jkwiens.com: Answers

Answer: Balance the Chemical Equation

Saturday, 16 of February , 2008 at 11:05 am

Question: Balance the chemical equation using Linear Algebra.

$CO + CO_2 + H_2 ~~\rightarrow ~~ CH_4 + H_2O$

Answer: In order to balance the chemical equations, we need to find the five unknowns (ab,c,d,e).

$aCO + bCO_2 + cH_2 ~~\rightarrow ~~ dCH_4 + eH_2O$

Since the number of chemical elements are conserved, we know that:

$\therefore$ we need to solve the following linear system.

$\left( \begin{array}{ccccc}1 & 1 & 0 & -1 & 0 \\1 & 2& 0 & 0 & -1 \\0 & 0 & 2 & -4 & -2 \end{array} \right)\left( \begin{array}{c}a \\b\\c \\d \\e \end{array} \right) = \left( \begin{array}{c}0 \\0\\0 \end{array} \right)$

This system of equations can easily be solved with Maple.

> with(Student[LinearAlgebra]):
> A := Matrix([[1, 1, 0, -1, 0, 0],
		[1, 2, 0, 0, -1, 0],
		[0, 0, 2, -4, -2, 0]]):
> ReducedRowEchelonForm(A)

According to maple, we can reduce the linear system to:

$\left( \begin{array}{ccccc}1 & 0 & 0 & -2 & 1 \\0 & 1& 0 & 1 & -1 \\0 & 0 & 1 & -2 & -1 \end{array} \right)\left( \begin{array}{c}a \\b\\c \\d \\e \end{array} \right) = \left( \begin{array}{c}0 \\0\\0 \end{array} \right)$

According to the above equation, we don’t have a unique solution. However, we know that the linear system will have a solution $\forall d,e \in \mathbb R$ where

Since we know that $a,b,c,d,e \in \mathbb N$ (where $0 \notin \mathbb N$ ), we can add additional constraints to our linear system. If we look at the above equations, it is easy to see that the following constraint must hold:

Using the linear system and the above constraint, we find the following solutions…

$CO + CO_2 + 7H_2 ~~\rightarrow ~~ 2CH_4 + 3H_2O$
$2CO + CO_2 + 10H_2 ~~\rightarrow ~~ 3CH_4 + 4H_2O$
$CO + 2CO_2 + 11H_2 ~~\rightarrow ~~ 3CH_4 + 5H_2O$
$3CO + CO_2 + 13H_2 ~~\rightarrow ~~ 4CH_4 + 5H_2O$
$2CO + 2CO_2 + 14H_2 ~~\rightarrow ~~ 4CH_4 + 6H_2O$
$CO + 3CO_2 + 15H_2 ~~\rightarrow ~~ 4CH_4 + 7H_2O$
$4CO + CO_2 + 16H_2 ~~\rightarrow ~~ 5CH_4 + 6H_2O$
…
…

As you can see, there is still an infinite amount of solutions for this chemical equations. Some can be reduced, however majority of the equations can’t be reduced

Answer: Cubes of three Consecutive Integers

Saturday, 9 of February , 2008 at 7:05 pm

Question: Show that the sum of the cubes of any three consecutive integers is divisible by 9.

Answer: In order to solve this problem, I am going to define the questions a little more formally.

Prove that there $\exists$ a $k \in \mathbb Z$ where given

and $n \in \mathbb Z$

I will be solving this problem using induction.

Base Case:
First we need show that the problem is true for the base case ()

Therefore, there $\exists$ a $k \in \mathbb Z$ (which would be ) where .

Induction Step (for positive values):
In order to prove that the problem holds for $n \in \mathbb N$ , we need to show:

If $f(n) = 9k \Longrightarrow \exists~a~j \in \mathbb Z$ where where $n,k \in \mathbb Z$

In order to prove this statement, we need to reduce .

Since $(n^2 +3n + 3 + k) \in \mathbb Z$ , we have shown if it implies that $\exists~a~j \in \mathbb Z$ where .

Induction Step (for negative values):
Since we already proved by induction that the problem holds for all natural numbers, all we need to do to extend this proof to all integers is to show that:

If $f(n) = 9k \Longrightarrow \exists~a~j \in \mathbb Z$ where where $n,k \in \mathbb Z$

In order to prove this statement, we need to reduce .

Since $(-n^2 - n - 1 + k) \in \mathbb Z$ , we have shown if it implies that $\exists~a~j \in \mathbb Z$ where .

Therefore, we have proven by induction that the sum of the cubes of any three consecutive integers is divisible by 9.

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Category: Algebra, Answers

Answer: Voltage and Current of a Circuit

Saturday, 2 of February , 2008 at 12:45 pm

The circuit in this question is an example of a Wheatstone Bridge (for the most part). In order to make this circuit a real Wheatstone Bridge the center resistor has to be replaced with a galvanometer. This circuit was introduced to me in a lab for my first E&M course in university which we used to accurately measure the resistance of an unknown resistor.

Question: Calculate the current and voltage difference between points D and B in the following circuit.

Answer: The easiest way to solve this problem is to use Kirchhoff’s Laws. We will start off by using Kirchhoff’s first law which is defined as:

The current entering a junction must equal the current out of the junction.
$\sum I_{in} = \sum I_{out}$

In order to use this law, we first need to define the direction the current is traveling. It really doesn’t matter how you define the current direction. If you picked the wrong direction, the current will just be a negative value which means that it flows the opposite way. This is how I defined the current direction for the circuit.

If we apply Kirchhoff’s first law to junction B, we get the following:

$I_{DB} + I_{AB} = I_{BC}$

If we this law again on junction D, we get:

$I_{AD} = I_{DB} + I_{DC}$

Next, we will need to use Kirchhoff’s Second law which is defined as:

The potential differences across all elements around any closed circuit loop must be zero.
$\displaystyle\sum_{Closed~Path} \Delta V = 0$

If we apply Kirchhoff’s 2nd law and Ohm’s Law to the loop ADB, we get:

$V_{AD} + V_{DB} - V_{AB} = 0$

$RI_{AD} + RI_{DB} - 3RI_{AB} = 0$

$I_{AD} + I_{DB} - 3I_{AB} = 0$ because R = 1~Ohm

If we do the same thing to the loop DBC, we get:

$V_{DB} + V_{BC} - V_{DC} = 0$

$RI_{DB} + RI_{BC} - 2RI_{DC} = 0$

$I_{DB} + I_{BC} - 2I_{DC} = 0$ because R = 1~Ohm

Now, we will use these laws on the loop ADC, we get:

$5 - V_{AD} - V_{DC} = 0$

$5 - RI_{AD} - 2RI_{DC} = 0$

$I_{AD} + 2I_{DC} = 5$ because R = 1~Ohm

Therefore, we have the following system of linear equations.

$I_{DB} + I_{AB} - I_{BC}= 0$

$I_{AD} - I_{DC} - I_{DB} = 0$

$I_{AD} + I_{DB} - 3I_{AB} = 0$

$I_{DB} + I_{BC} - 2I_{DC} = 0$

$I_{AD} + 2I_{DC} = 5$

which we can put in matrix form.

$\left( \begin{array}{ccccc}0 & 1 & 1 & 0 & -1 \\1 & 0 & -1 & -1 & 0 \\1 & -3 & 1 & 0 & 0 \\0 & 0 & 1 & -2 & 1 \\1 & 0 & 0 & 2 & 0 \end{array} \right)\left( \begin{array}{c}I_{AD} \\I_{AB}\\I_{DB} \\I_{DC} \\I_{BC} \end{array} \right) = \left( \begin{array}{c}0 \\0\\0 \\0\\5\end{array} \right)$

This system of equations can easily be solved with Maple.

> with(Student[LinearAlgebra]):
> A := Matrix([[0, 1, 1, 0, -1, 0],
		[1, 0, -1, -1, 0, 0],
		[1, -3, 1, 0, 0, 0],
		[0, 0, 1, -2, 1, 0],
		[1, 0, 0, 2, 0, 5]])


> ReducedRowEchelonForm(A)

Therefore, this means that the current going through the point D to B is $\frac{25}{29}~Amps$ . If we use Ohm’s law, we can also determine the voltage difference between the points D and B to be $\frac{25}{29}~Volts$ .