Problem 15 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Water flows into a tank at the rate of \(R(t)=1+t^{2} / 5 \mathrm{~m}^{3} / \mathrm{min}\) for \(0 \leq t \leq 3\) minutes and the concentration of salt in the water is \(C(t)=3.5 e^{-t} \mathrm{~g} / 1\) at time \(t\). Write an integral that is the total amount of salt that flowed into the tank. Confirm that the units on the integral are grams of salt.

Step-by-Step Solution

Answer

The integral \( \int_{0}^{3} (1 + \frac{t^2}{5}) \cdot 3.5 e^{-t} \, dt \) represents the total salt in grams.

Step 1: Identify the Rate Function

The rate of water flow into the tank is given by \( R(t) = 1 + \frac{t^2}{5} \text{ m}^3/\text{min} \). This represents the volume of water entering the tank per minute at time \( t \).

Step 2: Identify the Concentration Function

The salt concentration in the water is given by \( C(t) = 3.5 e^{-t} \text{ g/m}^3 \). This shows how many grams of salt are present in each cubic meter of water at time \( t \).

Step 3: Establish the Expression for Amount of Salt

To find the salt being added at time \( t \), multiply the rate of water entering the tank by the concentration of salt: \( R(t) \times C(t) = \left(1 + \frac{t^2}{5}\right) \times (3.5 e^{-t}) \text{ g/min} \).

Step 4: Formulate the Integral for Total Salt

The total amount of salt added over time from \( t = 0 \) to \( t = 3 \) is given by the integral: \[\int_{0}^{3} (1 + \frac{t^2}{5}) \cdot 3.5 e^{-t} \, dt\]This integral represents the total salt amount in grams.

Step 5: Confirm the Units of the Integral

The integrand has units of grams per minute (\( (\text{m}^3/\text{min}) \times (\text{g}/\text{m}^3) = \text{g/min} \)). Integrating this rate over time (in minutes) will yield the total amount of salt in grams. Therefore, the units of the integral are indeed grams.

Key Concepts

Rate Functions

Rate functions are a crucial concept in calculus, especially in scenarios involving fluid dynamics like the one considered in the exercise. The rate function, often denoted by \( R(t) \), describes how quickly a quantity changes over time. Here, \( R(t) = 1 + \frac{t^2}{5} \) is a rate function representing the flow of water into a tank in cubic meters per minute. This function tells us that at any moment \( t \), the tank receives some volume of water, which changes as time progresses.
Understanding rate functions helps us predict future scenarios by giving us a detailed picture of change. In practical terms, it means we can determine how much resource—in this case, water—enters a system over a given period. Analyzing this function is essential for tasks like scheduling refills or understanding water tank capacity over time.
To apply rate functions effectively, consider their practical implications:

• Determine the initial rate of input: At \( t = 0 \), the rate is \( 1 \) m\(^3\)/min.
• Recognize growth: As time increases, the rate of water flow rises due to the \( t^2 \) term.
• Use integration to find cumulative totals: Integrating \( R(t) \) over a time interval gives the total volume added.

By recognizing these elements, students can accurately model real-world situations with rate functions.

Concentration in Solutions

Concentration refers to the amount of a particular substance present in a mixture or solution. In chemical solutions, concentration defines how much solute is dissolved in a given solvent quantity. For this exercise, \( C(t) = 3.5 e^{-t} \) expresses the concentration of salt in the incoming water stream, measured in grams per cubic meter.
This exponential model indicates how concentration changes over time, potentially due to factors like dilution or reaction with other substances. Understanding concentration is key to chemistry and related fields since it determines reaction rates and product formation.
Key points about concentration include:

• Initial concentration: When \( t = 0 \), the initial concentration is \( 3.5 \) g/m\(^3\).
• Exponential decay: The \( e^{-t} \) term reflects how concentration decreases over time, a common scenario when a substance disperses or reacts.
• Impact on calculations: The concentration function is multiplied by the rate function to derive the salt input rate into the tank, categorically linking these measurements.

Understanding concentration helps students predict mixture behaviors and solve exercises involving chemical solutions effectively.

Unit Conversion in Integrals

Unit conversion is essential when working with integrals, especially in problems like determining the total amount of a substance based on rate and concentration functions. In this exercise, we deal with units of time, volume, and mass:

• Volume is measured in cubic meters (m\(^3\)), while time is measured in minutes (min).
• Concentration is given in grams per cubic meter (g/m\(^3\)).
• Hence, the multiplication \( R(t) \times C(t) \) results in g/min.

We integrate the compound function \( (1 + \frac{t^2}{5}) \cdot 3.5 e^{-t} \) over time from \( t = 0 \) to \( t = 3 \). The integral effectively sums up these rates to give a total mass of salt in grams.
It's important to consistently account for units during integration. Integrating over time (min) converts a rate measured in g/min into a total quantity of grams. This is verified by:

• Checking the integrand units: g/min, which upon integration results in grams.
• Evaluating the integral correctly ensures dimensional consistency and solution validity.

By learning to check units at each step, students can confidently address integration problems involving multiple units.