Problem 1 - Active Calculus

Question

A tank contains \(2860 \mathrm{~L}\) of pure water. A solution that contains \(0.04 \mathrm{~kg}\) of sugar per liter enters the tank at the rate \(6 \mathrm{~L} / \mathrm{min}\). The solution is mixed and drains from the tank at the same rate. (a) How much sugar is in the tank at the beginning? \(y(0)=\) (include units) (b) With S representing the amount of sugar (in \(\mathrm{kg}\) ) at time \(\mathrm{t}\) (in minutes) write a differential equation which models this situation. \(S^{\prime}=f(t, S)=\) Note: Make sure you use a capital S, (and don't use \(\mathrm{S}(\mathrm{t})\), it confuses the computer). Don't enter units for this function. (c) Find the amount of sugar (in kg) after t minutes. \(S(t)=\square\) (function of t) (d) Find the amount of the sugar after 78 minutes. \(S(78)=\square \quad\) (include units)

Step-by-Step Solution

Answer

a) 0 kg, b) \(\frac{dS}{dt} = 0.24 - \frac{6S}{2860}\), c) \(S(t) = 114.4 (1 - e^{-\frac{6t}{2860}})\), d) 17.26 kg

Step 1: Determine the Initial Amount of Sugar

At the beginning, the tank contains only pure water, so the initial amount of sugar, denoted as \(y(0)\), is zero. Given: Water in the tank: \(2860 \mathrm{~L}\), Sugar concentration: \(0 \mathrm{~kg/L}\) Thus, \( y(0) = 0 \mathrm{~kg} \).

Step 2: Write the Differential Equation

We need to model the change in the amount of sugar \(S\) in the tank over time. The solution with \(0.04 \mathrm{~kg/L}\) sugar enters at a rate of \(6 \mathrm{~L/min}\), so the sugar entering per minute is \(0.04 \times 6 = 0.24 \mathrm{~kg/min}\). Since the solution drains at the same rate it enters, the sugar leaving per minute is \(\frac{S}{2860} \times 6 \). Thus, the differential equation is: \[ \frac{dS}{dt} = 0.24 - \frac{6S}{2860} \].

Step 3: Solve the Differential Equation

We solve the differential equation: \[ \frac{dS}{dt} = 0.24 - \frac{6S}{2860} \]This is a first-order linear differential equation: \[ \frac{dS}{dt} + \frac{6}{2860}S = 0.24 \]Using an integrating factor: \[ \mu(t) = e^{\int \frac{6}{2860} dt} = e^{\frac{6t}{2890}} \]Multiplying through by \(\mu(t)\): \[ e^{\frac{6t}{2860}} \frac{dS}{dt} + e^{\frac{6t}{2860}} \frac{6}{2860} S = 0.24 e^{\frac{6t}{2860}} \]The left side is the derivative of \( S e^{\frac{6t}{2860}} \): \[ \frac{d}{dt} \left( S e^{\frac{6t}{2860}} \right) = 0.24 e^{\frac{6t}{2860}} \]Integrate both sides with respect to \(t\): \[ S e^{\frac{6t}{2860}} = \int 0.24 e^{\frac{6t}{2860}} dt + C \]\[ S e^{\frac{6t}{2860}} = 0.24 \frac{2860}{6} e^{\frac{6t}{2860}} + C \]\[ S e^{\frac{6t}{2860}} = 114.4 e^{\frac{6t}{2860}} + C \]Divide through by \( e^{\frac{6t}{2860}} \): \[ S(t) = 114.4 + C e^{-\frac{6t}{2860}} \]. Using the initial condition \( S(0) = 0 \): \[ 0 = 114.4 + C \rightarrow C = -114.4 \]. Thus, the solution is: \[ S(t) = 114.4 (1 - e^{- \frac{6t}{2860}}) \]

Step 4: Evaluate the Amount of Sugar After 78 Minutes

Substitute \( t = 78 \) into our solution for \( S(t) \): \[ S(78) = 114.4 (1 - e^{-\frac{6 \times 78}{2860}}) \] \[ S(78) = 114.4 (1 - e^{-0.1636}) \] Calculate: \[ S(78) \approx 114.4 (1 - 0.8492) \] \[ S(78) \approx 114.4 (0.1508) \approx 17.26 \mathrm{~kg} \].

Key Concepts

Differential Equations

Differential equations play a crucial role in understanding how quantities change over time. In our exercise, we're dealing with changes in the amount of sugar in a water tank. A differential equation connects these changes to the rates at which the sugar enters and leaves the tank. Formally, a differential equation represents relations involving derivatives of a function and the function itself.
In our example, the differential equation was derived as follows:
\[\frac{dS}{dt} = 0.24 - \frac{6S}{2860}\]
Where:

• \(S\) is the amount of sugar in kg
• \(t\) is time in minutes
• \(0.24\) kg/min is the rate at which sugar enters the tank
• The term \(\frac{6S}{2860}\) represents the rate at which sugar leaves the tank

This differential equation helps us model the dynamic changes in sugar content.

Linear Differential Equations

The differential equation we derived is classified as a linear differential equation. A linear differential equation is one in which the dependent variable and its derivatives appear to the first power and are not multiplied by each other.
Our equation after rearrangement:\[\frac{dS}{dt} + \frac{6}{2860}S = 0.24\]
This equation reveals the linearity because both \(S\) and its derivative are linear terms. Linear differential equations can be conveniently solved using integrating factors or other methods that simplify integration.

Rate of Change

The rate of change is a central concept in differential equations. It tells us how a quantity is changing over time. In our exercise, the rate of change of sugar content in the tank is influenced both by the sugar entering the tank and the sugar leaving it.
In our differential equation:\[\frac{dS}{dt} = 0.24 - \frac{6S}{2860}\]

• The term \(0.24\) represents the rate at which sugar is entering the tank.
• The term \(\frac{6S}{2860}\) represents the rate at which sugar is leaving the tank.

To understand how much sugar is currently in the tank, we need to consider both rates of change: the addition and the removal.

Integration

Integration is the process of finding the antiderivative or the original function given its derivative. In solving our linear differential equation, we used integration to find the function that describes the amount of sugar over time.
Starting with the equation:\[\frac{dS}{dt} + \frac{6}{2860}S = 0.24\], we used an integrating factor to solve it.
The integrating factor was:\[\mu(t) = e^{\int \frac{6}{2860} dt} = e^{\frac{6t}{2890}}\].
By multiplying through by this integrating factor and integrating both sides, we obtained the solution:
\[S(t) = 114.4 (1 - e^{-\frac{6t}{2860}})\].
This function describes the sugar content over time after including the effects of both the entering and exiting rates.

Initial Conditions

Initial conditions determine the specific solution to a differential equation by providing values for the function and possibly its derivatives at a particular point. For our sugar tank problem, the initial condition was given as the amount of sugar in the tank at the beginning.
Initially, the tank contains pure water, so the initial amount of sugar, \(y(0)\), was zero:\[S(0) = 0\].
Using this initial condition in our solution:\[S(t) = 114.4 (1 - e^{- \frac{6t}{2860}})\], we calculated the constant of integration. By substituting \(t = 0\), we found:
\[0 = 114.4 + C \rightarrow C = -114.4\].
This specific initial condition allowed us to complete the solution and accurately describe the sugar content over time in the tank.