[H220151129IAAV] Integration for an Eclipsed Ellipse

Problem

Introduction

     Here we have a challenging H2 Mathematics question that tests students on the technique of substitution, finding area and volume of solid of revolution.  Note that the diagram is not drawn to scale and the line actually intersects the ellipse (oval shape) at the point (-2, 2).  Also the region  R  is the shaded area, but the label “R” is put outside of it.  Students should do their due diligence in ascertaining the intersection point themselves.

Solution

Remarks

     Since the word “exact” is not used in the instructions to part (iii), one can also use the Graphing Calculator to obtain the approximate answer  3.14  to 3 significant figures.  Part (iii) can also be done by another method of integration called the Shell Method, but this is not in the H2 syllabus.  Nevertheless, the student who uses it would not be penalised, unless explicitly forbidden in the rubric.

     The word “ellipse” means an oval shape, and is not to be confused with “eclipse” which means to occlude or hide (e.g. eclipse of the sun, eclipse of the moon).  By the way, the moon orbits around the earth in an ellipse and planets revolve around the sun in ellipses (ovals).
     Although not part of the syllabus, it may be advantageous to know that the area of a right ellipse is  pab,  where  a  and  b  are the semi-axes.  Imagine a circle of radius  a  being stretched by  ^b/_a.  Then its area  pa²  will be multiplied by the same factor  ^b/_a.  So this is not too difficult actually.  With the formula, one can verify one’s answer obtained by integration.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards        [ e.g. calculating new limits for substitution ]

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* GCE ‘A’ Level H2 Mathematics

* IB Mathematics HL

* AP Calculus BC

* University / College calculus

* other syllabuses that involve applications of integration

* whoever is interested

[S1_20151124AESR] Slanted Rectangle does not need Pythagoras

Question

Introduction

     This is another “Bonus Question” at a secondary level from somewhere that the question poser did not mention, but I guess it is most likely an Integrated Programme school in Singapore.  It is a beautifully crafted question.  The presence of a slant line seems to necessitate the usage of Pythagoras’ Theorem.  However, we have seen that Pythagoras’ Theorem can actually be avoided even in Primary (Elementary) School problems.  So a 10 year old kid with a rudimentary knowledge of algebra could do this.  Can you spot a short cut?

Making Observations

     Stare at the diagram for a while.  What do you observe?

Solution

             area of  DDB_nC_n =  ½  of the area of  AB_nC_nD.

              area of  DDB_nC_n =  ½  of the area of  DB_nPQ.

        \  area of DB_nPQ  =  area of AB_nC_nD = n cm².

HeuristicsUsed

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Lower Secondary Mathematics

* challenge for Primary school Olympiad

* other syllabuses that involve areas and a tiny bit of algebra

* anyone game itching for a challenge

[IB-HL H&H_8G Q16] Sum of Squares of some Binomial Coefficients

Question

Introduction

     This problem is taken from the Haese textbook for International Baccalaureate, 3^rd Edition, page 262.  It looks pretty daunting doesn’t it?  Where do we even begin?  The key to solving this problem is to realise that the binomial coefficients are coefficients of (numbers attached to) certain powers of  x  in the expansion.  The question is:  which power or powers?

     Before we go into that, let us review some important relevant facts.

Reminders

Solution

Final Remarks

     This problem was solved by using the symmetry property and treating binomial coefficients as coefficients of certain powers of  x.  We also worked backwards by noting that the RHS of the equation to be proven is the coefficient of  xⁿ.  This suggests that we compare this with the coefficients of  xⁿ  on the LHS.

HeuristicsUsed

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* International Baccalaureate Mathematics (HL)

* GCE ‘A’ Levels H2 Mathematics

* other syllabuses that involve complex numbers and polynomials

[Pri20150503SMT] Seeing through the Area of Mess

Question

Introduction

     This looks like a challenging problem regarding area.  The diagram looks very confusing.  The shaded area consists of many convex pieces.  Furthermore, the unshaded regions comprise triangle-like pieces of which we do not know all the dimensions.  The only dimensions we know is related that of the large triangle  DACB.  What shall we do?

     The key to solving this problem is to appropriately cut up the figure so as to be able to “see” it properly.  We can cut up the figure like this:-

     You might note that the pink triangles and green triangles are right-angled triangles, because they are in semi-circles.  Yeah!  Smart!  But how can this be useful?  We do not know their individual bases and heights.  We only know their longest sides.  Hmmmmm ...  Ah!  However, the pink triangles and green triangles all add up to the large triangle DACB.  This is a key observation.

     Note that the convex parts can be viewed as semi-circles with either a pink triangle or green triangle taken away.  There are two pairs of semi-circles: one large and one small.  Therefore, the shaded area is the total of one small circle plus one large circle minus the total of the areas of the pink and green triangles (which is the same as the area of triangle DACB).  As you know, the area of the triangle DACB is ½ ´ base ´ height, in which  ½ ´ base = 5 cm.  Once you understand all these, the calculation is very easy.

Solution

   total area of shaded regions

= area of large circle + area of small circle – area of triangle DACB

= [p (5)² + p (⁵/₂)² – 5 ´ 10] cm²

= 48.18 cm²

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels
* Primary School Mathematics
* other syllabuses that involve area of triangles and circles

[Pri20150429PPP] Pythagoras for Primary Pupils?

Question

Introduction

     This primary (elementary) school mathematics examination question created quite a stir among some parent support groups on Facebook.  The issue is that the height of the triangle seems to have been omitted.

Pythagoras’ Theorem

     Some participants who know secondary school mathematics were quick to suggest the use of Pythagoras’ Theorem to find the height of the triangle, which works out to be  12 cm.  This leads to the answer (2) 114 cm²,  which is correct.  The problem is that pupils are not taught Pythagoras’ Theorem until secondary school, and so it would seem an unfair test for the pupils.  So the discussion turned to thinking of various methods by which a primary school pupil may uncover the answer without resorting to advanced knowledge.

Elimination and Educated Guessing

     Mr Teo Kai Meng, a tutor who regularly participates in the support groups, offered some insightful observations.  Assuming that the height measurement is a whole number of centimetres, only options (2) and (4) need to be considered as they were divisible by  19, which the area had to be under the said assumption.  [ Another tutor, Melissa Song had a similar idea by observing that since the triangles DKLN and DKMN  have the same height, the ratio of their areas is the ratio of their bases LN : MN = 19 : 16. ]  We can ignore choices (1) and (3).  As we know,  area = ½ ´ base ´ height .  Since teachers like to catch students for being careless in forgetting to multiply by ½  (or dividing by 2), it is quite likely that option (4) was set up as a booby trap.  Thus one may intelligently surmise that option (2) should be the answer.

     Mr K L Chua, a tutor who calls himself “Mathematics Specialist”, used a similar reasoning.  He worked backwards from each of the four choices to get the heights and chose the most plausible answer.  Of the two whole-number answers, (4) was eliminated and (2) was chosen since from the diagram the height should roughly be near to  16 cm  even though the diagram was not drawn to scale.

Scale Drawing

     Another participant suggested doing a scale drawing to estimate the height.  Indeed this can be done, and is a good tactic too, since this could be done quickly with a ruler and pencil.

A Visual Solution

     Assuming no knowledge of Pythagoras’ Theorem, it is possible to construct a visual solution.  First we take four copies (indicated in orange/light-orange) of the right-angled triangle  DLNK  and arrange them to surround a square of side  37  cm (indicated in green),  which is the same as the longest side (hypotenuse) of the said triangle.     This green area is  1 369 cm².  Now we rearrange the triangular pieces as indicated by the red arrows.  The areas of the orangey triangles do not change when you shift them.  Neither will the green area change, since the total area everything in the containing square (orangey plus green areas) remains the same.

With pairs of right-angled triangles joined together along their longest sides, we now obtain two green squares, the larger of which has side  35 cm.  This gives an area of  1 225 cm².  The total area of the two green squares is the same as the area of the large green sqaure before the shifting, namely  1 369 cm².  Hence the area of the small square is  (1 369 – 1 225) cm² = 144 cm².

From here we quickly deduce that the unknown height is  12 cm,  12 being the square root of  144.  Hence we conclude that the area of  DKLM  is  114 cm².

Remarks

     I have shown that it is theoretically possible for primary school pupils without knowledge of Pythagoras’ Theorem to derive the answer in an exact manner.  By the way, the method of shifting triangular pieces as indicated above can be generalised to give a proof of Pythagoras’ Theorem.

     Some parents expressed fear that this is another one of those Cheryl-like or olympiad type of problems.  Is there a conspiracy by the school teachers to purposely set difficult questions and make life difficult for pupils, disadvantaging those who cannot afford private tutors?  In this case, could this just have been an oversight on the part of the teacher who set the question?

     Entrepreneur John Low Jiayong and tutor John Lim sourced for and managed to obtain faithful copies of the original question.  It turned out that some school-paper vendors had inadvertently erased the 12 cm measurement.  Thus in the original question, the height  KN  was given as 12 cm, and the measurement of  37 cm for the hypothenuse  LK  was purposely given as extraneous information to distract students.

     Some parents observed that this is after all just a multiple-choice question that carries a credit of only one mark.  If this were an exam situation with the  12 cm omitted, it would be best to either sacrifice the  1  mark and move on, or to use tactics like elimination, educated guessing or estimation with scale drawing.
     Anyway, it had been quite a fruitful discussion, as adults (parents and tutors) attacked this problem from many angles in a purposeful way.  Many parties put in concerted effort and contributed in an engaged matter.  It would be good if this type of rich discussion were enacted in our classrooms everyday among pupils and teachers, perhaps enabled by social technology.  For then, pupils would deeply learn and perhaps would not be so stressed out, nor be needing so much extra external help.  Private tutors could then move on to focus on value-added  mentorship  for 21^st Century Learning, instead picking up the tab where school teachers have left off.

Heuristics Used / that can be Used

H01. Act it out (e.g. scale drawing)

         (as a class learning activity, pupils can use scissors to cut out paper triangles

          and physically move them around)
H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H07. Use guess and check

H08. Make suppositions

H09. Restate the problem in another way

Suitable Levels
* Primary School Mathematics
* Lower Secondary Mathematics
* any precocious / gifted pupil who wants to learn
* any person of any age interested in creative problem solving

[EM_20150412SCA] Is your Airport Design career “taking off”?

Question

An Airbus 380 has constant acceleration of  1 m/s2.  Its takeoff velocity is 280 km/h.  How long must the runway be at a minimum to allow the plane to take off?

Introduction

     A practical question for airport design, perhaps?  I present two solutions.  The first uses a graphical method (speed-time graph) which is in the (“Elementary”) Mathematics syllabus and the other uses formulas for motion under constant acceleration taught in Physics.  Whichever method is used, remember to convert from km/h to m/s.  The target velocity is  ⁷⁰⁰/₉ m/s,  and the time to achieve this is  ⁷⁰⁰/₉ s  starting from rest,  since the acceleration is  1 m/s².

Solution 1 [“Elementary” Mathematics, speed-time graph]

     The speed-time graph is very useful because it is able show the acceleration (as the gradient or slope of a straight line) and at the same time the area under the graph gives the numerical value of distance travelled.  If the acceleration is constant, we usually we get a trapezium.  But since the aeroplane starts from rest, we get a triangle (see diagram below).  All we need to do is to calculate the area under the graph and get the answer.
     It is interesting to observe that if we used the average speed  ⁷⁰⁰/₁₈ m/s,  we would also get the answer because the area under the graph (yellowish green rectangle) is the same as the area of the triangle.  This trick works for constant acceleration, but it may not work in other situations.

Solution 2 [Physics, constant acceleration]

     We use the important formulas  v = u + at  and  s = ut + ½at².  In our example,  u = 0  because the initial velocity is zero (the airplane starts from rest).  This makes our calculations very easy.  If we compare the two methods, you find that the calculations are very similar, and we get the same answer.  Remember that speed = |velocity|  the magnitude of velocity.  In this relatively easy problem, the velocity means the same as speed because we are going in a straight line and in one direction only.  In other situations, this may not be so.

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels
* GCE ‘O’ Level “Elementary” Mathematics
* GCE ‘O’ Level Physics
* other syllabuses that acceleration, speed and distance
* precocious kids who always want to learn more

[Pri20150220FSA] A Fishy Shaped Area

Question

Plan of Attack

     This problem looks difficult because the shaded area does not seem to look like any regular shape.  Is it a fish whose head is pointing in the top left direction and whose tail is in the bottom left direction?  Fortunately, this is not a Rorschach ink-blot test.

     As with all “area” problems in primary (elementary) school, we try to break down the unfamiliar shape into regular shapes (e.g. parts of circles, squares, triangles, rectangles).  It is basically a divide-and-conquer strategy (using heuristics H10 & H11).  If we look carefully, we realise that the required area consists of a semi-circle less a funny horn-shaped area, which I call ‘F’.  F is for funny, for want of a better description.  So I am going to find the area of the semicircle (which is half of a circle), then subtract the area of F.  We’ll worry about finding the area of F later.

Solution

     Area of semi-circle [in cm²]

= ½ ´ p ´ radius²

= ½ ´ p ´ (5) ² = ²⁵/₂ p   

     It is good to leave the calculation with  p  until the last step.

     OK, we are done with the first part.  [Heuristic H11]  Let us us tackle the next part, which is to find the area of  F.  Note that this is a 45°-45°-90° isosceles triangle minus a 45°-degree sector (which is one-eighth of a circle, because ^45°/_360° = ¹/₈).

     Area of F  [in cm²]

= Area of triangle – area of sector

= ½ ´ 10 ´ 10 – ¹/₈ ´ p ´ radius²

= ½ ´ 10 ´ 10 – ¹/₈ ´ p ´ (10) ²

= 50 – ²⁵/₂ p

     Let us combine our answers.  We need to subtract 50 – ²⁵/₂ p  from the area of the semi-circle.  If we subtracted 50 from ²⁵/₂ p, we would have over-subtracted.  So we need to add back ²⁵/₂ p.  Hence

    Required Area [in cm²]

= Area of semi-circle – area of F

= ²⁵/₂ p  – (50 – ²⁵/₂ p)

= ²⁵/₂ p  – 50 + ²⁵/₂ p

= 25p  – 50

Using the calculator’s value of p,  we obtain

     Required Area = 28.54 cm²  (to 2 decimal places)

Heuristics Used

H10. Simplify the problem

H11. Solve part of the problem

Commentary

     This difficult problem was solved by dividing the problem into smaller pieces and tackling each piece one at a time.  We break down a complicated shape into familiar shapes.  That is the secret.

Please refer to this similar problem.