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Assume an initial nutrient amount of \(I\) kilograms in a tank with \(L\) liters. Assume a concentration of \(c \mathrm{~kg} / \mathrm{L}\) being pumped in at a rate of \(r \mathrm{~L} / \mathrm{min}\). The tank is well mixed and is drained at a rate of \(r \mathrm{~L} / \mathrm{min}\). Find the equation describing the amount of nutrient in the tank.

Short Answer

Expert verified
The nutrient amount at time \( t \) is \( A(t) = Lc(1 - e^{-\frac{rt}{L}}) + Ie^{-\frac{rt}{L}}. \)

Step by step solution

01

Understand the System

We have a tank with an initial nutrient amount of \( I \) kilograms and a volume of \( L \) liters. Nutrients are being added to the tank at a rate of \( r \) liters per minute with a concentration of \( c \) kg/L. The tank is well mixed, and nutrients are removed at the same rate \( r \) liters per minute.
02

Express Inflow and Outflow Rates

Nutrients are introduced into the tank by the inflow stream. The rate of nutrient input in kilograms per minute is \( r \times c \) because \( c \) is the concentration and \( r \) is the flow rate. Meanwhile, the outflow rate of nutrients in kilograms per minute is proportional to the concentration in the tank. If \( A(t) \) is the amount of nutrient at time \( t \), then the concentration in the tank is \( \frac{A(t)}{L} \). Thus, the outflow rate in kilograms per minute is \( r \frac{A(t)}{L} \).
03

Formulate a Differential Equation

The rate of change of the nutrient amount \( A(t) \) in the tank over time can be described by the equation: \[ \frac{dA}{dt} = \, \text{inflow rate} - \text{outflow rate} = rc - r\frac{A(t)}{L}. \] Simplified, it becomes: \[ \frac{dA}{dt} = rc - \frac{rA(t)}{L}. \]
04

Solve the Differential Equation

This is a first-order linear differential equation which can be solved using the integrating factor method. The integrating factor \( \mu(t) \) is \( e^{\frac{rt}{L}} \). Multiply through the differential equation by the integrating factor to get: \[ e^{\frac{rt}{L}} \frac{dA}{dt} + \frac{r}{L} e^{\frac{rt}{L}} A(t) = rc \, e^{\frac{rt}{L}}. \] The left-hand side is the derivative of \( e^{\frac{rt}{L}} A(t) \), so integrating both sides with respect to \( t \), we have: \[ e^{\frac{rt}{L}} A(t) = \int rc \, e^{\frac{rt}{L}} dt. \] This simplifies to: \[ A(t) = \left( \frac{Lc}{e^{\frac{rt}{L}}} \right)(e^{\frac{rt}{L}} - 1) + Ce^{-\frac{rt}{L}}, \] where \( C \) is a constant determined by initial conditions.
05

Apply Initial Condition

At \( t = 0 \), the nutrient amount \( A(0) = I \). Substituting in this condition, we have \( I = C \) because the initial exponential terms cancel each other. Substituting \( C = I \) into the equation gives: \[ A(t) = Lc(1 - e^{-\frac{rt}{L}}) + Ie^{-\frac{rt}{L}}. \] This equation describes the amount of nutrient in the tank over time.

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Key Concepts

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

Mixing Problems
Mixing problems are a common type of differential equation problem that involve substances being mixed at specific rates within a system, such as a tank. Imagine a tank where a nutrient solution is introduced at a certain rate, while at the same time, the solution is also drained at the same constant rate. This creates a dynamically changing chemical environment within the tank.
  • The concentration of incoming and outgoing substances needs careful consideration to predict changes over time.
  • The tank is assumed to be 'well-mixed,' meaning the substances are uniformly distributed throughout the tank at all times.
  • Such problems are ideal for modeling with differential equations because they allow us to express the rate of change of the nutrient in the tank as a function of time.
A key goal is to determine the amount of a particular substance within the tank at any given moment, which involves balancing the inflow and outflow of the substance. Understanding these dynamics is essential for accurately modeling real-world scenarios like pollutant dilution, medication dosage in the bloodstream, or even the rate of a chemical reaction in a lab setting.
First-Order Linear Differential Equation
A first-order linear differential equation is a type of equation that involves the rate of change of a function. In mixing problems, such as the one involving nutrients in a tank, these equations describe the system efficiently.The general form of a first-order linear differential equation is \[ \frac{dy}{dt} + P(t)y = Q(t), \] where \( y \) represents the function we want to find, \( P(t) \) and \( Q(t) \) are given functions of \( t \).
  • This type of equation is called 'linear' because it involves the first power of the unknown function and its derivatives.
  • The objective is to find \( y(t) \), given initial conditions and the behavior of the system over time.
  • In the context of mixing problems, \( y(t) \) might represent the nutrient concentration or amount in the tank.
Understanding and solving first-order linear differential equations is fundamental for problems where systems are changing at a rate proportional to the existing state and external inputs.
Integrating Factor Method
The integrating factor method is a powerful technique used to solve first-order linear differential equations. It simplifies the process by transforming the given equation into a form that can be easily integrated. In tackling the mixing problem, this method helps us find a functional expression that describes how the nutrient amount evolves over time.Here’s how it works:
  • Identify the equation in the form \( \frac{dy}{dt} + P(t)y = Q(t) \).
  • Compute the integrating factor \( \mu(t) = e^{\int P(t) \, dt} \).
  • Multiply through the entire differential equation by \( \mu(t) \) to ensure the left-hand side becomes the derivative of \( \mu(t) y \).
  • Integrate both sides of the equation with respect to \( t \) to find \( y(t) \).
In the case of the tank problem, the integrating factor was derived as \( e^{\frac{rt}{L}} \), leading us to integrate and finally solve for the nutrient amount \( A(t) \). This technique is particularly helpful as it provides a straightforward path to finding solutions, making it a key tool for engineers, scientists, and mathematicians dealing with dynamic systems.

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