{"id":822,"date":"2025-06-20T17:15:44","date_gmt":"2025-06-20T17:15:44","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=822"},"modified":"2025-09-10T15:08:13","modified_gmt":"2025-09-10T15:08:13","slug":"separation-of-variables-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/separation-of-variables-learn-it-2\/","title":{"raw":"Separation of Variables: Learn It 2","rendered":"Separation of Variables: Learn It 2"},"content":{"raw":"
Applications of Separation of Variables<\/h2>\r\n
Separable equations appear frequently in real-world situations. Let's examine two common types of problems where this method proves especially useful.<\/p>\r\n\r\n\r\n
Solution concentrations<\/h3>\r\nConsider a tank being filled with a salt solution. We want to determine the amount of salt present in the tank as a function of time. The separation of variables method provides an effective way to solve this type of concentration problem.\r\n\r\n\r\n
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A tank containing [latex]100\\text{L}[\/latex] of a brine solution initially has [latex]4\\text{kg}[\/latex] of salt dissolved in the solution. At time [latex]t=0[\/latex], another brine solution flows into the tank at a rate of [latex]2\\text{L\/min}\\text{.}[\/latex] This brine solution contains a concentration of [latex]0.5\\text{kg\/L}[\/latex] of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of [latex]2\\text{L\/min}[\/latex], so that the level of liquid in the tank remains constant (Figure 2). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.<\/p>\r\n\r\n<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"] Figure 2. A brine tank with an initial amount of salt solution accepts an input flow and delivers an output flow. How does the amount of salt change with time?[\/caption]<\/figure>\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n
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First we define a function [latex]u\\left(t\\right)[\/latex] that represents the amount of salt in kilograms in the tank as a function of time. Then [latex]\\frac{du}{dt}[\/latex] represents the rate at which the amount of salt in the tank changes as a function of time. Also, [latex]u\\left(0\\right)[\/latex] represents the amount of salt in the tank at time [latex]t=0[\/latex], which is [latex]4[\/latex] kilograms.<\/p>\r\n
The general setup for the differential equation we will solve is of the form<\/p>\r\n\r\n
INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of [latex]2[\/latex] L\/min, and each liter of solution contains [latex]0.5[\/latex] kilogram of salt, every minute [latex]2\\left(0.5\\right)=1\\text{kilogram}[\/latex] of salt enters the tank. Therefore INFLOW RATE = [latex]1[\/latex].<\/p>\r\n
To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time [latex]t[\/latex] is equal to [latex]u\\left(t\\right)[\/latex]. Thus, the concentration of salt is [latex]\\frac{u\\left(t\\right)}{100}[\/latex] kg\/L, and the solution leaves the tank at a rate of [latex]2[\/latex] L\/min. Therefore salt leaves the tank at a rate of [latex]\\frac{u\\left(t\\right)}{100}\\cdot 2=\\frac{u\\left(t\\right)}{50}[\/latex] kg\/min, and OUTFLOW RATE is equal to [latex]\\frac{u\\left(t\\right)}{50}[\/latex]. Therefore the differential equation becomes [latex]\\frac{du}{dt}=1-\\frac{u}{50}[\/latex], and the initial condition is [latex]u\\left(0\\right)=4[\/latex]. The initial-value problem to be solved is<\/p>\r\n\r\n
The differential equation is a separable equation, so we can apply the five-step strategy for solution.<\/p>\r\n
Step 1. Setting [latex]1-\\frac{u}{50}=0[\/latex] gives [latex]u=50[\/latex] as a constant solution. Since the initial amount of salt in the tank is [latex]4[\/latex] kilograms, this solution does not apply.<\/p>\r\n
The solution to the initial value problem is [latex]u\\left(t\\right)=50 - 46{e}^{\\frac{\\text{-}t}{50}}[\/latex]. To find the limiting amount of salt in the tank, take the limit as [latex]t[\/latex] approaches infinity:<\/p>\r\n\r\n
Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is [latex]50[\/latex] kilograms, then it remains constant. If it starts at less than 50 kilograms, then it approaches 50 kilograms over time.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>Watch the following video to see the worked solution to the example above.
![\"A](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234119\/CNX_Calc_Figure_08_03_002.jpg\")
Figure 2. A brine tank with an initial amount of salt solution accepts an input flow and delivers an output flow. How does the amount of salt change with time?[\/caption]<\/figure>\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n\r\n