal, sets a starting point for the function's behavior.Solving such a problem usually involves:
  • Defining the initial state of the system (e.g., time-zero concentration).
  • Using this information to predict future states under particular conditions (e.g., after a number of hours).
In practice, solving an initial value problem gives us a roadmap of how a situation like the chlorination level will evolve under the current circumstances. This is particularly helpful for operational adjustments in water treatment facilities to maintain safe chlorine levels.

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Most popular questions from this chapter

Newton's law of cooling says that a hot object cools rapidly when the difference between its temperature and that of the surrounding air is large, but it cools more slowly when the object nears room temperature. Suppose a piece of aluminum is removed from an oven and left to cool. The following table gives the temperature \(A=A(t)\), in degrees Fahrenheit, of the aluminum \(t\) minutes after it is removed from the oven. $$ \begin{array}{|c|c|} \hline t=\text { Minutes } & A=\text { Temperature } \\ \hline 0 & 302 \\ \hline 30 & 152 \\ \hline 60 & 100 \\ \hline 90 & 81 \\ \hline 120 & 75 \\ \hline 150 & 73 \\ \hline 180 & 72 \\ \hline 210 & 72 \\ \hline \end{array} $$ a. Explain the meaning of \(A(75)\) and estimate its value. b. Find the average decrease per minute of temperature during the first half- hour of cooling. c. Find the average decrease per minute of temperature during the first half of the second hour of cooling. d. Explain how parts b and c support Newton's law of cooling. e. Use functional notation to express the temperature of the aluminum after 1 hour and 13 minutes. Estimate the temperature at that time. (Note: Your work in part c should be helpful.) f. What is the temperature of the oven? Express your answer using functional notation, and give its value. g. Explain why you would expect the function \(A\) to have a limiting value. h. What is room temperature? Explain your reasoning.

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What if interest is compounded more often than monthly? Some lending institutions compound interest daily or even continuously. (The term continuous compounding is used when interest is being added as often as possible - that is, at each instant in time.) The point of this exercise is to show that, for most consumer loans, the answer you get with monthly compounding is very close to the right answer, even if the lending institution compounds more often. In part 1 of Example 1.2, we showed that if you borrow \(\$ 7800\) from an institution that compounds monthly at a monthly interest rate of \(0.67 \%\) (for an APR of \(8.04 \%\) ), then in order to pay off the note in 48 months, you have to make a monthly payment of \(\$ 190.57\). a. Would you expect your monthly payment to be higher or lower if interest were compounded daily rather than monthly? Explain why. b. Which would you expect to result in a larger monthly payment, daily compounding or continuous compounding? Explain your reasoning. c. When interest is compounded continuously, you can calculate your monthly payment \(M=\) \(M(P, r, t)\), in dollars, for a loan of \(P\) dollars to be paid off over \(t\) months using $$ M=\frac{P\left(e^{r}-1\right)}{1-e^{-r t}}, $$ where \(r=\frac{A P R}{12}\) if the APR is written in decimal form. Use this formula to calculate the monthly payment on a loan of \(\$ 7800\) to be paid off over 48 months with an APR of \(8.04 \%\). How does this answer compare with the result in Example 1.2?

Total cost: The total cost \(C\) for a manufacturer during a given time period is a function of the number \(N\) of items produced during that period. To determine a formula for the total cost, we need to know the manufacturer's fixed costs (covering things such as plant maintenance and insurance), as well as the cost for each unit produced, which is called the variable cost. To find the total cost, we multiply the variable cost by the number of items produced during that period and then add the fixed costs. Suppose that a manufacturer of widgets has fixed costs of \(\$ 9000\) per month and that the variable cost is \(\$ 15\) per widget (so it costs \(\$ 15\) to produce 1 widget). a. Use a formula to express the total cost \(C\) of this manufacturer in a month as a function of the number of widgets produced in a month. Be sure to state the units you use. b. Express using functional notation the total cost if there are 250 widgets produced in a month, and then calculate that value.

Hubble's constant: Astronomers believe that the universe is expanding and that stellar objects are moving away from us at a radial velocity \(V\) proportional to the distance \(D\) from Earth to the object. a. Write \(V\) as a function of \(D\) using \(H\) as the constant of proportionality. b. The equation in part a was first discovered by Edwin Hubble in 1929 and is known as Hubble's law. The constant of proportionality \(\mathrm{H}\) is known as Hubble's constant. The currently accepted value of Hubble's constant is 70 kilometers per second per megaparsec. (One mega- \(\rightarrow\) parsec is about \(3.085 \times 10^{19}\) kilometers.) With these units for \(H\), the distance \(D\) is measured in megaparsecs, and the velocity \(V\) is measured in kilometers per second. The galaxy G2237 \(+305\) is about \(122.7\) megaparsecs from Earth. How fast is G2237 + 305 receding from Earth? c. One important feature of Hubble's constant is that scientists use it to estimate the age of the universe. The approximate relation is $$ y=\frac{10^{12}}{H} $$ where \(y\) is time in years. Hubble's constant is extremely difficult to measure, and Edwin Hubble's best estimate in 1929 was about 530 kilometers per second per megaparsec. What is the approximate age of the universe when this value of \(H\) is used? d. The calculation in part c would give scientists some concern since Earth is thought to be about \(4.6\) billion years old. What estimate of the age of the universe does the more modern value of 70 kilometers per second per megaparsec give?
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