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Chapter 1: Problem 1

A one-liter flask contains one liter of distilled water and \(2 \mathrm{~g}\) of salt. Repeatedly, 50 \(\mathrm{ml}\) of solution are removed from the flask and discarded after which \(50 \mathrm{ml}\) of distilled water are added to the flask. Introduce notation and write a dynamic equation that will describe the change of salt in the beaker each cycle of removal and replacement. How much salt is in the beaker after 20 cycles of removal?

Short Answer

Expert verified
Approximately 0.358 grams of salt remain after 20 cycles.

Step by step solution

01

Understand the Problem and Initial Notation

We begin by noting that we start with a one-liter flask containing 1 liter of distilled water and 2 grams of salt. We introduce notation: let \( x_n \) be the amount of salt in grams in the flask after the \( n \)-th cycle of solution removal and replacement.
02

Model the Removal Process

Each cycle involves removing 50 ml of solution from the flask. Since the total volume of the flask is 1000 ml, the concentration of salt in the solution after \( n \) cycles is \( \frac{x_n}{1000} \) grams/ml. Therefore, the amount of salt removed in 50 ml is \( 50 \times \frac{x_n}{1000} = \frac{x_n}{20} \) grams.
03

Update the Amount of Salt After Removal

After removing \( \frac{x_n}{20} \) grams of salt, the amount of salt remaining in the solution is \( x_n - \frac{x_n}{20} = x_n \left(1 - \frac{1}{20}\right) = \frac{19x_n}{20} \) grams.
04

Model the Addition Process

Since 50 ml of distilled water is added, the volume is restored to 1000 ml. However, adding distilled water does not change the amount of salt already calculated in Step 3. Thus, the amount of salt remains \( \frac{19x_n}{20} \) grams at this point.
05

Establish the Recurrence Relation

Using the calculations above, we can describe the amount of salt after \( n+1 \) cycles (denoted \( x_{n+1} \)) in terms of \( x_n \): \[ x_{n+1} = \frac{19}{20} x_n. \] This is the dynamic equation describing the change in salt after each cycle.
06

Calculate the Amount of Salt After 20 Cycles

We start with \( x_0 = 2 \) grams. Using the recurrence relation from Step 5, we find \( x_{20} \ = \left(\frac{19}{20}\right)^{20} \times 2 \approx 0.358 \) grams. Therefore, after 20 cycles of removal and addition, there are approximately 0.358 grams of salt left in the beaker.

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Key Concepts

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

Recurrence Relation
In dynamic systems, a recurrence relation helps us describe the evolution of a variable over time. In this exercise, we use a recurrence relation to model the changing amount of salt in the flask after each cycle of removing and adding solutions. A recurrence relation expresses how each term in a sequence is related to its preceding term. Here, we define the recurrence relation for the amount of salt, denoted as \( x_{n+1} \), where \( n \) represents the cycle number.

The recurrence relation given is \( x_{n+1} = \frac{19}{20} x_n \). This equation tells us that after each cycle, the amount of salt is reduced to 19/20 of the previous quantity. Each step depends on the outcome of the previous step, creating a chain of calculations. This type of relationship is common in dynamic systems where the current state influences the next state in a predictable manner.
Salt and Water Solution
A salt and water solution has two main components: the solvent (water) and the solute (salt). The initial setup involves a 1-liter flask, indicating the total volume of the liquid solution, with 2 grams of salt dissolved in distilled water.

When a portion of the solution is removed, both salt and water decrease, altering the solution's overall concentration. In this exercise, 50 ml of solution is removed per cycle, affecting the balance between salt and water. Once the same amount of distilled water is added, it resets the total volume but does not change the salt concentration left after the removal process. This repetitive process illustrates how the dynamics of a salt and water solution can be described through cycles of change, helping us understand how consistently introducing variables impacts the system.
Volume and Concentration
Volume and concentration are pivotal when analyzing the changes in a salt and water solution over multiple cycles. Volume refers to the amount of space the solution occupies, while concentration describes the amount of solute (salt) per unit volume of the solvent (water).

Initially, the salt concentration can be calculated using the formula \( \frac{x_n}{1000} \), where \( x_n \) is the amount of salt in grams. With each cycle, we remove some solution, meaning some salt is also taken out, reducing the concentration. The formula for salt removed each cycle is \( \frac{x_n}{20} \).

Despite restoring the volume each time by adding distilled water, the concentration of salt decreases with the cycle's progression, demonstrating how both volume changes (removal and addition) and the consistency of salt concentration affect the overall solution. Understanding these concepts is crucial for comprehending how systems evolve in dynamic processes.

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Most popular questions from this chapter

An egg is covered by a hen and is at \(37^{\circ} \mathrm{C}\). The hen leaves the nest and the egg is exposed to \(17^{\circ} \mathrm{C}\) air. After 20 minutes the egg is at \(34^{\circ} \mathrm{C}\). Draw a graph representative of the temperature of the egg \(t\) minutes after the hen leaves the nest. Mathematical Model. During any short time interval while the egg is uncovered, the decrease in egg temperature is proportional to the difference between the egg temperature and the air temperature. a. Introduce notation and write a dynamic equation representative of the mathematical model. b. Write a solution equation for your dynamic equation. c. Your dynamic equation should have one parameter. Use the data of the problem to estimate the parameter.

Write a solution equation for the following initial conditions and difference equations or iteration equations. In each case, compute \(B_{100}\). a. \(\quad B_{0}=100 \quad B_{t+1}-B_{t}=0.2 B_{t}+5\) b. \(\quad B_{0}=138 \quad B_{t+1}-B_{t}=0.05 B_{t}+10\) c. \(\quad B_{0}=138 \quad B_{t+1}-B_{t}=0.5 B_{t}-10\) d. \(\quad B_{0}=100 \quad B_{t+1}-B_{t}=10\) e. \(\quad B_{0}=100 \quad B_{t+1}=1.2 B_{t}-5\) f. \(\quad B_{0}=100 \quad B_{t+1}-B_{t}=-0.1 B_{t}+10\) \(\begin{array}{lll}\text { g. } & B_{0}=100 & B_{t+1} & =0.9 B_{t}-10 \\\ \text { g. } & B_{0}=100 & B_{t+1} & =-0.8 B_{t}+20\end{array}\)

Suppose \(a, b,\) and \(c\) are numbers and $$ P_{t}=a t^{2}+b t+c $$ where \(t\) is any number. Show that $$ Q_{t}=P_{t+1}-P_{t} $$ is linearly related to \(t\) (that is, \(Q_{t}=\alpha t+\beta\) for some \(\alpha\) and \(\left.\beta\right)\), and that $$ R_{t}=Q_{t+1}-Q_{t} $$ is a constant.

Two kilos of a fish poison that does not decompose are mixed into a lake that has a volume of \(100 \times 20 \times 2=4000\) cubic meters. A stream of clean water flows into the lake at a rate of 1000 cubic meters per day. Assume that it mixes immediately throughout the whole lake. Another stream flows out of the lake at a rate of 1000 cubic meters per day. a. Write a mathematical model that describes the daily change in the amount of poison in the lake. b. Let \(P_{0}, P_{1}, P_{2}, \cdots\) denote the amounts of poison in the lake, \(P_{t}\) being the amount of poison in the lake at the beginning of the \(t \underline{h}\) day after the poison is administered. Write a dynamic equation representative of the mathematical model. c. What is \(P_{0} ?\) Compute \(P_{1}\) from your dynamic equation. Compute \(P_{2}\) from your dynamic equation. d. Find a solution equation for your dynamic equation.

Determine the doubling times of the following exponential equations. (a) \(y=2^{t}\) (b) \(y=2^{3 t}\) (c) \(y=2^{0.1 t}\) (d) \(y=10^{t}\) (e) \(y=10^{3 t}\) (f) \(y=10^{0.1 t}\)
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