Problem 44 - Calculus Single Variable

Question

A \(1000 \mathrm{~L}\) tank initially contains a \(200 \mathrm{~L}\) solution in which \(24 \mathrm{~kg}\) of salt is dissolved. Beginning at time \(t=0,\) an inlet valve allows fresh water to flow into the tank at the constant rate of \(12 \mathrm{~L} / \mathrm{min},\) and an outlet valve is opened so that \(10 \mathrm{~L} / \mathrm{min}\) of the solution is drained. When the tank has been filled, how much salt does it contain?

Step-by-Step Solution

Answer

When filled, the tank contains approximately 3.84 kg of salt.

Step 1: Identify Initial Conditions

The tank initially contains 200 L of solution with 24 kg of salt. The volume (\(V(t)\)) of solution changes over time due to inflow and outflow rates.

Step 2: Set Up Volume Equation

The volume of solution in the tank at time \(t\) is given by \(V(t) = 200 + (12 - 10)t = 200 + 2t\) L. The tank will be full, i.e., 1000 L, when \(200 + 2t = 1000\). Solve for \(t\): \(t = \frac{800}{2} = 400\) minutes.

Step 3: Set Up Salt Content Differential Equation

Let \(S(t)\) denote the amount of salt in the tank at time \(t\). Since only pure water is added, the rate of salt removal depends on the concentration and the outflow, given by \(\frac{dS}{dt} = -\frac{10S}{V(t)}\).

Step 4: Solve the Differential Equation

The differential equation \(\frac{dS}{dt} = -\frac{10S}{200 + 2t}\) is separable. Rearrange terms: \(\frac{dS}{S} = -\frac{10}{200 + 2t}dt\). Integrate both sides: \(\ln |S| = -5\ln |200 + 2t| + C\). Exponentiate: \(S(t) = \frac{A}{(200 + 2t)^5}\).

Step 5: Find Constant of Integration

Using the initial condition \(S(0) = 24\), substitute into the equation \(24 = \frac{A}{200^5}\), giving \(A = 24 \times 200^5\).

Step 6: Calculate Salt Content at\(t = 400\) Minutes

Substitute \(t = 400\) into \(S(t) = \frac{24 \times 200^5}{(200 + 2(400))^5} = \frac{24 \times 200^5}{1000^5}\). Simplify to get \(S(400)\) or the remaining salt content.

Key Concepts

Separable Differential Equations

Separable differential equations are a specific type of first-order differential equations. They are called "separable" because the variables can be separated on different sides of the equation. For instance, if you have an equation like \( dy/dx = g(x) h(y) \), it can be rewritten as \( h(y) \, dy = g(x) \, dx \), making it possible to integrate both sides separately. This is how we solved the salt content equation in the tank problem. To manage this, we transformed the differential equation \( \frac{dS}{dt} = -\frac{10S}{200 + 2t} \) into a separable form: \( \frac{dS}{S} = -\frac{10}{200 + 2t} \, dt \). Then, we integrate each side with respect to its own variable.

Initial Value Problem

The initial value problem involves solving a differential equation with a given starting value, or initial condition. In our tank problem, the initial condition is that at time \( t = 0 \), there are 24 kilograms of salt in the tank. The goal is to find how this initial amount changes over time, under the influence of the rates of inflow and outflow. After deriving the form of \( S(t) \), the function showing salt amount over time, we used the initial value \( S(0) = 24 \) to determine the integration constant. This solution allows us to predict the amount of salt at any given time, given the specific conditions.

Rate of Change

The concept of rate of change is pivotal in differential equations and this exercise. Rate of change explains how quickly or slowly a quantity changes over time, which is often expressed as a derivative. In the tank problem, the rate of salinity change is described by \( \frac{dS}{dt} = -\frac{10S}{200 + 2t} \), representing how the quantity of salt decreases due to the outflow of the solution. This mathematical expression indicates that the change in salt amount depends on both the current salt concentration and the volume of the solution. Understanding these dependencies is crucial for predicting the system's behavior.

Integration

Integration is a mathematical process used to solve differential equations and analyze accumulative changes. It’s particularly useful when dealing with continuous changes over time, such as in our tank problem. Once we have separated the variables in our differential equation, the next step is integration. We perform integration on both sides: \( \int \frac{dS}{S} = \int -\frac{10}{200 + 2t} \, dt \). Solving these integrals leads to \( \ln |S| = -5\ln |200 + 2t| + C \), from which we extract the general solution. Integration facilitates the understanding and prediction of how quantities evolve under specific conditions.