Salt mixture A tank initially contains 100 gal of brine in which 50 lb of salt are dissolved. A

step 1 All right, we've got a system here, a tank with 100 gallons of brine. We've got Brian entering a

Salt mixture A tank initially contains 100 gal of brine in...

Key Concepts

Mass Balance Principle

This principle involves accounting for the total amount of a substance entering and leaving a system. In mixing problems, it is used to set up an equation representing the rate of change of the substance’s mass in the tank based on the rates of inflow and outflow. The principle ensures that the model accurately reflects the conservation of mass within the system over time.

Initial Value Problem

An initial value problem (IVP) in differential equations consists of a differential equation together with a specified value at an initial time. In mixing problems, the IVP includes the initial concentration or amount of salt in the tank. Solving the IVP provides the particular solution that describes the system's behavior from the known starting condition.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives, capturing the rate at which quantities change. In the context of mixing problems, a first order linear differential equation is used to model how the salt concentration changes over time as a result of inflows and outflows, making them essential for predicting system dynamics.

Mixing Problems

Mixing problems are a class of problems that involve the mixing of substances, such as salt in water, within a tank or other container. They require the application of mass balance and differential equations to model the process over time. The goal is often to determine concentrations or amounts of the substance after a certain period given the input and removal rates.

Integrating Factor Method

The integrating factor method is a technique used to solve first order linear differential equations. This method involves multiplying the differential equation by an integrating factor, which simplifies the left-hand side into the derivative of a product, making the equation easier to integrate and solve for the unknown function.

Time-Dependent Volume

In many mixing problems, the total volume within the tank changes over time due to different rates of inflow and outflow. This time-dependence affects the concentration of the substance being mixed, and it must be accounted for in the differential equation model. The changing volume adds complexity to the problem, requiring careful formulation of the rate at which the substance leaves the system.

Transcript

00:02 All right, we've got a system here, a tank with 100 gallons of brine.

00:06 We've got brian entering at a steady rate.

00:09 We've also got it exiting the tank.

00:11 So now we've got a series of questions that we need to answer with respect to this situation and the scenario within this system.

00:21 So let's start with part a here.

00:23 It asks at what rate pounds per minute does salt enter the tank at time t? so we can take our rind entering, which is two gallons, sorry, two pounds per gallon of salt entering, multiplied by the rate, which we can see it says it's five gallons per minute.

00:48 These gallons will cancel out, and you're looking at 10 pounds per minute.

01:00 That is the rate at which salt enters the tank at time t.

01:09 All right.

01:11 And now the second part of the question, part b, is asking us to find a function of volume with respect to time of the brine in the tank.

01:28 So volume as a function of time.

01:38 We know that initially there was 100 gallons of brine inside the tank.

01:46 We know five gallons are being added per minute.

01:52 Okay.

01:53 So we want to include time there and four are exiting okay and if we simplify this and have t gallons we can have a hundred plus t gallons okay this right here will be the function of volume with respect to time so as time increases and this t will be in minutes that's how many gallons and volume will have a brine in the tank.

02:38 Part c is asking us to determine at what rate pounds per minute does salt leave the tank at time t.

02:49 Okay, in order for us to do that, first we need to determine the concentration at any time t.

02:57 So we would have, let's say, salt is s.

03:07 That that is a that might make things confusing the s variable but let's just use um let's just use x we'll use x assault uh you can use any variable but typically you want to stick with x and y's um over our volume function so we know salt over our volume will give us our concentration and um and we wanted to calculate the question asked us to calculate the rate at which the salt's leaving the tank.

03:46 So we take the concentration and we multiply it by the rate at which the brine was leaving the tank.

03:51 We know it says leaving at four gallons per minute.

04:04 And if we go ahead and multiply across we have 4x over 100.

04:23 We'll have this in gallons.

04:34 Wait.

04:39 This is also in gallons.

04:43 We want to make sure we're balancing out our our units as you can see that this was also gallons and then salt has to be in pounds so really we're left with pounds per minute so that was wrong units there very important to keep track of your units and make sure you're canceling out the correct time so we got 4x x being salt in pounds over 100 plus t which is our time minutes.

05:21 Part d, move up just a tad bit.

05:33 Part d asks, write down and solve the initial value problem describing the mixing process.

05:39 Okay, this is where we're going to get into our first order differential equations.

05:45 And we want to write out our initial problem as change of x over time.

05:54 Okay.

06:00 Will be equal to our rate at which salt enters the tank, which we know is 10, because we just saw for that in part a.

06:21 And we can subtract our rate at which salt leaves the tank.

06:34 We knew that was 4x plus 4x, sorry, 4x over 100 plus t.

06:45 This was the final answer here, by the way.

06:47 I got to box it there.

06:51 If we subtract it from this, that will give us our initial problem there, okay? that's a change of salt over time.

07:10 Okay? now, if we set our, if we solve for x at zero, i'm sorry, what x was at initial time, right? so let's say when x was at initial time, what is our value of salt? and we're told at the beginning of the problem that there's 50 pounds of salt in there.

07:40 Okay.

07:41 Therefore, what we can do is we can write out our problem where we substitute the initial conditions as 50.

08:13 Before we move forward there, i'm going to revisit first shorter differential equations.

08:19 Okay, we want to write out the equation so that all of our xs are on one side.

08:27 And our constant is on the other side.

08:31 That way we can separate the p and our q.

08:37 Okay, so let's go ahead and do that first.

08:42 So we'll do d, x over t, plus 4 of x, let me so 4x plus time.

09:00 And i'm going to go ahead and stop including units for a minute because it's actually just going to, complicate things here.

09:09 So let's leave them out for now.

09:11 We can include them as we go once we get closer to the end there.

09:17 And that will be equal to 10.

09:21 Okay.

09:21 And now we remember our first order equation, which will basically, it's just going to have dx, dt, which in this case would have been, which is actually d .y over dt, but we chose x as our variable.

09:37 So that's why we're going to say dx, dt.

09:40 And we now.

09:41 That it is p t multiplied by our y which what like i said in this case would be our x so we would say p t x is equal to q okay so that way we know that our p t value is equal to 4x is equal to 100 plus t and our qt is equal to 10.

10:23 Okay, and that's why we write that out in that way so that we can get those values.

10:31 Excuse me.

10:35 All right.

10:35 Now, what we can do before we go through and solve, because essentially what we're solving for is to get the solution, which is what is salt with respect to time, right? but before we get to that point, what we want to do is to make it easier, what we can do is we can solve for our integrating factor first.

10:56 And our integrating factor equation is going to be vt is equal to e, the integral of pt with respect to time.

11:17 Okay, so we're integrating with respect to t.

11:25 Now, let's go ahead and break this down...

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