## Calculus Tutorial

*Differential Equations*

#### Intro

**Handling of Materials**

Suppose that we have a tank containing water. Suppose also that we

have 2 pumps — one pump can drain the tank in 6 hours, and the

other can drain the tank in 5 hours. How long will it take to drain

the tank with both pumps running?

Since one pump can drain the tank in 5 hours, this pump can drain

of the tank in one hour (assuming constant time rate

of pumping). Using the same reasoning, the other pump can drain

of the tank in one hour. Therefore, with both pumps

running we can drain

(1)

of the tank in one hour. Therefore, to

find the amount of time it will take to drain 1 whole tank, we

divide 1 whole tank by the number of units of the tank that can be

drained in one hour;

(2)

Therefore, it will take approximately 2.73 hours to drain the tank

with both pumps running.

**Chemical Solutions**

In the branch of science known as chemistry, a *solution* is made

when we mix one substance, the *solute*, with another substance,

the *solvent*. Up to a certain *concentration* (mass of

solute per unit volume of solvent), the volume of the final solution

is that of the solvent.

Suppose we have at our disposal two salt water solutions — one

solution has a concentration of 0.5 kilograms salt per cubic meter

of water, and the other has a concentration of 0.2 kilograms salt

per cubic meter of water. Suppose also that we wish to make 0.7

cubic meters of salt water solution of concentration 0.3 kilograms

salt per cubic meter of water. How much of each salt water solution

should be mixed together to yield the desired amount and

concentration of the final solution?

Let the volume of 0.5 kilograms salt per cubic meter of water

solution be and the volume of 0.2 kilograms salt per cubic

meter of water solution be . From the problem statement, we

know that

(3)

(4)

Multiplying both members of Equation 4 by 2, we

have

(5)

Subtracting from the left and right members of Equation

3 the left and right members of Equation

5, respectively, we have

(6)

From Equations 3 and 6, we have

(7)

Therefore, from Equations 6 and 7,

the approximated desired amounts of the 0.5 and 0.2 kilograms salt

per cubic meter of water solutions are, respectively, 0.233 and

0.467 cubic meters.

Suppose we have a tank of volume and containing a solution with

an initial mass concentration of kilograms per cubic meter.

Let us also suppose that one pump is pumping in, at a rate of

cubic meters per second, another solution, of the same type, with a

mass concentration of kilograms per cubic meter. The tank

is also constantly stirred, to maintain uniform mass concentration,

and is pumped out at the same rate of cubic meters per second.

The problem is to find how the mass concentration , of the

contents of the tank, is related to the amount of time that both

of the pumps are in operation.

To begin to solve this problem, let us consider the time rate of

mass of dissolved substance being pumped in, ,

the time rate of mass of dissolved substance being pumped out,

, and the rate of change of the total mass of

dissolved substance in the tank, . Quite clearly,

these three quantities are related such that

(8)

Since at ;

(9)

(10)

Equation 10 is the

solution to our problem.

#### Sample Problem

# About The Author

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I have written my own book on the subject of the calculus, and its prerequisites, which rigorously proves all theorems, as well as other topics in mathematics and physics. Besides a rigorous understanding of the topic matter, I will also be able to instill analytical problem solving skills in the l... |