Mathematics, 12.02.2020 04:49 froyg1234

A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon enters the tank at a rate o 5 gal/s, and the well-mixed brine in the tank flows out at the rate of 3 gal/s. How much salt will the tank contain when it is full of brine?

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Answer from: irenemonte

The amount of salt in the tank when it is full of brine is 393.75 pounds.

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x

(concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)

Let y(t) be the amount of salt (in pounds) in the tank at time t (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

$\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)} \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)} \:\frac{pound}{s}$

V(t) is the volume of brine in the tank at time t. To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

V(t)=100 + 2t

We can then write the initial value problem:

$\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50$

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

$\frac{dy}{dx}+P(x)y =Q(x)$

where P and Q are continuous functions on a given interval.

In our case, we have that

$\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50$

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, $\mu \left( x \right)$ , using $\mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}$

$\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}$

Step 2: Multiply everything in the differential equation by $\mu \left( x \right)$ and verify that the left side becomes the product rule $\left( {\mu \left( t \right)y\left( t \right)} \right)'$ and write it as such.

$\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}$

Step 3: Integrate both sides.

$\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C$

Step 4: Find the value of the constant and solve for the solution y(t).

$50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000$

$y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}$

Now, the tank is full of brine when:

$V(t) = 400\\100+2t=400\\t=150$

The amount of salt in the tank when it is full of brine is

$y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75$

A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing

Answer from: Quest

answer: learning objective(s)

· describe the creation and use of systems of equations.

· graph a system of linear equations on the coordinate plane and identify its solution.

introduction

sometimes graphing a single linear equation is all it takes to solve a mathematical problem. other times, one line just doesn’t do it, and a second equation is needed to model the situation. this is often the case when a problem involves two variables. solving these kinds of problems requires working with a system of equations, which is a set of two or more equations containing the same unknowns.

let’s take a look at systems of equations, and see what the graphs of individual equations within a system reveal about the mathematical relationship of the variables.

systems of equations

a system of equations contains two or more linear equations that share two or more unknowns. to find a solution for a system of equations, we must find a value (or range of values) that is true for all equations in the system.

the graphs of equations within a system can tell us how many solutions exist for that system. look at the images below. each show two lines that make up a system of equations (in the graph on the right the two lines are superimposed and look like a single line). how many points in common does each system of lines reveal?

one solution

no solutions

infinite solutions

if the graphs of the equations intersect, then there is one solution that is true for both equations.

if the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.

if the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

remember, the graph of a line represents every point that is a possible solution for the equation of that line. so when the graphs of two equations cross, the point of intersection lies on both lines, meaning that it is a possible solution for both equations. when the graphs of two equations never touch, there are no shared points and there are no possible solutions for the system. when the graphs of two equations lie on top of one another, they share all their points and every one is a possible solution.

graphing as a solution method

graphing equations in order to identify a specific point of intersection is usually not a precise way to solve systems because it is often difficult to see exactly where two lines intersect (unless you are using a computer-based graphing program that allows you to zoom in on a point). however, the graph of a system of equations can still give a good idea of what type of solution, if any, exists. let’s graph a system, and see how it works.

graph the system y = 3x and x + 2y = 4. how many solutions does this system have?

a graph of the two lines y = 3x and x + 2y = 4 shows us that the lines intersect, meaning that there is a single (x, y) value that satisfies both equations. looking at the graph does not tell us exactly where that point is, but we don’t need to know that information here, because the question only asks for the number of solutions.

so a system made of two intersecting lines has one solution. now let’s look at a different situation:

how many solutions exist for the system y -0.5x = 7 and 2y = x − 3?

plotting both equations, it looks like there is no solution—the lines are parallel. to check this finding, we can compare the slopes of the equations. to make that easy, we’ll rearrange the equations into slope-intercept form, or y = mx + b. that gives us the equations y = 0.5x + 7 and y = 0.5x − 1.5. yes, the slope of both lines is 0.5, which means the lines really are parallel. they never intersect, so there is no point that lies on both lines, and no solution to the system.

step-by-step explanation:

Answer from: Quest

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A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing...

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