Water flows into a tank at the variable rate of R in = 20/(1+t) gal min and out at the constant

Water flows into a tank at the variable rate of R in =...

Water flows into a tank at the variable rate of $R_{\text { in }}=$ 20$/(1+t)$ gal min and out at the constant rate $R_{\text { out }}=5$ gal min. Let $V(t)$ be the volume of water in the tank at time $t .$ (a) Set up a differential equation for $V(t)$ and solve it with the initial condition $V(0)=100$ . (b) Find the maximum value of $V$ . (c) CAS Plot $V(t)$ and estimate the time $t$ when the tank is empty.

Water flows into a tank at the variable rate of $R_{\text { in }}=$ 20$/(1+t)$ gal min and out at the constant rate $R_{\text { out }}=5$ gal min. Let $V(t)$ be the volume of water in the tank at time $t .$ (a) Set up a differential equation for $V(t)$ and solve it with the initial
condition $V(0)=100$ . (b) Find the maximum value of $V$ . (c) CAS Plot $V(t)$ and estimate the time $t$ when the tank is empty.

Key Concepts

Modeling with Differential Equations

This concept involves representing a changing system, such as the volume of water in a tank, using an equation that relates the rate of change of a quantity to other factors in the system. In this context, the rate at which water enters and exits the tank is used to form a differential equation describing how the volume changes over time.

Net Rate of Change

The net rate of change is determined by the difference between the inflow and outflow rates of the system. It represents the overall effect of competing processes (inflows and outflows) on the volume. Formulating the differential equation requires calculating this net rate, which is fundamental in setting up the model correctly.

Initial Value Problems

Initial value problems involve solving differential equations with a prescribed initial condition. The initial condition provides a specific starting point for the system, allowing the unique determination of the solution. This is essential for real-world scenarios, where the state of the system at the beginning influences its future behavior.

First Order Linear Differential Equations

This is a class of differential equations that involve the first derivative of the unknown function and the function itself, typically expressed in a linear form. They often appear in modelling natural processes and can be solved using standard techniques such as the integrating factor method.

Integrating Factor Technique

The integrating factor method is a systematic procedure used to solve first order linear differential equations. It involves multiplying the differential equation by a specially chosen function (the integrating factor), which simplifies the equation into an exact derivative, making it possible to integrate both sides and find an explicit solution.

Optimization of Functions

Optimization techniques are used to determine maximum or minimum values of a function. In this problem, it is applied to find the maximum volume reached by the system over time. This usually involves taking the derivative of the function with respect to time, setting it equal to zero, and solving for the critical points.

Transcript

00:01 All right, so we are dealing with a problem here, and we have a tank of water.

00:04 We aren't told any specifics in it other than the rate of our water in is 20 divided by 1 plus our time in gallons per minute, and our rate out is in 5 gallons per minute.

00:17 So what we need to do is we actually need to set up an equation.

00:21 Now, because this is what we know, the only equation we can actually set up here is looking at our weight in divided by the weight out.

00:31 So just setting up how our water is changing.

00:37 So to set that one up, we would have 20 over 1 plus our time minus 5.

00:45 So rate in, divided by our rate out.

00:49 So if i looked at that and i wanted to then set up my equation, i can kind of think of that as a integral.

01:01 So here's my rate in, my rate out, and i'm going to set that up, and i'm going to just make sure that when i do do this, i end up with a constant as well, so minus five plus.

01:14 So working backwards here to figure out my equation, i would have my natural log of 1 plus t.

01:21 That would take care of that.

01:24 I still have that 20 that i need to be there as well, though, so i would have a 20 times that.

01:29 And i'd have minus my five times my time.

01:33 And then again, i have my constant along for the ride.

01:37 So my equation here to find my volume would be looking at my antiderivative for my rate.

01:46 Now, we want to figure out our constant.

01:50 It did tell us that our volume at zero is 100.

01:55 So solving that, i'd have 100 is equal to 20 times my natural.

02:00 Log of 1 plus 0 minus 5 times 0 plus my constant.

02:08 Natural log of 1 is 0, 0 times 20 is 0, 5 times 0 is 0.

02:13 So i'm left with my constant has to be 800.

02:17 So my general solution here, for my equation would be 20 times natural log 1 plus my time, minus 5 times my time, plus my time, plus 100.

02:37 So for the next one, it does ask us to figure out when is our max volume and what could it be...

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