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Chapter 4: Problem 40

Chemistry Salt water with a concentration 0.1 pounds of salt per gallon flows into a large tank that initially holds 100 gallons of pure water. If 5 gallons of salt water per minute flows into the tank, show that the concentration of salt in the tank is given by $$ c(t)=\frac{t}{200+10 t} $$ where \(t\) is measured in minutes. What is the rate of change of \(c\) with respect to \(t ?\)

Short Answer

Expert verified
The rate of change of concentration \( \frac{dc}{dt} = \frac{200}{(200 + 10t)^2} \).

Step by step solution

01

Determine Initial Condition

Initially, the tank contains 100 gallons of pure water, meaning there is 0 pounds of salt in the tank at time \(t = 0\).
02

Calculate Salt Inflow

Salt water with a concentration of 0.1 pounds per gallon flows into the tank at 5 gallons per minute. Thus, \(5 \times 0.1 = 0.5\) pounds of salt flow into the tank per minute.
03

Determine Volume Over Time

The volume of the water in the tank increases by 5 gallons each minute. So, at time \(t\), the volume \(V(t)\) is given by \(V(t) = 100 + 5t\).
04

Calculate Salt Amount Over Time

The amount of salt in the tank at time \(t\) is given by \(0.5t\) since 0.5 pounds of salt are added each minute.
05

Find Concentration of Salt

The concentration \(c(t)\) is the amount of salt divided by the volume of water, \[ c(t) = \frac{0.5t}{100 + 5t} = \frac{t}{200 + 10t} \]. This matches the given expression.
06

Rate of Change of Concentration

To find the rate of change of concentration \( c(t) \), we need to differentiate \( c(t) = \frac{t}{200 + 10t} \) with respect to \( t \). Using the quotient rule, \[ \frac{dc}{dt} = \frac{(200 + 10t)(1) - t(10)}{(200 + 10t)^2} = \frac{200}{(200 + 10t)^2} \].

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Key Concepts

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

Concentration Calculation
In the context of a saltwater mixing problem, concentration refers to the amount of salt present per unit of liquid, usually expressed as pounds per gallon. When calculating concentration, two components are essential: the amount of solute (here salt) and the total volume of solution (here water).
  • The initial condition starts with 100 gallons of pure water and hence an initial concentration of 0 since there is no salt present.
  • The problem involves a continual inflow of saltwater, adding 0.5 pounds of salt every minute (calculated from 0.1 pounds per gallon at 5 gallons per minute).
  • The formula for concentration thus evolves over time, becoming a function of both the amount of salt and the time-dependent total volume of water.
The expression for concentration over time is given by\[ c(t) = \frac{t}{200 + 10t} \]where 0.5t represents the pounds of salt after t minutes, and 100 + 5t is the new volume in gallons. The concentration gradually increases as both salt and total volume change over time.
Rate of Change
The rate of change refers to how the concentration changes with respect to time. Understanding this concept involves differentiation, used to find how quickly concentration increases or decreases as time progresses.
  • Here, we need to determine how \( c(t) \), given by \( \frac{t}{200 + 10t} \), changes with respect to time \( t \).
  • This process requires taking the derivative \( \frac{dc}{dt} \) of the concentration formula.
Finding this derivative will give the instantaneous rate at which salt concentration in the tank changes as more saltwater is added - a crucial aspect in understanding the efficiency of mixing processes.
Quotient Rule
The quotient rule is a technique used in calculus to differentiate functions that are expressed as a quotient of two other functions. For a function \( f(t) = \frac{g(t)}{h(t)} \), the derivative \( f'(t) \) is calculated using:\[ f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{(h(t))^2} \]
  • In our problem, \( g(t) = t \) and \( h(t) = 200 + 10t \), making the derivative \( \frac{dc}{dt} \) crucial for determining the rate of concentration change.
  • Applying the quotient rule to these functions, \( g'(t) = 1 \) and \( h'(t) = 10 \), allows us to find the rate of change of concentration over time.
Ultimately, this gives:\[ \frac{dc}{dt} = \frac{(200 + 10t)(1) - t(10)}{(200 + 10t)^2} = \frac{200}{(200 + 10t)^2} \]This derivative indicates how the concentration changes with each passing minute.
Saltwater Mixing Problem
The saltwater mixing problem is a classic example in differential equations where you evaluate how variables such as concentration change over time as a result of various conditions. Here, water dynamics come into play with the inflow of saltwater into a tank originally containing pure water.
  • Such problems require understanding not just how much salt is added, but how the interaction with constantly increasing total volume affects concentration.
  • Key to solving these problems is setting up differential equations that consider both inflow rates and changes in volume.
This scenario allows students to see practical applications of calculus and differential equations in tracking real-life mixing scenarios, demonstrating how theoretical mathematical techniques can be practically applied to understand everyday processes such as dilution and concentration changes over time.

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