## Saturday, June 1, 2013

### Making Sense of Variables

Some of us have no trouble with writing expressions or equations for word problems.  But we are in the minority.  Earlier today I was reading from a rather old (1980s) study on this very difficulty.  One might have hoped that in the 30 plus years since then, a pedagogical solution to the problem would have been found, but any algebra teacher can tell you that making sense of word problems remains traumatic for many children.
The following question (taken from Translation difficulties in Learning Mathematics, American Mathematical Monthly, v88 n4 p286-90 Apr 1981) was given to 47 (non-science) algebra students and 150 calculus students.
Write an equation for the statement, ‘There are six times as many students as professors at this university.’ Use S for the number of students and P for the number of professors.
I was not surprised that more than half (57%) of the algebra students got this question wrong. But a large minority (37%) of the calculus students also had an incorrect answer.   According to the authors, the most common mistake was 6S = P.  When asked to explain their thinking, some students drew a picture like this:

The mathematics teacher might be frustrated that the requirement (or warning? or advice?) “Use S for the number of students…” was not followed.   Otherwise, the drawing should look like this:

The number of students = 6.
The number of professors   = 1
Obviously, 6 1.  However,

Replace “The number of students” with S and “The number of professors” with P to obtain:
Make ‘S’ the subject of the equation by multiplying both sides by P.
Finally, obtain, S = 6P.
Is this too much work for something that’s obvious? Perhaps the joke is on us:
The professor is giving a lecture and has made an assertion as part of his presentation. A student, not understanding the basis for the assertion asks why it is true. The professor responds that "It is obvious." Then the professor steps back, stares at the board and ponders for several minutes. Then he turns and walks out of the lecture hall. He is absent for a fairly long time  … Finally, just before the class is scheduled to end the professor reappears, and announces "Yes, it is obvious."

## Saturday, April 20, 2013

### Visualisation versus Algebra?

Here is a Primary 5 (Grade 5) word problem from Yan Kow Cheong in Singapore:
There are 100 chickens and rabbits altogether. The chickens have 80 more legs than the rabbits. How many chickens and how many rabbits are there? (http://www.singaporemathplus.net/1/post/2013/03/the-chickens-and-rabbits-problem.html ).    In ‘the West’ this sort of problem would not be attempted until the students have been taught algebra (i.e., y = mx + c, etc) so it would probably be a Grade 9 problem!   In a comment on another problem involving chickens and ducks, Kow-Cheong states that “the model heuristic is nothing but algebra in disguise. Instead of using the variable x, we use a unit, part or line to represent some unknown quantity ..[which] … allows us to use visualization to solve higher-order word problems.

The same author asks if we can use the bar method, or the Sakamoto method, to solve the following problem:  Mr. Yan has almost twice as many chickens as cows. The total number of legs and heads is 184. How many cows are there?

I must admit that my preferred technique for this sort of problem is algebraic.
Let the number of cows be k.
Let the number of chickens be h.
Then h = 2k – x, where x is a ‘slack’ variable such that 0 < x << k.
(4k + 2h) + (k + h) = 184
Substitute for h.
(4k + 2(2k – x)) + (k + (2k – x)) = 184
4k + 4k – 2x + k + 2k – 2x = 184
11k – 3x = 184
11k = 184 + 3x.
Apparently, k < 20 and x is much less than k, so x might be 1, 2, 3 or 4.
Assuming x = 1
11k = 184 + 3 = 187
k = 187/11 = 17
Then h = 2k – x = 2(17) – 1 = 33.
Note that if x = 2, 3 or 4, k cannot be an integer but we obviously expect there to be a whole number of cows so the problem is solved.  Mr Yan has 33 chickens and 17 cows.  I do look forward to seeing the Sakamoto solution.

## Wednesday, February 27, 2013

### An Introduction to the Vector Equation of a Line

Given two different points we can always draw a line segment between them.  We can also sketch a line given just one point and a direction.  How?  The instructions are ‘hiding in plain view’ in the equation y = mx + c. Take, for example y = ½ x – 3.
Please do not make a table of values for x and y.

Tables are slow and inefficient.  Remember, we only need one point!  Which point? The y-intercept.  For the equation y = ½ x – 3, start at the point (0, c) = (0, -3).   Now what?  From the gradient, m = ½,  we find that the rise is 1 and the run is 2.
If the gradient is written as a fraction (½, 23/7, -4/3, …) the numerator is the ‘rise’ and the denominator is the run.  If, however, the gradient is written as a decimal number (0.5, 1.6, - 3.9, …) the given number is the ‘rise’ and the run is always ‘1’. Positive runs go to the right; negative runs to the left.  Positive rises go up; negative rises go down.

With preliminaries out of the way, here is the line: Start at the y-intercept, sketch the run (running parallel to the x-axis) and then sketch the rise (rising parallel to the y-axis).  Label the end point, P.   Take a ruler and connect the y-intercept and point P.

You can explore with the Geogebra file “mx_plus_c” to vary intercept c and gradient m using the respectively labeled sliders.  Turn on the trace function for point P by right clicking on the point and selecting “Trace On”.  Move the c and m sliders to choose your desired y intercept and gradient. After you have chosen these parameters, go the pull-down menu, View and click “Refresh Views”. Finally, go to the xP slider and drag that button.  What happens?

The rise and run are continually refreshed, so the screen only shows the latest values, but we see all the previous values of point P.  The locus of point P is the required line. Ta Dah!
Conclusion:
For many of you this essay merely re-stated the obvious.  However, the next installment will use very similar ideas to explain the vector equation of a line:
p = q + kv .

Appendix:  Geogebra Construction
 No. Name current value 1 Number, m a slider variable m = 1.7 2 Number, c a slider variable c = -1 3 Point, O the origin O = (0, 0) 4 Point, I (0, c) I = (0, -1) 5 Vector, vectorc Vector[point, point] = Vector[O, I] vectorc = (0, -1) 6 Number, xp a slider variable xp = 2.8 7 Vector run Vector[point, point] = Vector[(0, c), (xp, c)] run = (2.8, 0) 8 Vector rise Vector[point, point] =Vector[(xp, c), (xp, c + xp times m)] rise = (0, 4.76) 9 Point P vectorc  + run + rise P = (2.8, 3.76)
Created with

A copy of the applet is stored on http://www.geogebratube.org/

## Saturday, February 23, 2013

### The Pentagram

The pentagram is an ancient symbol (http://mathworld.wolfram.com/Pentagram.html) , recently popularized by Dan Brown (http://www.danbrown.com/the-davinci-code/).   Mathematicians have analysed its properties since the time of Pythagoras.  (www.math.tamu.edu/~dallen/history/pythag/pythag.html)
One such question is what is the area of the blue star below?

Johnny decides to find the area of the large pentagon (ABCDE) and to subtract the area of the five (congruent isosceles) triangles, each coloured light brown above. ABF is one such triangle.
The exterior angle of a regular pentagon is 360°/5 = 72°. Hence each interior angle is 180° - 72° = 108°.  Angle EAB is one such angle so <EAB = 108°.  Call the centre of the pentagon K. Segment KA bisects angle EAB. Hence Angle KAB = ½ ×108° = 54°.

The area of the pentagon is 5 times the area of the green triangle.  The height  of the required triangle is 5 tan 54° = 6.88190…. Hence the area of the triangle is ½ (10)( 6.88190….) = 34.40954. The area of the entire pentagon is 5×34.40954. = 172.047.
Each base angle of the highlighted isosceles triangle is ½ (180° -108°) = 36°.Hence, the height of the highlighted triangle is 5 tan 36° = 3.632712. Consequently  the area of the triangle is ½ (10)( 3.632712….) = 18.163563.  The area of all five triangles  is 5×34.40954. = 90.81781.

The area of the star is the area of the area of pentagon (ABCDE) minus the area of the five triangles; 172.047 - 90.817 = 81.2 units2 to 3 sig figs.
Mary takes a different approach.  She decides the area of the star is five times the area of the dart KAFB (shaded grey below).  A dart is simply an unusual kite.

The area of a kite is half the product of the diagonals = ½ (AB)(KF). Note that diagonal KF is inside the polygon and diagonal AB is outside the polygon; nevertheless the formula still holds. Diagonal AB is given (10 units). Hence, Mary only needs the length of segment KF =  5 tan 54°-  5 tan 36° = 5(tan 54°-   tan 36°) = 6.88190 - 3.632712 = 3.249188.
Mary calculates the area of the star = 5×½ (10)( 3.249188)= 25×3.249188 = 81.2 units2 to 3 sig figs.
Conclusion: Johnny and Mary both used the five-fold symmetry of the shape, both used the trigonometric ratio ‘tangent θ’, and both used the same angles (54° and 36°).  Hence their algebra is essentially the same. They only differed in the physical meaning they gave to the algebra. Johnny calculated the area of two triangles (ABK and ABF) whereas Mary calculated the area of the dart KAFB.
Can you find another way to find the area of the star? Is it possible to find the area but avoid using trigonometry?  What about the mysterious number phi (http://en.wikipedia.org/wiki/Golden_ratio ).

## Wednesday, February 13, 2013

### Trigonometry Quiz

Give answers which are lengths correct to three significant figures.  Give answers in degrees correct to one decimal place.

1. In the right-angled triangle MNO,   MO = 4.5 metres and angle MON = 32o.  Calculate the length of MN.

2.            GHI is a right triangle.  GH  = 8 cm and HI = 7 cm.   Angle H = 90°. Calculate the size of angle HIG.

3.   UVWX is a dart.

a) Calculate the length of side UV.          nnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
b) Calculate the size of angle UXW.

4. Square-based pyramid PQRST is shown in the diagram below.  All the edges are 3 cm in length.

a) Calculate the length of PR.
b) Calculate the vertical height, MT.
c) Calculate the angle between PT and the base PQRS.

1. In the right-angled triangle MNO, MO = 4.5 metres and angle MON = 32 o.  Calculate the length of MN.

Trigonometry usually begins with a sketch of a right angled triangle and definitions of the hypotenuse and the opposite and adjacent ‘legs’ for a given angle. Next up, of course, are the sine, cosine and tangent functions.

Consequently we begin by identifying the side opposite the 32 o angle, the side adjacent to the 32 o angle and the hypotenuse.   We write down the mnemonic ‘SOH CAH TOA’ to help us recall that which function is which.
As we are given the hypotenuse and are asked for the opposite side, our best choice is the sine function.  (If you choose the cosine function you will calculate the length of segment ON.  You can still find MN by following on with Pythagoras (a2 + b2 = c2) but that option is more time-consuming.   If you go with ‘tan θ’ however, you are at a dead end.
sine 32 o = MN/MO  = MN/4.5
MN = 4.5* sine 32 o = 2.384…
To get an answer with three significant figures, we ought to jot down four figures. The digit on the right (alternatively, the digit for the thousandths place) will be the ‘decider’. As the rule for ‘4’ is to round down, we write MN = 2. 38 metres (to 3 significant figures).
If the teacher wanted to make the question a little bit harder s/he could change the orientation of the triangle.  Or the student could be given the opposite side and required to find the hypotenuse.
2.           GHI is a right triangle;  GH  = 8 cm and HI = 7 cm;  angle H = 90° Calculate the size of angle HIG.
Again this question follows the standard development of trigonometry.  After learning to add, we learn to subtract; after learning to multiply, we learn to divide.  After learning the standard ‘trig’ functions, we learn the inverse functions: arc-sine, arc-cosine and arc-tangent.

Your calculator and your text probably have the notation sin– 1,   cos– 1   and tan – 1. The latter notation emphasizes that these are inverse functions, whereas the somewhat old-fashioned ‘arc’ notation emphasizes that the output is an angle.
tan (θ) = 8/7
tan – 1(tan (θ)) = tan – 1 (8/7)
θ = 48.81°
After giving the required rounding, angle HIG = 48.8°  (to one decimal place).
3. UVWX is a dart.
a) Calculate the length of side UV.     nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
This question is best solved with the cosine rule, c2 = a2 + b2 – 2ab cos C.  The cosine rule is an extension of Pythagoras.  The last term disappears when C = 90°, since cos 90° = 0.   The cosine rule is (in my opinion) most easily proved with vectors, so for today we will simply take it as it is.  To apply the formula we will re-label triangle UVW as follows:

c2 = a2 + b2 – 2ab cos C
c2 = 3.92 + 5.72 – 2*3.9*5.7 cos 124°
c2 = 3.92 + 5.72 + 24.86… (Note that cos 124° is negative so the last term becomes positive.)
c2 = 72.5617…
Some students, carelessly, write c 2 to 3 figures.  The table below shows that their final answer will be marked wrong and a very ‘easy’ mark has been lost as a result.
 If c 2 is written down as then ‘c’ is calculated as and ‘c’ written down to 3 significant figures as 72.56217… 8.518343 8.52 72.561… 8.518274 8.52 72.5… 8.514693 8.51

b) Calculate the size of angle UXW.
A syllabus certainly has more topics than a quiz (or a test or an exam) can cover.  The examiner is unlikely to ‘waste’ a precious resource by asking the same question twice.  (Of course if your examiner has a question bank from which a computer program samples randomly, all bets are off.  I once had a 25 question multiple choice paper for which ‘oxytocin’ was the correct answer, thrice!)
Yes, you could find the angle by doing another cosine rule calculation.  But the givens are two sides and one angle so it is much easier to use the sine rule, instead.  The formula is shorter, and there is no need to take a square root at the end.
sin x/5.7 = sin 108°/8.2
sin x = 0.661100…
x =   41.4°  (to one decimal place).
Once again, it is advisable to keep many more figures than required, and round off as your final step. The sine rule is proved quite easily from the definition of sine and the formula for the area of a triangle, but that is a story for another day.

4.                    Square-based pyramid PQRST is shown in the diagram below.  All the edges are 3 cm in length.

This is a so-called 3-D problem, but in reality three points (not all on a straight line) define a triangle and a planar (or 2-D) surface.  The skill is in finding the correct triangle.  With 6 points (P, Q, R, S, T and M) we can find 6C3 = 20 different triangles, where (for example) PQR and PQS are different, but PQR and RQP are the same.  So guesswork will take too long.   You need to become proficient at looking at a fairly cluttered diagram and extracting just the bits you need.

a) Calculate the length of PR.

To calculate PR we only need the base PQRS.  In fact, we only need the isosceles right triangle PQR.  It is a right triangle because angle Q is the vertex of a square, hence 90°.  It is isosceles, because PQ and QR are the sides of a square so both are the same length, i.e., 3 cm.

Notice that the sketch on the right is very rough.  There is no attempt made to make angle Q look like a right angle , but it is labelled as a right angle, which is enough.
Since the triangle is isosceles, both angle P and angle R = 45°.  In keeping with the theme (trigonometry) we can calculate PR with sine or cosine.  For example
sin 45° = 3/PR  ß à PR = 3/(sin 45°) = 3/(1/2) = 32 = 4.14 cm (to 3 ‘sig figs’)
Undoubtedly, most students will use Pythagoras instead, with the same result.

b) Calculate the vertical height, MT.

As the length of PR was 32, the length of PM is 1.5*2 or 2.12 cm.   Point M is vertically under T.  As the square is symmetrical, point M is where the diagonals PR and QS intersect, i.e. M is the midpoint of PR.
We could use the inverse cosine function to find angle TPM and with the angle we could use any of sin, cos or tan to solve for h.   Or we could use Pythagoras, base2 + height2 = hypotenuse2 à hypotenuse2  –  base2 = height2.
If we do part c) before part b), we get another isosceles right triangle, which from the symmetry of the situation should not be terribly surprising.  In other words, the height h is the same as the base, 1.5*2 or 2.12 cm.

c) Calculate the angle between PT and the base PQRS.
As already indicated in part b), we have hypotenuse and the side adjacent to the required angle, so this is an inverse cosine problem.

cos (θ) = 2.12/3
cos – 1(cos (θ)) = cos – 1 (0.70666…)
θ = 45.0°  (to 3 sf). Had you calculated with the exact value, ( 1.5*2)/3 = 1/2, you would realise that indeed θ is exactly 45°.

What else could have gone into this trigonometry quiz?  Reading graphs of the sine or cosine function, bearings and scale drawings are the likely candidates.