## Engineering Topics – Differential Equations

A differential equation is an equation that has a derivative in it. A derivative is a rate of change, like velocity. So if you are driving in a car where your velocity from a starting point is v which is some function of time, this can be solved to find your position from a starting point at any time. There are lots of techniques to solve these equations and the study of this scares many students. But the fact is, if you drive a car, pick up a glass of water, throw a ball at a target, your mind subconsciously handles the differential equations that model these activities quite well.

Let’s stick with the car example. Suppose you are 100 metres away from a stop sign or a red traffic light. So you apply the brakes. Let x(t) be your distance from the stop sign in metres at a time t, t be the time in seconds, and (t) be your velocity at time (t). Now you want your distance to the stop sign to decrease from 100 to 0 metres comfortably in say 15 seconds. What about this linear way:

It does stop at the stop sign, but is it comfortable? This equation of a line has a constant slope (rate of change) of −100/15 = −6.67 m/s = −24 km/hr. This means that at the stop sign, you are going 24 km/hr when you hit the brakes hard to stop. The passengers drinking coffee at the time, would not appreciate that. Also, if you are going 100 km/hr at 100 meters away, to follow this profile, you have to slam on the brakes to suddenly get at a speed of 24 km/hr. Well maybe there would be no coffee left to spill at the stop sign.

So this shows that we need to be aware of our speed and our distance to do this comfortably.

What about stopping following this red curve:

This starts at 100 metres and ends at 0 like the linear graph but has a better rate of change profile. The rate of change (that is the velocity) varies on this curve. It is visually seen as the slope of the line tangent to the graph at a point. The grey line shown is an example of a tangent line. Notice that the gradient at the beginning of the curve is high (in the negative direction) but at the end, it is near zero (the slope of a horizontal line is zero). This would be a much smoother stop than a linear approach.

But your mind during this action is not just seeing your distance from the stop sign, it is also sensing your velocity and adjusting it as you get closer to the stop sign. The following is an equation that relates the velocity and position:

$\dot{x}(t)=-0.3[x(t)+1.11]$

If you were to solve this equation for x(t) using differential equation techniques, you would get the equation seen in the graph above. If you were to design a control system (which is what your mind is when performing this action) you would use the above differential equation to control both your position and your velocity.

But even this stopping profile has flaws. Notice that the deceleration at the beginning is quite steep (the slope of a tangent line at t = 0). Perhaps a better profile would be:

This starts with a more gentle deceleration, increases the deceleration until you get closer to the stop sign, then the deceleration decreases until you come to a full stop at the sign.

Regardless of the stopping profile used, your mind controls the braking action to conform to a desired profile based on your current speed (the slope of the tangent line) and your distance from the stop sign. People who are designing driverless cars, robotic arms, aircraft autopilots, etc, use differential equations. And because they are working in three dimensions, these equations can be in the form of matrix and/or vector equations. And the solutions will use complex numbers: all of these topics were covered in my last few posts.

So besides the basic algebraic skills you may be studying or have studied, more advanced topics like this one or those covered in in my last few post are the heart of engineering.

## Engineering Topics – Matrices

If you looked at my last post on complex numbers, the example I used was 1-dimensional, there was only one variable of interest. However, if you include time, the problem is 2-dimensional and we can plot the result on a 2-dimensional coordinate system. But our world is 3-dimensional (4 if you include time). In engineering, there are frequently many more than 3 variables needed to solve a problem. To handle these kind of problems, matrices are needed. Matrices come with their own set of algebra rules, but you don’t need to know these to follow this post. Let’s look at some examples.

### System of Equations

Perhaps the simplest example that looks intuitively correct is using matrices to solve a set of m linear equations with m unknowns. In year 10, students are taught how to solve a system of 2 equations and 2 unknowns. They learn about two methods to solve these: substitution and elimination. These methods can be used for higher number of unknowns but this quickly becomes unwieldy as the number of unknowns increases.

An engineering example using 3 unknowns comes from my experience as an astronautical engineer. I was tasked to model the output of a mechanical gyroscope. A gyroscope is used to measure rotation about a specific direction (axis). This information is used in an inertial navigation system to determine an object’s orientation and velocity. There are several kinds of gyroscopes: mechanical, laser, semiconductor. The one in your smartphone is a semiconductor one. Mechanical ones are still used in aircraft and spacecraft navigation systems because of their long-term stability. However, their electronic output is not directly proportional to the rotational input, there are errors in the output. In an ideal world, the output, v, will be kω = v where ω is the rotation detected (radians/second), v is the output voltage, and k is the conversion factor needed to convert radians/second to volts. But as our world is not ideal, there are errors in the signal produced by the gyroscope. These errors need to be subtracted from the output before it is sent to the navigation system.

The model I used had many more terms in it, but for purposes of this post, I will simplify it to

$k_1\omega_1+k_2\omega_2+k_3\omega_3=v$

where k1ω1 is the desired output in direction 1, and the other two terms are errors introduced from rotations about the other 2 perpendicular directions in 3-dimensional space. For a particular gyroscope, I had to find the k‘s so that the navigation system would know what the actual rotation about direction 1 is.

So the gyroscope was placed on a very accurate test platform where its orientation with respect to the earth’s rotation was accurately known and the platform could also rotate an accurately known amount. So if the gyroscope was subjected to three different orientations/rotations and the output measured at each position, three equations in the 3 k unknowns could be generated. Actually, many more measurements were made. There are errors in the measurements of the outputs and the inputs so I actually used a least squares matrix process (yes, statistics is used in engineering as well) to find the best estimates of the k‘s. But again, for purposes of this post, let’s assume we have perfect knowledge of the inputs and outputs so that only three measurements are needed.

Using different rotation rates about the primary measurement axis and the two perpendicular ones, we generate the following set of equations:

$3k_1+2k_2+1k_3=2.51\\5k_1-5k_2+7k_3=3.82\\6k_1-6k_2-7k_3=4.43$

There is a matrix version of this system of equations:

$\begin{bmatrix}3&2&1\\5&-5&7\\6&-6&-7\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}2.51\\3.82\\4.43\end{bmatrix}$

Without going into the rules of matrix algebra, I think you can see how each of the objects in the above equations were assembled: the array (matrix) of numbers are the coefficients of the unknowns, listed in the same order as in the system of equations. The vertical matrix next to it (also called a vector as it only has the one column) lists the unknowns, and the matrix on the right side is the numbers on the right side of the system. If we let A be the matrix of coefficients, k be the matrix of unknowns, and b be the matrix of the right side numbers, the matrix equation and its solution is:

$\textbf{Ak}=\textbf{b}\Longrightarrow\textbf{k}=\textbf{A}^{-1}\textbf{b}$

On a CAS calculator, this is solved very quickly. The answer will be the matrix k and you just pick off the elements for each ki in the same order as in the setup of the matrix equation. Doing this on my CAS, I get the answer :

$\textbf{k}=\begin{bmatrix}0.8\\0.05\\0.01\end{bmatrix}$

So the output model for this gyro is

$0.8\omega_1+0.05\omega_2+0.01\omega_3=v$

Given the corrected outputs of the other two gyros, the navigation system knows what the true rotation about axis 1 is.

### Rotation matrix

Another “rocket scientist” application of matrices associated with navigation are rotation matrices. Supposed your spacecraft is halfway between earth and the moon and a course correction is needed. In which direction do you burn your rocket engines? If you just point your spacecraft toward the moon and fire, you will miss by a lot. This is because you are not taking your spacecraft’s and the moon’s motion into account. And when I say “motion”, I mean with respect to an inertial (that is static) reference coordinate system, called an inertial reference frame. Depending on the scenario, this could be a system at the centre of the earth or the sun that does not rotate with the earth or sun.

So calculations are made with respect to the inertial frame but the spacecraft’s navigation system only knows its reference frame when it comes to firing the engines. Rotation matrices is how an inertial direction is converted to a spacecraft’s reference frame.

To simplify this a bit, I will limit the coordinate systems to be 2-dimensional, but the concept can easily be extended to 3 dimensions.

Consider the two coordinate systems below:

where the x-y system is the inertial frame and the x’-y’ system is the spacecraft reference frame. Now I know that the origins of these two frames will be physically separated, but it turns out that this does not matter. Only the angle (angles for 3-dimensions) between the two frames matter. I draw them together so you can more easily see the result.

Now suppose that, with respect to the inertial frame which does take into account the motions of the spacecraft and the moon, it is calculated that the direction the engines should fire for the course correction is v as indicated in the diagram. This direction has an x value and a y value which corresponds to where the arrow of v is. The spacecraft has to convert this to its coordinate system and it does it with the following matrix multiplication:

$\begin{bmatrix}x’\\y’\end{bmatrix}=\begin{bmatrix}\text{cos}(\theta)&\text{sin}(\theta)\\-\text{sin}(\theta)&\text{cos}(\theta)\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$

The angle 𝜃 is known from the various navigation systems that are being used to track the spacecraft, including an inertial navigation system that has gyroscopes that we just modelled. Now the spacecraft knows in what direction it has to fire its engines.

### Other uses

In the previous example, the navigation computers needed to have a model of all the forces acting on the spacecraft and the resulting motions, that is its dynamics, in order to calculate the vector v. This model is a matrix differential equation that is constantly being numerically solved. To do this, the computer needs to keep track of the spacecraft’s position in 3-dimensional space (3 variables) as well as the velocity in each of those dimensions (3 more variables). The matrix needed for that model has to have the same number of rows and columns as the number of variables, so that is a 6 × 6 matrix (36 elements).

Matrices are used in electronics, optics, quantum mechanics, cryptography, robotics – you get the point.

## Engineering Topics – Complex Numbers

The world moves and so do many things that occur naturally or are made by humans. Because of this, when engineers or scientists want to mathematically model a system or process, they frequently need to identify not only the variables of interest (like position), but the rate of change of these variables (like velocity). These mathematical models are equations that relate the variables and their rates of change. Equations like this are called differential equations and how to solve these equations is usually introduced to students after they have studied calculus. And solutions to these equations frequently have complex numbers in them.

What is a complex number? A complex number is defined using the definition of the imaginary unit, i:

$i=\sqrt{-1}$

This may look like a crime against mathematics as through much of our maths education, we were told that you cannot take the square root of a negative number. This is reinforced by trying to take the square root of a negative number on many calculators resulting in an error. But it turned out that this invention had some usefulness in maths.

Numbers like bi where b is a real number are called imaginary numbers. If you add a real number a to this, a + bi, you get a complex number, where a is the real part and b is the imaginary part. There is a lot of theory surrounding complex numbers, but I will only cover what is necessary for this post.

Complex numbers satisfy many of the properties you are familiar with using real numbers including the rules involving exponents. So using a complex number as an exponent to the natural base e (a number, like 𝜋, which is frequently used in engineering), the expression can be split into two parts:

$e^{a+bi}=e^ae^{bi}$

You are familiar with ea, but what do we do with ebi? This is actually a complex number as well and can be put into a standard form using Euler’s formula:

$e^{bi}=\text{cos}\,b+i\,\text{sin}\,b$

In textbooks, the right side of the above equation is abbreviated as cis b. So

$e^{a+bi}=e^a(\text{cos}\,b+i\,\text{sin}\,b)=e^a\,\text{cis}\,b$

So what can imaginary numbers tell us about the real world? Well, the “useless” things you were taught about quadratic equations, are about to become useful.

### The Damped Harmonic Oscillator

There are many things that oscillate, but not forever: electronic circuits, your car when it hits a bump, aircraft when they hit an air pocket. A simple example is a mass on a spring with a damper attached:

There’s a lot of physics happening here. But all you need to know is that there are 3 main forces affecting the motion of the mass. These forces are created by: gravity (which creates an acceleration downward), spring (proportional to the position of the mass), and damper (proportional to the velocity of the mass). What is the position of the mass at any time t?

Using Newton’s second law, F = ma, the following differential equation can be generated. A differential equation is an equation that relates a variable (in this case, the position x) with its rates of change (in this case velocity and acceleration):

$m\ddot{x}+b\dot{x}+kx=0$

where x is the position of the mass relative to a reference point, b is the damping coefficient (how strong is the damper), k is the spring constant (how strong is the spring), x with one dot above it is the rate of change of x with respect to time (commonly known as velocity), and x with two dots is the rate of change of velocity with respect to time (commonly known as acceleration).

Solving differential equations is a whole university course, but for this type of equation, the solution will be of the form:

$x(t)=Ae^{𝜆t}$

where A is the initial position of the mass at t = 0. So the problem reduces down to finding 𝜆. It turns out that 𝜆 is the solution to the corresponding algebraic equation (called the characteristic equation):

$m\lambda^2+b\lambda+k=0$

So yes, here is an example where you use the quadratic skills you learned. Using the quadratic formula:

$\lambda=\frac{-b\pm\sqrt{b^2-4mk}}{2m}$

The discriminant b2 – 4mk, dictates the type of solutions for 𝜆. In this post, I am interested in the case where b2 – 4mk < 0, the under damped case which is graphically shown in the animation above.

With a little bit of algebra and using the definition of i to factor out the −1 inside the square root (and what remains in the square root is positive):

$\lambda=\frac{-b\pm\sqrt{(-1)(4mk-b^2)}}{2m}=\frac{-b}{2m}\pm i\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}$

If we let

$\alpha=\frac{b}{2m}\text{ and }\omega=\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}$

then 𝜆 = -⍺ ± ωi. So the solution is

$x(t)=Ae^{(-\alpha\pm \omega i)t}=Ae^{-\alpha t}e^{(\pm\omega ti)}=Ae^{-\alpha t}(\text{cos}(\pm \omega t)+i\,\text{sin}(\pm\omega t))$

Now it looks like we still have an imaginary part in the answer. We need a real solution that fits the real world. As I said before, solving differential equations is a separate subject usually studied at uni. In that subject, you would learn about the superposition principle where any linear combination of two separate solutions of a differential equation will also be a solution. Notice that we do in fact, have two solutions above: one using the + and the other using the −. Using relationships that exist for circular (trig) functions for sine and cosine, we can add these two solutions together and the imaginary part will cancel out, leaving only a real solution:

$x(t)=Ae^{-\alpha t}\text{cos}(\omega t)$

As t grows, the cosine part of the solution just bounces up and down between ±A. But the exponent of e gets more negative as t grows making e⍺t smaller, starting at 1 when t = 0. This generates the following curve:

Notice how the exponential part of the solution is an envelope that the cosine curve must fit into.

Instead of a mass hanging on a spring, the differential equation we started out with could represent your car’s suspension system which has springs and shock absorbers (dampers). Notice that we can change the parameters, the strengths of the springs and shock absorbers, to change the way a car handles bumps. The response in the graph above may be too loose and we may want to change the parameters to make the car settle down more quickly. This is engineering.

## Engineering Topics – Vectors

As an ex-engineer, I know that almost all of the math skills I have learned in school have been used in my career. However, students generally do not appreciate this when, for example, they are factoring a quadratic. It becomes easier to demonstrate the usefulness of math to students when they are being taught skills that have a direct application. So I thought I would present a few topics in maths with some applications in engineering. The first of these topics is vectors.

A vector is an object with a magnitude and a direction. If you are travelling at 60 km/hr, that number is just a magnitude. But if you are travelling north at 60 km/hr, that has a directions as well and can be represented as a vector.

There are a lot of new math skills associated with vectors, but I will only cover what is necessary to present this application of vectors.

Vectors can be added algebraically but I will just be talking about how to add vectors graphically.

Here is a typical vector:

It’s magnitude can be its physical length or a labelled quantity like 60 km/hr or 100 N (newtons) of force. Its direction though, needs a reference. This will usually be a coordinate system like the Cartesian coordinate system:

This vector represents a velocity of 60 km/hr in the direction 30° from the positive x-axis. Now this direction is completely dependent on the coordinate system used. We could have made the x-axis line up with the vector itself. But once established, questions regarding the vector will be answered with respect to the coordinate system used.

A vector can be the sum of two or more vectors. There are many variations to this but the scenario applicable to this post is where the vector is the sum of two vectors that are perpendicular to each other. Given the coordinate system we have used, the vector V can be broken up into two components, Vx and Vy. That is, V = Vx + Vy. Graphically, this looks like:

Now we already know the direction with respect to this coordinate system of each of the component vectors Vx and Vy, but what are their magnitudes?

Looking at the above figure, you can see that the vector V splits the rectangle formed by the x and y axes and the dotted lines into two similar right triangles. If the length of the hypotenuse of these triangles is 60, then

$\left|{\boldsymbol{V_x}}\right|=60\text{ cos}(30°)=51.96 \text{ km/hr}$

where |vector| is notation to represent the magnitude of a vector. Similarly, you can see that

$\left|{\boldsymbol{V_y}}\right|=60\text{ sin}(30°)=30 \text{ km/hr}$

If you were given the perpendicular vector components, you can use the Pythagorean theorem to find the magnitude. Let’s use the components found above to see if they give back the original vector:

$\left|{\boldsymbol{V}}\right|=\sqrt{\boldsymbol{V_x}^2+\boldsymbol{V_y}^2}=\sqrt{51.96^2+30^2}\approx60\text{ km/hr}$

This is approximately 60 because the 51.96 number used is a two decimal place approximation to the actual irrational number.

So here is a practical application of this:

A cable supports a beam at an angle of 29.745°. A weight at the end of the beam provides a force of 500 N. Assume that the weight of the beam is negligible. What is the tension (force) seen by the cable?

Now before I tackle this, a bit of background. First, the “N” in “500 N” stands for “newtons”. This is the unit of force in the SI system of units. SI is the International System of Units that is usually used in science and engineering. Though your mass may be 80 kg, the force you exert on the earth is 784.8 N. In return, the earth exerts a force of 784.8 N against you in the opposite direction. Second, this problem is a statics problem, that is, nothing is moving. That means that all involved forces (vector quantities that have direction) must add to zero. The force you exert on the earth and the opposite force that the earth exerts on you add to zero, so there is no motion up or down. Third, tension and compression in a structural member is a force that the member exerts on neighbouring objects in response to forces that cause the tension or compression.

Let’s isolate the end of this structure where the force is being applied:

As there is no motion, there must be an opposite but equal force of 500 N:

What is causing this opposite force? It’s the tension in the cable. This tension must be great enough to cause a force of 500 N in the y (up) direction. This 500 N up is the y component of the force from the cable:

Knowing the y component, the opposite side of the right triangle shown, we can use trigonometry to find the hypotenuse, the value of the tension in the cable T:

$\left|{\boldsymbol{T}}\right|=\frac{500}{\text{sin}(29.745°)}=1000.78 \text{ N}$

I leave it to you to review the definition of the sine of an angle in a right triangle.

Even though this was not asked, the tension in the cable creates a force in the x direction (horizontal direction). Again using trigonometry, the value of this force is 875.00 N. Where is the force that opposes this so that there is no motion? It comes from the compression imposed on the horizontal member:

This is an engineering problem because if you were to design and build this structure, you would have to ensure that the materials used could handle the tension and compression loads with an additional safety factor. Before reading this post, if you just thought that the cable needs to handle 500 N, you would soon be looking for a new job.

## Transformations 4

Let’s do a couple of examples using the knowledge from my last 3 posts and show how the order of transformations make a difference.

The image of the equation that follows will be generated by the following transformations:

a. Dilate along y-axis by factor 1/2
b. Reflect along y-axis (across x-axis)
c. Dilate along x-axis by factor 3
d. Translate along x-axis +4 units
e. Translate along y-axis -2 units

$y=\sqrt{3x-4}$

In the previous post, I showed that given an equation y = f(x), its image under a general transformation is given by

$f(x)\Rightarrow af\left[\frac{1}{n}(x-h)\right]+k$

where
a = dilation\reflection factor along y-axis (from x-axis)
n = dilation\reflection factor along x-axis (from y-axis)
h = translation along x-axis
k = translation along y-axis

This assumes that dilations/reflections are done first. So if

$f(x)=\sqrt{3x-4}$

then

$f(x)\Rightarrow af\left[\frac{1}{n}(x-h)\right]+k=-\frac{1}{2}\sqrt{3\left [ \frac{1}{3}(x-4)\right]-4}-2$

This can be simplified to

$y=-\frac{1}{2}\sqrt{ x-8}-2$

In this form, the pre-image (the equation that this one came from before the transformation) is lost. The question could be asked, what are the transformations required to go from

$y=-\frac{1}{2}\sqrt{ x-8}-2\Rightarrow y=\sqrt{3x-4}$

One would expect that if we just do the opposite of the transformations above, we would get the original equation. Let’s see. Let’s do the following:

a. Dilate along y-axis by factor 2
b. Reflect along y-axis (across x-axis)
c. Dilate along x-axis by factor 1/3
d. Translate along x-axis -4 units
e. Translate along y-axis +2 units

These undo the previous transformations. Putting these in our model

$f(x)\Rightarrow af\left[\frac{1}{n}(x-h)\right]+k$

We get

$f(x)\Rightarrow -2f\left[3(x+4)\right]+2=\sqrt{3(x+4)-8}+4+2=\sqrt{3x+4}+6$

Not exactly what we wanted. What went wrong? Well, the model we used assumes that dilations go first. If we want to undo the previous transformations, not only do we use the values we just used, but they must be applied in reverse order as well: the translations go first then the dilations/reflections. Otherwise the dilations affect the translations before they are applied.

I’ll leave this as an exercise for the reader, but the model for transforming y = f(x) assuming that translations go first is

$y=af\left( \frac{x}{n}-h \right)+ak$

So undoing the original transformations in reverse order:

a. Translate along y-axis +2 units
b. Translate along x-axis -4 units
c. Dilate along x-axis by factor 1/3
d. Reflect along y-axis (across x-axis)
e. Dilate along y-axis by factor 2

gives the result

$f(x)\Rightarrow -\frac{1}{2}f\left(3x+4\right)\Rightarrow-\frac{1}{2}\sqrt{3x+4-8}-2+2\Rightarrow\sqrt{3x-4}$

which is the original equation.

So the order of transformations steps will change the final result.

I hope this series of posts helps you better understand transformations.

## Transformations 3

In my last post, I took a general point (x,y) and through the series of transformations:

A. (x,y)
B. Dilate/reflect by factor a along y-axis
C. Dilate/reflect by factor n along x-axis
D. Translate k units along y-axis
E. Translate h units along x-axis

the point was changed to:

A(xy) —-> B(xay) —-> C(nxay) —-> D(nxay+k) —->
E(nx+hay+k).

Now let’s restrict the original point to be one that satisfies the equation y = x2. How does this equation change so that all of its points are transformed correctly?

Think of the transformation process as changing old points to new ones. So the old points, (xy), are changed to the new points (nx+hay+k) = (x‘, y‘), where the apostrophe ‘ is used to distinguish the new x‘s and the new y‘s from the old ones. So under this transformation, x‘ = nx+h and y‘ = ay+k.

Now the equation y = x2 is the equation that the old points satisfy. To find the equation that the new points satisfy, we need to find what the old variables are in terms of the new ones. Solving for the old x and y in the above equations, we get

$x’=nx+ h\Longrightarrow x = \frac{1}{n}\left(x’-h\right)$
$y’=ay+ k\Longrightarrow y = \frac{1}{a}\left(y’-k\right)$

If we replace the old variables in the equation with the right side of the above equations, then we will get an equation with the new variables, which can be cleaned up with a little algebra:

$\frac{1}{a}(y’-k)=\left[\frac{1}{n}(x’-h)\right]^2\Longrightarrow y’=a\left[\frac{1}{n}(x’-h)\right]^2+k$

Now that we have the new equation, we don’t need the ‘ anymore. So the new equation is:

$y=a\left[\frac{1}{n}(x-h)\right]^2+k$

Notice that if we were given this new equation and were asked, “what are the transformations that generated this equation from y = x2 ?”, then we can pick off the dilations and the translations. Be aware though, that we must assume that the dilations occurred first. Textbooks will tell you this to keep answers consistent. With this caution, if a = 2, n = 3, h =-5, and k =4, then the new equation would be

$y=2\left[\frac{1}{3}(x+5)\right]^2+4$

The graph below shows the effect of these transformations on the original curve:

So given a set of transformations, the steps that I did above for the particular equation y = x2, can be done for any equation.

What if the problem is reversed: given the end result, what is the set of transformations that created the new equation from and original one? Using functional notation, notice that:

$f(x)={x}^{2}\Longrightarrow af\left[\frac{1}{n}(x-h)\right]+k=2{\left[\frac{1}{3}(x+5)\right]}^{2}+4$

If you were given this final result, you can pick off the dilations along both axes (These are assumed to be first) and the translations along both axes. If either the a and/or the n were negative, then there would be reflections as well.

## Transformations 2

In my last post, I took specific points and transformed them using dilations/reflections and translations. The goal is to transform an equation into a new equation. But before we get there, let’s again look at a single point. But this time, this will be a general point (x, y).

So starting with general point A(x, y), let’s do the same transformation as I did in my last post:

A. (x,y)
B. Dilate by factor 2 along y-axis
C. Dilate by factor 1/3 along x-axis
D. Reflect across x-axis
E. Reflect across y-axis
F. Translate 3 units up
G. Translate 3 units to the left

In this order, the new transformed points become:

A(x, y) —-> B(x, 2y) —-> C(x/3, 2y) —-> D(x/3, -2y) —->
E(-x/3, -2y) —-> F(-x/3, -2y+3) —-> G(-x/3-3, -2y+3)

From my last post, if you use the particular point A(3,1), and use these values in the result above, you get the same point G(-4,1).

This can be generalised more by using letters to represent the dilation factors and the translation amounts:

a = dilation/reflection factor along y axis
n = dilation/reflection factor along x axis
h = translation along x axis
k = translation along y axis

Note that I have combined dilations and reflections. This is because a reflection can be viewed as a negative dilation. If a is negative, then this is a reflection along the y axis (across the x axis) as well as a dilation. If n is negative, then this is a reflection along the x axis (across the y axis) as well as a dilation. This does slightly restrict the flexibility of transformations as we cannot separate the order of a dilation and a reflection, but for most problems, this is not an issue. Also, if h is negative, that is a translation to the left. If k is negative, that is a translation down.

Now I’ll apply this to the same set of transformations as above except that I will combine the dilation and reflection steps:

A. (x,y)
B. Dilate/reflect by factor a along y-axis
C. Dilate/reflect by factor n along x-axis
D. Translate k units along y-axis
E. Translate h units along x-axis

This leads to the new point:

A(x, y) —-> B(x, ay) —-> C(nx, ay) —-> D(nx, ay+k) —->
E(nx+h, ay+k)

If these transformations are reversed:

A. (x,y)
B. Translate h units along x-axis
C. Translate k units along y-axis
D. Dilate/reflect by factor n along x-axis
E. Dilate/reflect by factor a along y-axis

you get the new point:

A(x, y) —-> B(x+h, y) —-> C(x+h, y+k) —-> D[n(x+h), y+k] —->
E[n(x+h), a(y+k)]

This is different and you will generally get a different ending point for a specific point and transformation values.

Some terminology is needed here. We have an original point (x,y) and then under a set of transformations, a new point is created. In our first set of transformations, this new point was (nx+h, ay+k). Point (nx+h, ay+k) is the image of (x,y) under this transformation. Or (x,y) is the pre-image of (nx+h, ay+k).

It may help to think of this as a black box where (x,y) enters and is magically transformed to (nx+h, ay+k). Though we know it is not magic at all but just maths:

As mentioned, the goal here is to transform a set of points, usually defined by an equation, to a new set (new equation). The above mental image (pun intended) will help keep in focus what is being done.

In my next post, I will transform equations and show how that is done.

## Transformations 1

From my experience teaching mathematics, I think that the topic most students find the most difficult is circular functions (trigonometry). The next most difficult topic is transformations: transforming a function to another via dilations, reflections, and translations. This is the first of several posts to address this topic.

First, by way of looking at individual points in a Cartesian coordinate system, let’s do some transformations and define some terms.

Consider the following points A, B, C, and D:

The first transformation will be to dilate these points. This means that we will change their distance from the x or y axes by multiplying the appropriate coordinate by a number. This number is called the dilation factor. If we dilate along the x-axis, we will change the x coordinate of each point by multiplying it by the dilation factor. Be aware that this is also called “dilating from the y-axis” because this dilation changes the point’s distance from the y-axis

So let’s dilate these points along the x-axis by a factor of 2. This makes each point twice as far from the y-axis. So point A becomes A1(6,1), B becomes B1(-4,3), C becomes C1(-8,-1), and D becomes D1(2,-2):

If we had used a factor of 1/2, then the new points would be half the distance from the y-axis than the original points.

If the factor we use is negative, not only is the dilation happening, but the points are also reflected across the y-axis. So if the dilation factor is -2 along the x-axis, the new points are A1(-6,1), B1(4,3), C1(8,-1), and D1(-2,-2):

Another transformation is translation: moving a point left or right or up or down. So if you translate a point 3 units to the right (along the x-axis), then the new points are A1(6,1), B1(1,3), C1(-1,-1), and D1(4,-2):

Now all of these transformations can be done along the y-axis (from the x-axis) as well. So doing all transformations along the y-axis, I will dilate point A by a factor of 2, reflect point B with no dilation, translate point C up 3 units, and dilate point D by a factor of 1/2. The new points will then be A1(3,2), B1(-2,-3), C1(-4,2), and D1(1,-1):

The previous example just changed the x or the y coordinate of each point. Let’s look at point A and mix up these transformations. This will also show that the order of the transformations can affect the final result. Consider the following sequence of transformations of point A:

A. (3,1)
B. Dilate by factor 2 along y-axis
C. Dilate by factor 1/3 along x-axis
D. Reflect across x-axis
E. Reflect across y-axis
F. Translate 3 units up
G. Translate 3 units to the left

The following graph shows the sequence of these transformations .The new point’s letter refer to the result of the transformation’s letter above:

Now let’s do the previous steps in reverse:

A. (3,1)
B. Translate 3 units to the left
C. Translate 3 units up
D. Reflect across y-axis
E. Reflect across x-axis
F. Dilate by factor 1/3 along x-axis
G. Dilate by factor 2 along y-axis

We end up at a very different place. Also notice that if a point is on the y-axis, reflections across the y-axis and dilations along the x-axis, have no effect.

I will generalise this for a generic point (x,y) in my next post.

## Graphing Circular Functions

The following file is for year 12 students covering trigonometric functions sine and cosine. It presents an orderly method to graph functions of the form:

f(x) = a sin[n(xh)] + k, x₁ ≤ xx₂ or f(x) = a cos[n(xh)] + k, x₁ ≤ xx₂. It assumes knowledge of trig functions (or circular functions) and how to use the unit circle.

## Word Problems – 7

This one is a year 12 problem involving calculus.

A company initially provides a service to 1000 customers for $5 per month. The marketing department says that for every 10¢ reduction in price, they could get 100 more customers. What price would give the company the maximum revenue per month and what would that revenue be? Let’s let x be the monthly price for the service. Then the revenue, R(x), would be x times the number of customers. The number of customers is the initial 1000 plus 100 times the number of 10¢ increments below$5 that is charged. The number of 10¢ increments below $5 is (5 – x)/0.1, so the revenue is $R\left(x\right)=x\left[1000+100\frac{\left(5-x\right)}{0.1}\right]=x\left[1000+1000\left(5-x\right)\right]$ $=1000\left(6x-x^2\right)$ Looking at this function, you can recognise that this is an upside down parabola because of the minus sign in front of the x² term. So the maximum would be at the top of the parabola. This makes sense because there is a balancing act going on between a lot of customers and too low a price. The revenue will rise until the price is too low to increase the revenue. To find that point that is the maximum revenue, we need to find the derivative of R(x) and set that equal to 0, that is find the stationary point that is the top of the parabola. $R’\left(x\right)=1000\left(6-2x\right)=0$ $\Longrightarrow\ x=3$ So the price that maximises revenue is$3, and R(3) = \$9000. The number of customers is 1000 + 1000(2) = 3000.