Matrices, Part 3

In my last post, I multiplied a 2 × 2 matrix by another 2 × 2 matrix. Now let’s multiply a 2 × 2 matrix by a 2 × 1 matrix. If you were paying attention last time, this is possible because the inside dimensions are the same (2= 2) and the resulting matrix will be a 2 × 1 matrix, that is, the outside dimensions.

This is done exactly the same way as illustrated in my last post, only there is only 1 column in the second matrix to multiply with the first matrix:

\[
\left[{\begin{array}{cc}{1}&{{-}{2}}\\{3}&{{-}{4}}\end{array}}\right]\hspace{0.33em}\times\hspace{0.33em}\left[{\begin{array}{c}{5}\\{6}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{(}{5}\times{1}{)}{+}{(}{6}\times{(}{-}{2}{))}}\\{{(}{5}\times{3}{)}{+}{(}{6}\times{(}{-}{4}{))}}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{-}{7}}\\{{-}{9}}\end{array}}\right]
\]

Now let the second matrix be composed of variables. This does not change the method at all. It just means that the result is a matrix with algebraic expressions:

\[
\left[{\begin{array}{cc}{1}&{{-}{2}}\\{3}&{{-}{4}}\end{array}}\right]\hspace{0.33em}\times\hspace{0.33em}\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{(}{x}\times{1}{)}{+}{(}{y}\times{(}{-}{2}{))}}\\{{(}{x}\times{3}{)}{+}{(}{y}\times{(}{-}{4}{))}}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{x}{-}{2}{y}}\\{{3}{x}{-}{4}{y}}\end{array}}\right]
\]

Please keep this example in mind for my next post when I use matrices to solve a system of equations.

The last skill I need to present is matrix division. When using matrices, you do not actually divide a matrix by another matrix. Rather, you multiply by the inverse of a matrix.

In scalar arithmetic, you can think of dividing a number say 4, by another number, say 2, as multiplying the 4 by the inverse (reciprocal) of 2:

\[
\frac{4}{2}\hspace{0.33em}{=}\hspace{0.33em}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}{2}
\]

The same thing is done with matrices. However, finding the inverse of a matrix is a little involved and I will not cover that in this set of posts. Rather I will just give you the result when needed. However I will say a few things about the properties of matrix inverses.

In scalar arithmetic, multiplying a number by it’s reciprocal (inverse) equals 1:

\[
\frac{2}{2}\hspace{0.33em}{=}\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}{1}
\]

The same thing is true with matrices, only what is “1” in the matrix world?

The equivalent “1” for matrices is the Identity Matrix. This is a square (rows = columns) matrix with 1’s down its diagonal and 0’s everywhere else:

\[
\left[{\begin{array}{cc}{1}&{0}\\{0}&{1}\end{array}}\right]
\]

This is the identity matrix for a 2 × 2 matrix. The inverse of a matrix is that matrix where multiplying it by the original matrix, results in the identity matrix. The inverse of a matrix A, is denoted as A-1.

\[
{\mathbf{A}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathbf{A}}^{{-}{1}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{{-}{2}}&{1}\\{\frac{3}{2}}&{{-}\frac{1}{2}}\end{array}}\right]
\] \[
\begin{array}{l}
{{\mathbf{A}}\times{\mathbf{A}}^{{-}{1}}{=}\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\times\left[{\begin{array}{cc}{{-}{2}}&{1}\\{\frac{3}{2}}&{{-}\frac{1}{2}}\end{array}}\right]{=}\left[{\begin{array}{cc}{{(}{-}{2}\times{1}{)}{+}{(}\frac{3}{2}\times{2}{)}}&{{(}{1}\times{1}{)}{+}{(}{-}\frac{1}{2}\times{2}{)}}\\{{(}{-}{2}\times{3}{)}{+}{(}\frac{3}{2}\times{4}{)}}&{{(}{1}\times{3}{)}{+}{(}{-}\frac{1}{2}\times{4}{)}}\end{array}}\right]}\\
{{=}\hspace{0.33em}\left[{\begin{array}{cc}{1}&{0}\\{0}&{1}\end{array}}\right]}
\end{array}
\]

And it turns out that when multiplying a matrix by its inverse, order does not matter: A × A-1 = A-1 × A.

In my next post, I will put all this talk about matrices in practice and use them to solve a system of equations.

Matrices, Part 2

In order to set up a system of equations using matrices, you need to understand how matrices multiply one another. Not all matrices can be multiplied together – they need to be compatible with one another. Not only that, unlike scalar (single number) arithmetic, multiplication does not commute, that is, the order of the multiplication will generally produce different results or one order may not even be possible. So what do I mean by compatible?

Let’s start with an example:

\[
\begin{array}{l}
{\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\times\left[{\begin{array}{cc}{5}&{6}\\{7}&{8}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{{(}{5}\times{1}{)}{+}{(}{7}\times{2}{)}}&{{(}{6}\times{1}{)}{+}{(}{8}\times{2}{)}}\\{{(}{5}\times{3}{)}{+}{(}{7}\times{4}{)}}&{{(}{6}\times{3}{)}{+}{(}{8}\times{4}{)}}\end{array}}\right]}\\
{{=}\hspace{0.33em}\left[{\begin{array}{cc}{19}&{22}\\{43}&{50}\end{array}}\right]}
\end{array}
\]

To multiply these two 2 × 2 matrices, you take the first column of the second matrix and lay it over the top of the first matrix:

\[
\begin{array}{l}
{\left[{\begin{array}{cc}{5}&{7}\end{array}}\right]}\\
{\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]}
\end{array}
\]

Starting with the top row of the first matrix, Multiply the numbers in the same position together and add the result of each: (5 × 1) + (7 × 2) = 19. This result is the first row, first column number in the new matrix. Repeat this using the second row of the first matrix: (5 × 3) + (7 × 4) = 43. This is the first element of the second row of the new matrix. Now do the same with the second column of the second matrix:

\[
\begin{array}{l}
{\left[{\begin{array}{cc}{6}&{8}\end{array}}\right]}\\
{\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]}
\end{array}
\]

to get the second column of the new matrix. I will leave it as an exercise for you to confirm that if I reverse the order of the matrices, you will get a different result. That is,

\[
\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]\times\left[{\begin{array}{cc}{5}&{6}\\{7}&{8}\end{array}}\right]\hspace{0.33em}\ne\hspace{0.33em}\left[{\begin{array}{cc}{5}&{6}\\{7}&{8}\end{array}}\right]\hspace{0.33em}\times\hspace{0.33em}\left[{\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}}\right]
\]

So this method works for any size matrices as long as they are compatible. From this example, you see that this works only if the second matrix has the same number of rows as the number of columns in the first matrix. This is easy to see if you put the dimensions together: (2 × 2) × (2 × 2). The inside numbers need to be the same if multiplication is possible (2 = 2). The outside numbers give the dimensions of the resulting matrix (2 × 2).

So you can multiply a 3 × 2 matrix by a 2 × 4 matrix to get a 3 × 4 matrix, but you cannot reverse the order because the inside dimensions will not be equal. It’s interesting that if you multiply a 1 × (anything) matrix by a (same anything) × 1 matrix, you will get a 1 × 1 matrix which is just a number (a scalar).

This multiplication works even if some or all of the elements of the matrices are variables. I will illustrate this in my next post.

Matrices, Part 1

In my two-part series on equation systems, I mentioned that there is a third method using matrices. Before I use this method, I have to explain what a matrix is. There is a whole new algebra surrounding matrices (the plural of matrix), so I will only explain what is needed to solve a system of equations.

All of the algebra we have used so far is called scalar algebra. That is, it is used on scalars. A scalar is just a single number like 3, -2.7, or x. It just has a value and you cannot get any more information from it like direction. In most engineering problems, 3-dimensional space is the playground and engineers are also interested in not only the speed of an object, but also the direction it is travelling in. Scalar algebra will not suffice. Enter matrix algebra. This is sometimes equated to linear algebra, but there are differences. I will continue to use matrix algebra as that is the most appropriate at this level.

A matrix is simply an array of numbers. Here are some examples:

\[
\left[{\begin{array}{cc}{{-}{1}}&{0}\\{3}&{2.4}\end{array}}\right]{,}\hspace{0.33em}\left[{\begin{array}{c}{0}\\{\sqrt{7}}\\{{-}{3}{.}{23}}\\{1}\end{array}}\right]{,}\hspace{0.33em}\left[{\begin{array}{ccc}{2}&{{-}{7}}&{0}\end{array}}\right]{,}\hspace{0.33em}\left[{\begin{array}{ccc}{2}&{0}&{3.8}\\{{-}{1}}&{0}&{0}\end{array}}\right]
\]

Of course, you cannot talk about a whole new field of maths without definitions. The first definition is dimension. The dimension of a matrix is its size. By convention, the dimension always indicates the number of rows first then the number of columns. So the first matrix in the examples above, is a 2 × 2 matrix because it has 2 rows and 2 columns. The second matrix is a 4 × 1 matrix. Matrices with 1 as one of the dimensions, can be called vectors as well. The third matrix is 1 × 3 and the last is 2 × 3.

It is very common to use bold typeface when using a single letter to represent a matrix so that the reader knows that they are talking about a matrix and not a scalar:

\[
{\mathbf{A}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{{-}{1}}&{0}\\{3}&{2.4}\end{array}}\right]
\]

So I can now talk about A and you know that A is a matrix.

So there are the same types of operations for matrices as there are for scalars: adding, subtracting, multiplying, and dividing. But there are rules associated with these matrix operations that are specific to matrices.

Since I am directing the use of matrices to solving a system of equations, I will only discuss the multiplying and dividing operations as they relate to matrices. This is where I will begin in my next post.

System of Equations, Part 2

So last time I solved a system of two equations using the substitution method where the information from one equation is inserted into the other equation. This is the method of choice if it is easy to solve for one of the unknowns. However, the example that I used also lends itself well to the other method: Elimination.

The elimination method, like the substitution method, uses the two equations to generate one equation with one unknown which can be solved. The system I worked on last time was:

x + y = 108
xy = 38

The elimination method is simply adding the two equations together with the goal of eliminating one of the unknowns. So let’s add these two equations:

x + y = 108
xy = 38
2x = 146

Notice that I now have an equation with one unknown. Dividing both sides of this new equation by 2 gives:

2x/2 = 146/2 ⟹ x = 73

the same answer as before. Now use this value for x in one of the original equations. Let’s use the first one:

x + y = 108 ⟹ y = 108 – x = 108 – 73 = 35

So we get (thankfully) the same solution as before. But this example is rather contrived in that the y‘s were conveniently of opposite signs in the given equations. So let’s consider the following system:

2x + 3y = 51
3x + 2y = 49

Adding these two equations together will just give us another equation with two unknowns. But just like we do with single variable equations, we can modify one or both of these. Let me take the first equation, and multiply it by – 2. Why I am doing this will soon be revealed:

-2(2x + 3y) = 51(-2) ⟹ -4x -6y = -102

I will now multiply the second equation by 3:

3(3x + 2y) = 49(3) ⟹ 9x + 6y = 147

Now let’s repeat the system, replacing the equations with the new ones:

-4x -6y = -102
9x + 6y = 147

Notice that if I now add these equations, the y variable will disappear:

-4x -6y = -102
9x + 6y = 147
5x = 45 ⟹ x = 45/5 = 9

I will now substitute this partial solution into the original second equation:

3x + 2y = 49 ⟹ 3(9) + 2y = 49 ⟹ 27 + 2y = 49 ⟹ 2y = 49 – 27
⟹ 2y = 22 ⟹ y = 11

So x = 9 and y = 11 solves both of these equations.

Both methods, substitution and elimination, can either be used to solve a system of equations, but one method may be less work than the other.

System of Equations, Part 1

The equations that I have solved so far, have been equations with one unknown (variable). Suppose we have two unknowns in a problem and wish to solve for both. There is a rule in maths that you have to have the same number of equations as the number of unknowns if you are to solve for all unknowns. These equations need to be independent. what do I mean by that?

Suppose I need to find two numbers that add up to 108. In equation form, I want to find x and y such that x + y = 108. So I need another piece of information (equation) if I need a specific solution. Suppose you say “I can get another equation by just multiplying both sides of the given equation by 2: 2x + 2y = 216.” The problem here is that this second equation depends on the first equation. This is what I meant by “independent”. The second equation must be completely new information.

So let’s say that these two numbers must also have a difference of 38 : xy = 38. This is a completely new (independent) requirement. So we now have the two equations:

x + y = 108
xy = 38

So what two numbers satisfy these two requirements? There are two main methods for solving this: Substitution and Elimination. Actually, there is a third method which uses matrices, but I will explain what a matrix is and how to use it in a subsequent post. Let’s first talk about the substitution method.

The substitution method is basically inserting information from one of the equations into the other equation. Let’s solve the second equation for x. That is, use algebra to get x on one side and everything else on the other side. If I add y to both sides of the second equation, I get

xy + y = 38 + yx = 38 + y

Now put this definition of x into the first equation. That is, replace x in the first equation with what x is equal to from the second equation:

x + y = 108 ⟹ 38 + y + y = 108 ⟹ 38 + 2y =108

Notice that this new equation, which combines the information of the two equations into one, is a single equation with a single unknown. This can now be solved with the techniques we have seen before:

38 + 2y =108 ⟹ 2y = 108 – 38 = 70 ⟹ y = 70/2 = 35

So y is 35. You can now use this solution into any of the equations involving x and y, then solve for x. Let’s use x = 38 + y:

x = 38 + y = 38 + 35 = 73

So the two numbers are 35 and 73. They both add up to 108 and subtract to equal 38.

I’ll start the next post doing another example.

Newton’s Clock, Part 4

So here is the last of this series explaining the expressions on Newton’s clock:

We are now up to 8. Let’s look at

\[
\mathop{\prod}\limits_{{k}{=}{0}}\limits^{1}{{(}{2}{k}{+}{2}{)}}
\]

This is another excellent example of how concise the words in maths can be. The symbol “𝚷” is the capital version of 𝜋 which corresponds to the english “P”. The “P” here stands for “product” which is the result of multiplying two or more numbers. The expression on the clock means : “Take the expression 2k + 2 and successively replace the k with the number at the bottom of the 𝚷 symbol (in this case “0”), evaluate the expression to get a number, and increment k by 1 and repeat until you reach the number at the top of the 𝚷 symbol (in this case “1”). Then multiply all these numbers together.”

You can see why maths expressions are much more concise than English. So to evaluate this expression, we first replace the k with 0, then work out 2(0) + 2. This equals 2. Now increment the k by 1 to get 1, then work out 2(1) + 2. This equals 4. Since k is now at the number at the top of 𝚷, we are done increasing k. Now multiply these numbers together. 2 × 4 = 8 which is the correct number at this position on the clock.

Most of you now know what

\[
\sqrt{81}
\]

means. It means “what number multiplied by itself equals 81?” The answer, of course is 9 as 9 × 9 = 81.

The next hour is

\[
{\log}_{2}1024
\]

The basics of this expression have already been explained for position “2” on the clock. This expression is asking the question “what does the exponent of 2 have to be so that 2x = 1024?”. Hopefully, the answer is “10” and it is because 210 = 1024.

Now let’s look at B16. Remember when I explained position “7” on the clock: 01112? That was a number in the base 2 system of counting. Another common base used with computers is the base 16 counting system. We are familiar with the base 10 counting system that has 10 symbols used to count with: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The base 16 system needs 16 symbols. So what is used after 9 is reached? Well, we resort to the letters of the alphabet. The numbers up to 9 in base 16 correspond to the same numbers in base 10. The next number in base 16 is “A” which corresponds to 10 in base 10 and the next number is “B” which is 11 in base 10. So B16 = 11.

Finally, the last expression:

\[
\mathop{\sum}\limits_{{i}{=}{1}}\limits^{3}{{(}{3}{i}{-}{2}{)}}
\]

The Σ symbol is the Greek capital “sigma” and corresponds to the english “S”. This letter stands for “Sum” which is the addition of two or more numbers. This expression is just like the one in position “8” on the clock except that you add the resulting numbers together instead of multiplying them. So starting at i = 1, 3(1) – 2 = 1, 3(2) -2 = 4, 3(3) – 2 = 7, and we are done as i now equals the numbers on top of the Σ. So now add these numbers together: 1 + 4 + 7 = 12. It is now high noon and that completes Newton’s clock.

Newton’s Clock, Part 3

So now I’m up to “4” on Newton’s clock:

So the expression

\[
{\left({2\sin\frac{\mathit{\pi}}{2}}\right)}^{2}
\]

uses the sine function which has been talked about many posts before. Only this time, it is using radian measure of angles instead of degrees. If your calculator is in degree mode, you can substitute 90° in place of 𝜋/2 to get the same answer. The sine of 𝜋/2 radians or 90° is 1. So in the brackets we have 2 × 1 = 2. 2² = 4, hence its position on the clock.

Now let’s look at

\[
\sqrt[3]{125}
\]

This is the cube root of 125. This expression is asking the question: “What number multiplied 3 times equals 125?”. The answer to that is 5 because 5 × 5 × 5 = 125. So once again, the clock does not lie.

Now let’s look at 3! This is pronounced “3 factorial”. The factorial of a number is that number successively multiplied by a number which is 1 less. So 5! = 5 × 4 × 3 × 2 × 1 = 120. So 3! = 3 × 2 × 1 = 6. Factorials are used a lot in probability. I have touched on this before but perhaps there is another future post here.

Now let’s look at 01112. We are very familiar with decimal system way of counting. This system is a base 10 system because we use 10 distinct digits (symbols) to count: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. When we run out of digits, like when we count up to 9, we add another place holder to the right of the number and put the starting digit 0 there: 10. And then successively increase it’s digits until we get to 9 again. then we increase the left digit by 1 and start over again: 20, 21, … . There are other number systems based on numbers other than ten.

Computers are composed of switches based on two states, on or off. We mathematically say that off is 0 and on is 1. Computers essentially count with just o’s and 1’s: a base 2 system. Counting in base 2 is done exactly as we do in base 10, we just have fewer digits to work with.

So we if we start counting we get 0, 1, but we’ve ran out of digits so we add a place holder to the right and start again: 0, 1, 10, 11. Ran out of digits again so add another place holder and start over: 0, 1, 10, 11, 100, 101, 110, 111. If you are keeping track, 111 in base 2 is equal to 7 in base 10. It is a convention to subscript a number with its base when dealing with other base systems, so 01112 means 7 in base 10. The leading 0 doesn’t add to the value but in computer maths, base 2 numbers are typically written 4 digit places at a time.

Newton’s Clock, Part 2

Well I didn’t get very far around Newton’s clock last time because the expression for “1” took a while to explain. Today I will explain the expressions for “2” and “3” on the clock:

The expression log10(100) is a logarithm (logs). I’ve talked about logs before. They are exponents. This particular one is the english equivalent of asking “What is the exponent of 10 so that 10x = 100?”. Well, the answer to that is “2” because 10² = 100.

The next expression

\[
\mathop{\int}\limits_{1}\limits^{2}{2xdx}
\]

again, will take a bit of explaining.

This is a calculus expression. The ∫ symbol is an elongated “S”. It has a German origin but this symbol was used because the expression represents a sum (addition) of infinitesimal (that is, ridiculously small) things. For this particular expression, you can think of it as finding the area under the plot of the equation y = 2x from x = 1 to x = 2:

So this expression can be thought of as adding the areas of infinitesimally thin rectangles from x = 1 to x = 2, with a width of dx and a height of 2x. This sum will equal the total area under y = 2x from x = 1 to x = 2. Using calculus, this area is equal to 2² – 1² = 4 – 1= 3. So this is why this expression is in the “3” position.

I will leave this as an exercise for the reader to confirm that this is the correct area by using geometric formulas for area for triangles or trapeziums (trapezoids in the USA).

Newton’s Clock, Part 1

As I mentioned many posts before, maths is a language – a much more elegant and concise language when taking about maths stuff. This will be very apparent when I talk about the expressions in the clock I just bought for my office:

Each one of the maths expressions on the clock equal the number corresponding to its position on the clock. So I will be talking about each of these.

By far, the expression corresponding to “1” on the clock, illustrates the conciseness of maths expressions:

\[
{e}^{\mathit{\pi}{i}}\cos\mathit{\pi}\hspace{0.33em}{=}\hspace{0.33em}{1}
\]

This expression makes use of more mathematical knowledge than any of the other expressions. It uses the maths constants e and 𝜋, and uses radians, trigonometric (or circular) functions, exponentials, and Euler’s identity. The two parts of this expression are e𝜋i and cos 𝜋. Let me first talk about cos 𝜋.

Now I have talked about trig functions before, but I mainly talked about sine of an angle (abbreviated “sin”). The cosine of an angle (abbreviated “cos”) is similar. You can take the cosine of an angle on your calculator, but you need to tell the calculator whether you are measuring angles in “degrees” or “radians”. In this expression, we are taking the cosine of 𝜋 radians. If your calculator is in radians mode and if you have ‘𝜋’ button, taking the cosine of 𝜋 will show -1. If you do not have a 𝜋 button, take the cosine of 3.14179 and you will get an answer approximately equal to -1. So, cos 𝜋 = -1.

Now the e𝜋i part requires a bit more of an explanation. The numbers e and 𝜋 have been discussed before. They are both irrational numbers which means that you cannot write them down exactly using numbers. We just agree that the symbol e is exactly e and the symbol 𝜋 is exactly 𝜋. If you have an ex button on your calculator, you can get an approximate value for e by typing 1, then the ex key. I have talked about e before. The letter e is for Euler who did a lot of work with this constant. So what is i?

The number i is not a real number. It is actually called an imaginary number. It is formally defined in terms of its square:

i2 = -1

which means that i is the square root of -1. I’m not making this up.

Since in the realm of real numbers, you know that you cannot take the square root of a negative number. So defining i is creating a whole new realm of numbers called complex numbers. Actually, complex numbers are the union of two realms: the reals and the imaginaries. There is a lot more to say about this but this will have to be a topic for a future post.

So what is e𝜋i where i is in the exponent? There is an identity (a maths rule) called Euler’s identity that explains what e𝜋i is equal to:

e𝜋i + 1 = 0

So e𝜋i must equal -1 for Euler’s identity to work. Again, explaining anything more is the topic of several future posts.

So if e𝜋i = -1 and cos 𝜋 = -1, then

e𝜋icos 𝜋 = (-1)(-1) = 1 which is the 1 o’clock spot on the clock.

I haven’t even scratched the surface of explaining this expression, but it has already filled up this post. The other expressions will not take as long to explain.

Newton’s Laws, Part 7

Please refer to the previous posts in this series if this post is to make any sense.

Believe it or not, I made a small error in the equations from the last post. I used kilometers instead of meters in part of the equation for the force due to gravity. The two equations for the thrust phase and the coasting phase are:\[ \begin{array}{l} {{F}{=}\hspace{0.33em}{2}{,}{800}{,}{000}\hspace{0.33em}{-}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{9}{.}{8}{\left({\frac{6400000}{{6400000}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}}\\ {{=}\hspace{0.33em}{(}{16000}\hspace{0.33em}{-}\hspace{0.33em}{57}{.}{2222}{t}{)}\hspace{0.33em}\times\hspace{0.33em}{a}} \end{array} \]

and \[ \begin{array}{c} {{F}{=}{-}{9}{.}{8}{\left({\frac{6400000}{{6400000}\hspace{0.33em}{+}\hspace{0.33em}{x}{(}{t}{)}}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{a}} \end{array} \]

Now as mentioned before, these are rather difficult to solve. However, first class rocket scientists rarely solve equations by hand. They resort to numerical methods to solve them. Numerical methods means that they enter the equations into a program and then let the computer solve them.

I have done that and have plotted the results. Below is a plot of the distance the rocket is after t seconds:

Looks like we sent the rocket into space with quite a kick! At 270 seconds, the thrust stops and the rocket keeps going with no sign of slowing down. Its velocity when the thrust stops is greater than the escape velocity at that height. Escape velocity is the speed an object needs to break free of the earth’s gravity. The rocket never falls back to earth.

I also plotted the velocity of the rocket. See how the velocity remains relatively constant after thrust cutoff because gravity is too weak to slow it down:

So that’s all I will say about Newton’s laws, at least for this series of posts. Next week, what’s the deal with this clock?