Rate of change and integrating factor

• nysnacc
In summary, the problem involves adding brine with a concentration of 2 grams of salt per liter to a tank of initially 5 grams of salt dissolved in 20 liters of water. The tank is mixed well and drained at a rate of 3 liters per minute. The task is to determine how long the process must continue until there are 20 grams of salt in the tank. Using the equations for rate in and rate out, the change in concentration is calculated using the integrating factor method. The resulting solution for x(t) is 2 - 7/4 EXP (-3/20 t), which gives a value of t = 3.73 minutes for a concentration of 1 g/L. However, the correct answer is
nysnacc

Homework Statement

Initially 5 grams of salt are dissolved in 20 liters of water. Brine with concentration of salt 2 grams of salt per lter is added at a rate of 3 liters per minute. The tank is mixed well and is drained at 3 liters a minute. How long does the process have to continue until there are 20 grams of salt in the tank?

Homework Equations

Rate in / Rate out

Concentration = mass / Volume

The Attempt at a Solution

Rate in = 3L /min
Concentration in =2 g/L

rate out =3L /min
Concentration out =x g/L

I set change of concentration as x'
x' = lim Δt→0 (3 L/ min * 2 g/L *Δt - 3 L/ min * x g/L *Δt) / V *Δt
so x' = (3*2 - 3*x) / V where V =20 L (given)

x' = 3/10 - 3x/20

then use integrating factor, but I have different answer what am I doing wrong? thank

nysnacc said:

Homework Statement

Initially 5 grams of salt are dissolved in 20 liters of water. Brine with concentration of salt 2 grams of salt per lter is added at a rate of 3 liters per minute. The tank is mixed well and is drained at 3 liters a minute. How long does the process have to continue until there are 20 grams of salt in the tank?

Homework Equations

Rate in / Rate out

Concentration = mass / Volume

The Attempt at a Solution

Rate in = 3L /min
Concentration in =2 g/L

rate out =3L /min
Concentration out =x g/L

I set change of concentration as x'
x' = lim Δt→0 (3 L/ min * 2 g/L *Δt - 3 L/ min * x g/L *Δt) / V *Δt
so x' = (3*2 - 3*x) / V where V =20 L (given)

x' = 3/10 - 3x/20

then use integrating factor, but I have different answer what am I doing wrong? thank

What is your answer? Maybe it is not wrong at all, but we cannot tell without seeing it.

nysnacc
Ray Vickson said:
What is your answer? Maybe it is not wrong at all, but we cannot tell without seeing it.

I have x(t) = 2 - 7/4 EXP (-3/20 t)
which then gives t = 3.73 min
for 20 gram salt in the water (concentration 1 g/L)

But the answer said to be 40/3 ln(4/3) which is 3.84 min

What is the rate of change?

The rate of change is a measure of how much a quantity changes with respect to time. It can be calculated by finding the slope of a line on a graph or by using the formula: rate of change = change in quantity / change in time.

What is an integrating factor?

An integrating factor is a mathematical tool used to solve differential equations. It is a function that is multiplied by both sides of a differential equation in order to make it easier to solve.

Why is the integrating factor important?

The integrating factor is important because it allows us to solve certain types of differential equations that would be difficult or impossible to solve without it. It simplifies the equation and makes it easier to find a solution.

How do you find the integrating factor?

The integrating factor can be found by multiplying both sides of a differential equation by an appropriate function. The function is usually determined by looking at the coefficients and variables in the equation and using a specific method depending on the type of equation.

What are some real-world applications of rate of change and integrating factor?

Rate of change and integrating factor have many real-world applications, such as predicting the growth of populations, modeling the spread of diseases, and analyzing the behavior of chemical reactions. They are also used in engineering and physics to understand and predict changes in systems over time.

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