# Differential Equation Water Tank Word Problem

• aves
In summary, the problem involves a tank filled with 1000 liters of pure water, being continuously mixed with two incoming brine solutions containing 0.08 kg and 0.03 kg of salt per liter, respectively. The tank also drains at a rate of 18 liters per minute. The differential equation for this system can be represented as dS/dt = 0.72 + 0.27 - (18S/1000), where S(t) represents the weight of salt in the tank after t minutes.
aves
I am having trouble starting this problem:

A tank is filled with 1000 liters of pure water. Brine containing 0.08 kg of salt per liter enters the tank at 9 liters per minute. Another brine solution containing 0.03 kg of salt per liter enters the tank at 9 liters per minute. The contents of the tank are kept thoroughly mixed and the drains from the tank at 18 liters per minute.

A. Determine the differential equation which describes this system. Let S(t) denote the number of kg of salt in the tank after t minutes. Then:
dS/dt=?

Any help on how to start this would be appreciated

welcome to pf!

hi aves! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

I am not even sure how to make the differential equation out of the word problem. I can do the rest from there, but I just can't figure it out.

Hint: in a small interval of time $\Delta t > 0,$ how many kg of salt enters the tank? How many kg of salt leaves the tank?

RGV

aves said:
I am not even sure how to make the differential equation out of the word problem. I can do the rest from there, but I just can't figure it out.

the question helps you on this, telling you to use "t" for time, in seconds, and S for the total weight of salt, in kg

now try translating into an equation the effect of …
Brine containing 0.08 kg of salt per liter enters the tank at 9 liters per minute.

Would it follow the general equation: S(t)=S0*e^(kt)?
Where t is the time and S0 is the initial weight of salt (0.08 kg)?

don't solve it

just write it!

dS/dt = … ?

Would it be:
dS/dt=0.72+0.27-18S/1000
?

Last edited:

## 1. What is a differential equation water tank word problem?

A differential equation water tank word problem is a type of mathematical problem that involves describing the behavior of a water tank using a differential equation. It typically involves a tank that is being filled or drained, and the goal is to determine the rate of change of the water level over time.

## 2. Why are differential equations used in water tank word problems?

Differential equations are used in water tank word problems because they provide a way to model and understand the behavior of complex systems, such as a water tank. These equations can describe how different factors, such as inflow, outflow, and volume, affect the water level in the tank over time.

## 3. What are the key components of a differential equation water tank word problem?

The key components of a differential equation water tank word problem are the initial conditions, the differential equation itself, and the variables involved. The initial conditions typically describe the initial water level and inflow/outflow rates, while the differential equation represents the relationship between these variables.

## 4. How do you solve a differential equation water tank word problem?

To solve a differential equation water tank word problem, you first need to set up the differential equation based on the given information and initial conditions. Then, you can use various methods, such as separation of variables, to solve the equation and determine the behavior of the water level over time.

## 5. What real-world applications use differential equation water tank word problems?

Differential equation water tank word problems have many real-world applications, such as in engineering, physics, and environmental science. These problems can help to model and understand the behavior of systems involving fluids, such as water distribution networks, chemical reactions, and groundwater flow.

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