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


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.

\[\Longrightarrow\ x=3\]

So the price that maximises revenue is $3, and R(3) = $9000. The number of customers is 1000 + 1000(2) = 3000.

Word Problems – 6

Now on to a trig (circular functions) problem. This problem requires that you have covered year 11 trigonometry (called circular functions in some texts).

The water depth in a harbour on a particular day is modelled by the following equation:

\[D\left(t\right)=10+3\mathrm{sin}\left(\frac{\pi t}{6}\right),0≤t≤24\]


D(t) = depth of water
t = hours after midnight limited to be between 0 and 24 hours

A ship has to have at least 8.5 meters of water depth to use the harbour. At what times can the ship safely dock at the harbour?

Solving this algebraically, let’s first find the times when D(t) = 8.5.

\[10+3\mathrm{sin}\left(\frac{\mathrm{\pi\ t}}{\mathrm{6}}\right)=8.5\Longrightarrow\mathrm{sin}\left(\frac{\mathrm{\pi t} }{\mathrm{6}}\right)=-0.5\]

So we have the requirement that the sine of “something” has to equal -0.5. When we take a function of something, that something is called the argument of the function. From a table of common sines and knowing that sin(-𝜃) = -sin(𝜃), we get that 𝜋t/6 = -𝜋/6 ⟹ t = -1. We need another angle on the unit circle that has a sine of -0.5:

So 𝜋t/6=7𝜋/6 ⟹ t = 7. The period of this sine function is

\[\mathrm{Period}\ =\frac{2\pi}{\frac{\pi}{6}}=12\]

Now we add this period to our two core values of t, until we get t > 24:

7 + 12 = 19
-1 + 12 = 11
11 + 12 = 23

So the t values we have that are between 0 and 24 are 7, 11, 19, 23. As the sine function initially increase from 0, the value of t = 7, which is 7 AM, is when the depth is decreasing and is below 8.5 meters. The depth is increasing and above 8.5 at t = 11 which is 11 AM. The depth then decreases below 8.5 at t = 19 which is 7 PM and rises above 8.5 at t = 19 which is 11 PM. So the ship can dock between midnight and 7 AM or between 11 AM and 7 PM or between 11 PM and the following midnight.

I’ve plotted D(t) and the line D(t) = 8.5 to show these solutions.

In fact, doing a rough sketch of this function would help get a feel for the answers before we begin.

We do this by raising the standard sine curve up 10 units and draw 2 cycles from 0 to 24 since we can calculate that the period is 12 hours. Since the sine of anything goes from -1 to +1, the minimums are at 10 – 3 = 7 and the maximums are at 10 + 3 = 13. Drawing a line at D(t) = 8.5, we can see that we should get 4 points that will divide the intervals where the depth is below or above 8.5 meters. Once we have the points at D(t) = 8.5, we can easily see the periods where the depth is above 8.5 meters.

Word Problems – 5

Carbon Dating

This one is about carbon dating. I find it fascinating that someone figured out this ingenious method to determine the age of once living things. This person was Willard Libby, a physicist and chemist who proposed this method in 1946.

Before I get to the word problem, let’s go through an explanation on how and why carbon dating works.

Life on earth is carbon-based, which means that its chemistry is based on carbon atoms. Living things ingest, breath in, grow from molecules which mostly include carbon atoms. But carbon atoms come in several isotope forms which means that the number of neutrons in its nucleus varies. Carbon 12 (6 protons and 6 neutrons), is the primary isotope (98.89% of all carbon). Carbon 13 (6 protons and 7 neutrons) is next at 1.11%. Both of these isotopes are stable, that is, they are not radioactive and break down into other elements. A trace amount of earth’s carbon is carbon 14 (6 protons and 8 neutrons). This isotope is radioactive. It is created mainly from the cosmic ray bombardment in earth’s upper atmosphere.

Though the percentage has varied in the past (and this is taken into consideration when very accurate results are required), it is a good approximation to assume that the percentage of carbon 14 has remained constant while life has been on earth. And this percentage remains constant in a living thing while it is alive. However, once it dies, the carbon in it is no longer being refreshed and the carbon 14 in it decays away. When a bone or any other object that was once living is found, measuring the remaining carbon 14 in it can be used to estimate its age.

Exponential Decay

Radioactive materials have a half life. The half life of a radioactive material is the time it takes for half of an original amount to remain. The amount left of a radioactive material is modelled as an exponential decay:

\[A=A_0e^{kt}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\]


A = the amount of material left after t units

A0 = the original amount of material at time 0

e = an irrational number used frequently in science and engineering. e is approximately 2.71828

k = is a negative number (for exponential decay problems) that relates to the half life of a material and the units used for t

t = time in specified units. In our case, the units are years.

The Word Problem

The half life of carbon 14 is 5730 years. A bone found at an excavation site is found to have 30% of the carbon 14 it would have contained when it died. Approximately, how old is the bone?

The Solution

There doesn’t appear to be a lot of information here, but there is enough. Given the exponential decay equation (1), we have a lot of unknowns here: the original amount, the amount now, the parameter k, and of course, the answer to the question, t. But the first sentence is enough to solve for k.

If there is half of the original amount after 5730 years , then from equation (1):

\[0.5A_0=A_0e^{5730k}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(2\right)\]

If you divide both side by A0, then the equation becomes:

\[0.5=e^{5730k}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(3\right)\]

This can be solved using logarithms, with a CAS calculator, or our all knowing internet. Solving this, we get k = −0.00012. Isn’t it interesting that we do not need to know the original amount to get this far?

Now that we know k, we can answer the question. Using the same trick as before, if there is only 30% carbon 14 left, then:

\[0.3A_0={A_0e}^{−0.00012t}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(4\right)\]

The A0 again cancels out and solving this we get t = 9953 years.

That’s old. No bones about it!

Word Problems – 4

Continuing this series on word problems, let’s look at one that many year 11 or 12 students have seen if they have covered calculus. The first part of the question though, does not need calculus:

A 6 by 8 cm rectangular piece of metal has a square cut out of each corner:

The metal is then folded along the dashed lines to form a box of height x.

a) What is the volume of the box in terms of x?

b) What is the maximum volume the box can have and at what value of x does the maximum occur?

Let’s first redraw the rectangle, labelling what we know. If the height of the box is to be x, then that is the size of the cutout square. If an x is subtracted from each end of each side, then the length of each of the dotted lines (the base of the future box) is 6 – 2x and 8 – 2x:

After the metal is folded, we have a box like:

So to answer part a), the volume is height × length × width. So

V = x(8 – 2x)(6 – 2x)

For part b), we need a little calculus. But before we do that, just to get a mental image of what is going on (this is not needed to solve the problem though), let’s plot V as a function of x:

Notice that the volume is 0 at x = 0 and x = 3. This makes physical sense because when x = 0, there is no square cut out and the box would just be a flat sheet. If x = 3, there would be nothing left of the original 6 cm side and the two remaining flaps would just fold up against each other giving no volume. So the physical restriction on x here is 0 ≤ x ≤ 3.

The maximum volume then occurs at the peak of the curve between 0 and 3. To find what and where this maximum is, requires calculus, specifically finding a local maximum (a stationary point).

To find this point, let’s first expand the expression for V:

V = x(8 – 2x)(6 – 2x) = 4x3 – 28x2 + 48x

For the uninitiated, derivatives of a function give the gradients of lines tangent (that is just touching at one point) to it. At the maximum, the tangent line is horizontal which has a gradient of 0. So to find this, we find the derivative of our volume function and set that equal to zero and solve for x.

For those who have had calculus and know how to find derivatives of a polynomial function (like we have here):

V′ = 12x2 – 56x + 48 = 4(3x2 – 14x + 12)

So now we find the values of x where the derivative is 0:

V′ = 4(3x2 – 14x + 12) = 0 ⟹ x = 1.1315, 3.5352

To get that answer, the quadratic formula can be used. Or, if lazy like me, an equation solver on the internet.

We want the value between 0 and 3. So a square of sides 1.1315 cm will maximise the volume. The other value of x is where the minimum point is as seen in the plot. We find the maximum volume by putting x = 1.1315 in the original V function:

V(1.1315) = 4(1.1315)3 – 28(1.1315)2 + 48(1.1315) = 24.258 cm3

As before, drawing pictures gets you started.

Word Problems – 3

This one is a little different. Recently, I have had a few students struggling with literal equations. These are equations with many letters or symbols. For science and engineering wannabes, you need to develop the skill to work with these. Below is an example from orbital dynamics, but first some background.

Two Body Problem

Accurately determining orbits in the real world, requires computers. However, a good approximation that makes orbital calculations possible by hand (OK, using calculators), is to assume that the only two bodies that exist in the universe are the bodies orbiting each other. The shape of an orbit of a body in orbit around another in this universe can have 1 of 4 shapes: circular, elliptical, parabolic, or hyperbolic. I won’t talk about the last three, but let’s consider the circular orbit.

Here is a picture of an circular orbit:

In this kind of orbit, the earth is at the centre and the satellite follows an circular path around the earth.

In an orbit, such as a satellite orbiting the earth, we want to know a lot of things about the orbit, but two primary things are the satellite’s distance from the earth and its position. To measure its position, an arbitrary axis is agreed upon and the satellite’s position is its angle 𝜃 measured from this axis. In orbital dynamics, the angle 𝜃 is called the true anomaly. I don’t know why it is called that, but a definition of anomaly is “a deviation from the normal”. If we consider the normal being on the reference axis, this term makes a bit more sense.

Now I chose a circular orbit because the equations describing it are simpler than for other shapes. For example, the time it takes for a body to complete one orbit, either circular or elliptical is

\[T=\frac{2\pi}{\mu^2}\left(\frac{h}{\sqrt{1-e^2}}\right)^3\ \ \ \ \ \ \ \ \ (1)\]


T = the period, the time it takes to complete one orbit

μ = gravitational parameter. It is a combination of the gravitation constant (a constant in the universe) and the masses of the bodies

h = angular momentum per unit mass. Since the body has mass and is rotating around the other body, it has angular momentum

e = the eccentricity of the orbit. A measure of how elliptical the orbit is.

The eccentricity of a circular orbit is 0 and this simplifies equation (1).

So now the word problem.

The Word Problem

Equation (1) applies to elliptical and circular orbits. For a circular orbit, e = 0. The radius of a circular orbit is r. There is a relationship between the angular momentum and the radius of a circular orbit:

\[r=\frac{h^2}{\mu}\ \ \ \ \ \ \ \ \left(2\right)\]

If a satellite is at 𝜃 = 0 at t = 0, the time it takes to travel 𝜃 radians is

\[t=\frac{\theta}{2\pi}T\ \ \ \ \ \ \ \ \left(3\right)\]

So using equations (1), (2), and (3), develop an expression for t in terms of r and 𝜃 and an expression for 𝜃 in terms of r and t. Remember that 𝜋 and μ are constants so they can be in these expressions.

The Solution

First, we can simplify equation (1) by substituting e = 0:

\[T=\frac{2\pi}{\mu^2}\left(\frac{h}{\sqrt{1-e^2}}\right)^3\Longrightarrow\frac{2\pi}{\mu^2}h^3\ \ \ \ \ \ \ \ \ \ \left(4\right)\]

We can rearrange equation (2) to get h in terms of r:

\[r=\frac{h^2}{\mu}\Longrightarrow h=\sqrt{r\mu}\ \ \ \ \ \ \ \ \ \ \ \ (5)\]

Now substitute h from equation (5) into equation (4):

\[T=\frac{2\pi}{\mu^2}h^3=\frac{2\pi}{\mu^2}\left(({r\mu)}^\frac{1}{2}\right)^3=\frac{2\pi r^\frac{3}{2}\mu^\frac{3}{2}}{\mu^2}=\frac{2{\pi r}^\frac{3}{2}}{\sqrt\mu}\ \ \ \ \ \ \ (6)\]

To follow the development in (6), you need to remember the exponent rules and how to convert between a fractional exponent and square root notation.

Now substitute this into equation (3):

\[t=\frac{\theta}{2\pi}T=\frac{\theta}{2\pi}\frac{2{\pi r}^\frac{3}{2}}{\sqrt\mu}=\frac{{\theta r}^\frac{3}{2}}{\sqrt\mu}\ \ \ \ \ \ \ \ \ \ \ (7)\]

which is one of the answers. Rearranging this to solve for 𝜃 gives the second answer:

\[t=\frac{{\theta r}^\frac{3}{2}}{\sqrt\mu}\Longrightarrow\theta=\frac{t\sqrt\mu}{r^\frac{3}{2}}\ \ \ \ \ \ \ \ \ \ \ (8)\]

Note that only basic algebra was used to find the answers. It doesn’t matter if you have equations with numbers or a lot of letters. The steps to find a solution are the same.

Word Problems – 2

I am writing posts of word problems to give students practice and underlying methods to solve these. My first post in this series provides some techniques to use to help you get to an answer. The most important thing is to not panic. Here is an example of a problem that could cause you to panic, but as you will see, calm logical thinking will get you to the answer.

A suspension bridge has been constructed over a river. The suspension cable is parabolic in nature. The distance between the two towers holding the cable is 180 metres and the minimum height of the cable above the road is 15 metres. At a point 40 metres from the vertex of the cable the height above the road is 30 metres.
How high above the road are the cables attached up the tower?

First, let me repeat from my last post, the things you should do to get started:

  1. Read through the entire problem.
  2. Draw a picture(s) if the problem lends itself to this, and label things with known values and use letters to label other things that you feel are useful to know.
  3. Read through the problem again and as you read, write equations that express mathematically what the words are saying. Use letters for unknown things. I find it helps to use letters that easily refer to the unknown thing. For example, use “C” to be an unknown price of a child’s ticket and “A” for the adult ticket. You can also use subscripts to differentiate unknowns. For example, use “tA” for the time to travel on path A and “tB” for the time to travel on path B.
  4. Identify the thing(s) that the problem is asking for, and look over the pictures and equations you have, to come up with a way to solve for the requested unknown(s).

Hopefully, you have already done the first step. Now let’s do step 2:

This picture has all the information provided in the problem.

Looking at step 3, I do not see any equations we can write, but there is a clue in the problem. The shape of the suspension cable is a parabola and this is the shape of a quadratic equation. But before we can write down equations, we need a coordinate system.

Welcome to your first engineering lesson! When engineers are presented with a problem, they frequently must come up with a coordinate system. But they are free to place the system anywhere they want to. You can choose one that makes the equations difficult or choose one that makes the equations as simple as possible. So where do we choose ours? There is no right answer but some answers are better than others.

Here are some possibilities:

I think we can agree that the random coordinate system 1 would be a poor choice. But 2, 3, and 4 look good because the unknown value would just be the y coordinate over an appropriate value of x. Any of these three systems can be used to solve the problem, but if you review the turning point form of a quadratic equation, you should see that system 3 would generate the simplest equation of the parabola.

I will now redraw the figure using coordinate system 3 and interpreting the distances given in the problem as coordinates:

Figure 1

You may notice the two points (-40,30) and (40, 30). As a parabola is symmetric, the statement “At a point 40 metres from the vertex of the cable the height above the road is 30 metres” actually defines two points.

Now the turning point form of a quadratic equation is

y = a(xh)2+k

where (h,k) are the coordinates of the vertex. So from figure 1, we can immediately write the equation

y = ax2+15

Notice how simple this equation is because of our choice of coordinate system. We can find the value of a by substituting the point (40,30) into this equation:

30 = a(40)2+15 ⟹ a = 3/320

So the equation of the suspension cable is

y = 3x2/320 + 15

Now we can find the height of the cable support by letting x = 90:

y = 3(90)2/320 + 15 = 90.9375 m

So you see that this was not too difficult of a problem.

Word Problems – 1

A constant complaint from my students is that they struggle with word problems. So I thought I would produce several posts showing examples of how to approach and solve these problems.

Now every word problem is unique and different approaches can apply (hence, why these problems are scary). But there are techniques to help you through this. I have found that, when confronted with a word problem, it helps to do the following:

  1. Read through the entire problem.
  2. Draw a picture(s) if the problem lends itself to this, and label things with known values and use letters to label other things that you feel are useful to know.
  3. Read through the problem again and as you read, write equations that express mathematically what the words are saying. Use letters for unknown things. I find it helps to use letters that easily refer to the unknown thing. For example, use “C” to be an unknown price of a child’s ticket and “A” for the adult ticket. You can also use subscripts to differentiate unknowns. For example, use “tA” for the time to travel on path A and “tB” for the time to travel on path B.
  4. Identify the thing(s) that the problem is asking for, and look over the pictures and equations you have, to come up with a way to solve for the requested unknown(s).


The digits of a 3-digit number add up to 13. The sum of the first two digits is 6. The number formed when the digits are reversed and 1 is subtracted is 3 times the original number. What is the original number?

I will get to this in a minute, but as you progress through maths, you will find that, more and more, you need knowledge from your past. This is an example of that. You may be tempted to use xyz to represent the unknown number, but that would be incorrect. In maths, xyz means x×y×z. If x = 1, y = 2, and z = 3, 123 does not equal 1×2×3 = 6.

From primary school (grade school in the USA), you learned that 123 means 1×100 + 2×10 + 3×1 because 1 is in the hundreds place, 2 is in the tens place, and 3 is in the units place. So let’s get to the problem.

So reading through the problem, you see that we need to find a 3-digit number. This problem does not lend itself to drawing pictures, but we can generate some equations. From the first sentence, we can write:

x + y + z = 13     (1)

From the second sentence, we can write:

x + y = 6     (2)

Now you may notice at this point, that we can use equation (2) in equation (1) as equation (1) has x + y in it and equation (2) says that we can replace this with 6:

x + y + z = 6 + z = 13 ⟹ z = 7

Wow! We haven’t finished going through the problem yet, but we have 1/3 of the problem solved. Now what about the third sentence? The numerical value of the number when the digits are reversed is:

100z + 10y + x

The third sentence says that if we subtract 1 from this, we get 3 times the value of the original number. A key word to look for in word problems is “is”. This means “=” in maths:

100z + 10y + x – 1 = 3(100x + 10y + z)     (3)

Since we now know that z = 7, this can be simplified to:

700 + 10y + x – 1 = 3(100x + 10y + 7)

⟹ 10y + x + 699 = 300x + 30y +21     (4)

Now we can get all the variables on the left side and the numbers on the right side, combine like terms to get:

-299x-20y = -678 ⟹ 299x+20y = 678     (5)

Equations (2) and (5) are a set of simultaneous equations in the variables x and y. You can manually use the substitution or elimination methods or use your CAS calculator to solve for x and y. I will assume that you can do these things to get x = 2 and y = 4. So the number that solves the problem is 247.

What is important here is to follow the process. Each word problem is unique but with practice, you will learn to look for key word and phrases that will help you to convert the word problem to a standard textbook problem. I will present more examples in future posts.

Learning to Count Again

Let’s define an event A, flipping heads on the flip of a coin for example. What is the probability of that event occurring? I will use the common notation of P(A) to represent the probability of an event A. The basic way to calculate a probability is to divide the number of ways A can occur by the number of ways anything can occur. That is,

\[\text{P(A)} =\frac{\text{Number of ways A can occur}}{\text{Number of ways anything can occur}}\]

In the case of getting a heads from a flip of a coin, there is only one way to get a heads and there are two possibilities. So the probability is 1/2.

If we flip two coins and we now let our event A be two heads, then the counting of the ways anything can happen takes just a little more thought. The possibilities are: HH, HT, TH, and TT. So there is only one way to get two heads but there are four possibilities. So the probability is 1/4. If the event is “at least one head”, there are three ways that can happen out of the total of four possibilities, so the probability is 3/4.

We can increase the complexity of our experiment and flip three coins or ask questions about choosing certain cards in a standard deck. The counting for the numerator and the dominator gets harder but it is still possible with a little more thought. But what if I asked “what is the probability of getting a flush (all cards of the same suit) in 5 cards randomly selected from a deck of cards?” or “A four-digit number (with no repetitions) is to be formed from the set of digits {1, 2, 3, 4, 5, 6}. Find the probability that the number is even.” Now the counting gets much harder. But fortunately, there are ways to handle this

Selection or Arrangement

Let’s say we have 8 people standing around just waiting for a math problem to show up. Someone wants to form a team of 5 people from these 8 and another person wants to arrange 5 of these people around a desk. How many different teams can be made and how many different settings around the desk can be done?

Notice that in the team question, it doesn’t matter what order you pick the players but in the table question, order does matter. Just using letters to represent the people, ABCDE and EDCBA are the same for the team question so would only be counted once but these are two separate arrangements in the table question. The table question is an arrangement question whereas the team question is a selection question. In textbooks, arrangements are often called permutations and selections are called combinations. It doesn’t matter what they are called, you need to know which type you have in order to count the number of possibilities correctly.


Let’s look at the table question first. You have 8 ways to choose the first person, but when you do, there are only 7 people left to choose as the second person. Then you only have 6 left for the third position, 5 left for the fourth and 4 left for the fifth. So the total number of arrangements is 8×7×6×5×4 = 6720. Multiplying numbers that sequentially decrease by one is a common thing when doing these kind of problems so, as is so common in maths, a shorthand notation was created. 8×7×6×5×4×3×2×1 is represented as 8! and is called “8 factorial”. But for our arrangement problem, we are missing the 3×2×1 part. Notice that 3×2×1 = 3!. So another way to show the solution is 8!/3!. Also notice that 8 – 5 = 3.

We can generalise this. If you have n things and want to arrange them r at a time, then the number of arrangements is n!/(nr)!. Once again, even this is too much to write for our lazy mathematicians, so this is given a shortcut notation where the P stands for permutation:

^{n} P_{r} =\frac{n!}{( n-r) !}

In a CAS calculator, the function “nPr” is used to calculate permutations, so for our problem nPr(8,5) = 6720.

By the way, what if we wanted to arrange all 8 people? Then using the permutation formula, we get 8!/(8-8)! = 8!/0!. This looks illegal but mathematicians foresaw this and defined 0! as being equal to 1. Must be nice to be able to make your own rules.


Now let’s look at the team question. Using letters again, we are now in the scenario where ABCDE and EDCBA are the same team and this should only be counted once. So we would expect that the number of selections using the same n and r would be smaller than the number of arrangements. In fact this is true. So for our team selection, if we use our arrangement formula, each team has 5! different arrangements and the calculated number is 5! too large. So if we divide our arrangement total of 6720 by 5! = 120, we would get the correct number for the number of teams possible, 6720/120 = 56. To generalise, if you have n things and want to select them r at a time without regard to order, then the number of arrangements is n!/r!(nr)!. Again, there is shorthand for this where the C stands for combination:

\[^{n} C_{r} =\frac{^{n} P_{r}}{r!} =\frac{n!}{r!( n-r) !}\]

In a CAS calculator, the function “nCr” is used to calculate combinations, so for our problem nCr(8,5) = 56.


So at the beginning of this post, I posed a hypothetical question about the probability of being dealt a flush hand of 5 cards out of a deck of 52 cards. As card order does not matter, this is a selection (combination) problem. We need to find the number of ways to get 5 card flush hands and the total number of possible 5 card hands. I assume you are familiar with the standard deck of cards that consists of 4 suits of 13 cards each.

First let’s find the total number of 5 card hands there are. Here n = 52 and r = 5:

\[^{52} C_{5} =\frac{52!}{5!( 52-5) !} =2,598,960\]

his will be our denominator to use in the probability formula. For the numerator, we need to find how many ways you can get a 5 card flush. You could get 5 clubs, 5 hearts, 5 diamonds, or 5 spades. Each one of these is a selection of 13 cards, 5 at a time or:

\[^{13} C_{5} =\frac{13!}{5!( 13-5) !} =1,287\]

We need to multiply this by 4 since there are 4 suits, 1287×4 = 5148. So the probability of being dealt a flush in 5 card poker is P(Flush) = 5148/2598960 = 0.00198 = 0.198% or about once in 505 hands. I wouldn’t bet on it.

The Circumference of the Earth

I often find it amazing that people long ago, made many discoveries that we would find difficult without today’s equipment and computing power. If asked “how long ago was the size of Earth accurately determined?”, what would you guess? Somewhere between 1500 CE and today? What about between 500 CE and today? It turns out that calculation was made almost 2300 years ago, around 245 BCE.


Eratosthenes was a Greek mathematician, astronomer, philosopher, poet, and music theorist. He was born in 276 BCE and died in 194 BCE. He noticed that in Syene (now Aswan Egypt), that a well with vertical sides did not cast any shadow midday at the summer solstice (when the sun is at its highest point in the sky as the earth revolves about the sun). However, in Alexandria, a vertical pole stuck into the ground, cast a shadow at the same time of the day and year. This happens because the earth is spherical and the direction of the sun at different latitudes is different.

So Eratosthenes measured the length of the pole above the ground in Alexandria and the length of the shadow. How can we use these measurements?

The Calculations

First let’s measure the angle the shadow in Alexandria makes with the stick. To get the same answer as Eratosthenes, I will assume the height of the stick above the ground in Alexandria was 1 meter and the length of the shadow was 12.63 centimeters:

From trigonometry, the tangent of the angle 𝜃, is the opposite side divided by the adjacent side. So if we use the inverse tangent on 0.1263m/1m (the angle whose tangent is 0.1263), we get 𝜃 = 7.2°. What does this angle have anything to do with the earth? Quite a lot actually.

Now let’s draw the earth underneath the stick:

As the sun is very far away, we can approximate the rays from the sun as being parallel at any spot on earth, so the line between the center of the earth and the well in Syrene is parallel to the edge of the stick’s shadow in Alexandria. This means that the angle between the stick and the well at the earth’s center is the same, 7.2°.

Now there are 360° to span the entire circumference of the earth, so the curved distance between Alexandria and Syrene divided by the earth’s circumference will be the same as 7.2°/360° = 0.02. But what is that curved distance.

Evidently, there were people called “bematists” who were trained to measure large distances by counting their steps. Well Eratosthenes employed them to measure the distance between Alexandria and Syrene. Looking at the map, the Nile River gets in the way of this at several places, but I’ll assume they found a way to handle this.

Converting their units to meters, they measured the distance to be 800 kilometers. So 800/(earth’s circumference) = 0.02, which means that earth’s circumference = 40,000 km.

This is a very accurate estimate of the earth’s circumference. As the earth is not a perfect sphere, the circumference measured around the poles is 40,008 km using satellite data and all sorts of equipment and calculation power not available to Eratosthenes. The circumference around the equator is 40,075 km.

Using the formula for the circumference of a circle, we can also get an estimate of the earth’s diameter: C = 𝜋d ⟹ d = 40000/𝜋 = 12,732 km compared to the technology derived value of 12,714 km.

I am impressed!

An Equation Transformation Example

One of the many skills maths students develop in high school is the ability to change the position of a graph. I sometimes need to remind my students that this is not just “busy” work. It is a skill used frequently in technical fields such as science and engineering. What follows is an example from celestial mechanics.

The Two-Body Problem

Calculating orbits and their characteristics is usually a computer-intensive exercise. For example, how far from the earth is an orbiting satellite at a particular time. However, a good first approximation of this is to pretend we are in the unrealistic universe where only two point masses exist. In this universe, the shape of orbits can be perfectly modelled with equations called conic sections. Why they are called conic sections is a bit beyond the scope of this post, but please free free to look that up.

The orbital shape I want to describe here is the ellipse. The equation of an ellipse which is centered at the origin is:\[\frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} =1\] where 2a is the length of the ellipse in the x direction and 2b is the length of the ellipse in the y direction. If a > b, the line along the a intercepts is called the major axis of the ellipse and the line along the b intercepts is called the minor axis:

\[\frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} =1\]

This is a possible shape of an orbit about one of the masses in our perfect two-mass universe. In reality, orbits of say satellites around the earth, approximate this shape but other factors like other objects in the solar system and the earth not being a point mass with an uneven distribution of its mass, make the orbital shape slightly different than an ellipse. Not only that, in our perfect universe, the orbit path is exactly contained in a plane and can be completely drawn on flat paper. In reality, orbits also go below and above the average plane of the orbit.

Now one of the masses, say m₂, is on the ellipse. The other mass, m₁, is not at the center of the ellipse, the origin in the above figure. So where is it?

An ellipse has two special points associated with it call focal points. These points have the property that any point on the ellipse has a total distance between it and the focal points is a constant. This constant is 2a if the longest dimension of the ellipse is along the x-axis. Otherwise, this constant length is 2b.

Total length of the red line = 2a regardless of where (x,y) is.

The other mass, m₁, is at one of these focal points. The coordinates of these points can be found using the Pythagorus theorem. Looking at the blue triangle above, you can see that \[c=\sqrt{a^{2} -b^{2}}\]

An Example

Suppose we want to analyze the relationship between two masses that are in orbit. In orbital mechanics, the shape of an orbit is given by its eccentricity, e. An elliptical orbit has an eccentricity between 0 and 1. By the way, a perfectly circular orbit has an eccentricity of 0. So let’s say we have an orbit with the major axis length 2a = 10 and minor axis length 2b = 6. This is an orbit with an eccentricity of 0.8. If we want to know the position of m₂ with respect to m₁, we need a coordinate system. A convenient one would be a Cartesian system aligning the major axis of the orbit along the x-axis and centering the ellipse at the origin. This way we can immediately write down the equation of the orbit:\[\frac{x^{2}}{25} +\frac{y^{2}}{9} =1\]

as a = 5 and b = 3. This means that the focal points are \[c=\sqrt{25 -9} =4\]from the center:

An example orbit

Two of the more basic pieces of information we would like to know (especially if this is a satellite orbiting the earth) is what is the distance r between m₁ and m₂ and what is the direction of m₂ with respect to m₁. We can define the direction as the angle 𝜃 of the line connecting the two masses with the positive x -axis. This is actually the reference used in much of celestial mechanics. This angle is called the true anomaly.

Well this is nice, but since we want to find relationships of m₂ with respect to m₁, it would be even more convenient to put the origin of our coordinate system at m₁. How do we do that?

Transformations to the Rescue

Somewhere in your year 10 or 11 maths (grades 10 or 11 in the States), you took the equation of a standard parabola, y = x², and replaced the x with xh to get y = (xh)². This moved the parabola h units to the left or right depending on the sign of h.

Effect of replacing x with (xh

It turns out that in any relationship between x and y, replacing the x with xh has the exact same effect on its graph. So in our example, if we move the ellipse 4 units to the left, m₁ would be on the origin of our coordinate system.

\[\frac{(x+4)^{2}}{25} +\frac{y^{2}}{9} =1\]

We can now calculate r and 𝜃 a bit more easily than before the transformation. Let’s look at where m₂ is in the above figure. Here, m₂ is at (-4,3). The following process would work with any point, but at this point, the numbers are “nicer”.

The figure below shows the answers. The distance r can be found using the Pythagorus theorem but once you see that the two sides are 3 and 4, then this is the standard 3-4-5 right triangle:

The solution

In this perfect universe, other orbital shapes are possible: circles, parabolas, and a hyperbolic. These have their own standard equations but they all can be transformed to move them to any place you wish on the coordinate system to make your calculations easier.