[P6_20160217RTTU] Books on Bookshelves

Problem

Introduction

     Here we have a numerically challenging problem that involves ratios, and it ultimately reduces to an algebraic problem with two unknowns.  Nevertheless, we are spoilt for choice as regards to methods of solution:-

     (1)   Bar Diagram Modelling

     (2)   explicit letter-symbolic Algebra

     (3)   “p” and “u”  (parts and units)

     (4)   Distinguished Ratio Units

     Despite the fact that Bar Diagram Modelling made “Singapore mathematics” famous, let us remember that it is only one of the ways of solving problem by diagramming, which is just one of the eleven Primary School heuristics recommended by the Singapore Ministry of Education.
     The methods have a lot in common, and they differ mainly in the form of presentation.  However, standard Bar modelling is impractical under high-stakes high-stress examination conditions for this problem, not least because one would have to cut the bars into many pieces.  One should not cut off one’s feet just so as to fit the shoes (削足适履), as one Chinese saying goes.  We need to be flexible and open-minded.  I present a solution using my own Distinguished Ratio Units.

Solution

Ans:  735 books

Commentary

     First off, we need to equalise the numerators of  ²/₅  and  1¹/₄ = ⁵/₄  and put them ratio form.   This is because the  “2”  in the  ²/₅  represents the same quantity as the  “5”  in  ⁵/₄.

We do this adjustment by multiplying the former through by  5  and the latter through by  2.  Thus we deduce that the original number of books in A and in B are  25  and  8  “heart” units respectively. 
     Next, we add on the  2  and  3  “triangle” units.  By doing a comparison, we can figure out that  1  “triangle” unit must be  45  more than  17  “heart” units.  So  2  “triangle” units must be equal to  34  “heart” units plus  90.  Replacing the  2  “triangle” units (shown in yellow) with their equivalent, we now know that  59  “heart” units plus 90 gives  444.  This allows us to figure out that  1  “heart” is actually  6.  Thus, we can work out what  1  “triangle” unit, and then what  5 “triangle” units are worth.

Final Remarks

     Due to the difficulty of the numbers, the solution presented above is about as streamlined as I can make it to be.  

     There is another variation that can be used – equalising the “triangle” units (akin to the technique of elimination in standard algebra).  What we do is we multiply the group with total  444  by  3  and to multiply the group with total  489  by  2.  This would give  6  triangle units on each side.  Then we can compare the “heart” units and continue from there.  This way of proceeding is not for those who fear 4-digit numbers.

     If there are nicer or more elegant ways to tackle this question, I would definitely love to hear from you.

Heuristics Used

H01. Act it out

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and ratios

* any problem solver who loves a challenge

[H2_VJC2015PromoQ10_IAXS] Volume of a Doughnut

Problem

Introduction

     This is a problem involving the calculation of the volume of solid of revolution of an enclosed region.   The junior colleges (or senior high schools) like to set this type of question. 

Technique

     If the axis of revolution is the  y-axis, the basic formula is   ò px² dy   with the appropriate lower and upper limits.  Notice that this only works for the region between one curve and the axis and when rotated, this will generate a solid with no hollow parts.  An enclosed region, however, consists of two curves.  In our case, when we make  x  the subject, we find that we have two choices.  One of them leads to a curve that is further away from the axis of rotation.  I call that the outer curve.  The other curve is the inner curve, and this is nearer the axis of rotation.  We need to subtract the volume generated by the inner curve from that generated by the outer curve.

Solution

Remarks

     In this example, the curve on the right happens to be the outer curve.  If the equation were

(x + 93)² + y² = 15²,  the circular region would be on the left of the  y-axis and the outer curve would be on the left.

     For your information, the above solid of revolution is a torus.  This is the shape of a doughnut, (or hoopla-hoop, circular tube, or Polo mint perhaps?).  It is the inner curve that gives the hole in the “doughnut”.

     You can imagine in your mind’s eye that as the circular disk revolves around the  y-axis, its centre traces out a circular path of  93 units.  By the Second Centroid Theorem of Pappus,

     volume = length of path of centroid × area of cross section = 2p(93) × p(15)² =  41850p²

In general, the volume of a torus with major radius  R  and minor radius  r  is

     volume = 2pR × pr² =  2Rr²p²

If you know this fact, you use it to check your calculations.  Although this is not in the H2 Syllabus, but it is something interesting to explore.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards                                [e.g. making  x  the subject]

H09. Restate the problem in another way  [symmetry: volume is twice of upper half]

H10. Simplify the problem                         [integration by substitution]

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* IB Mathematics HL (Applications of Integration)

* Advanced Placement (AP) Calculus AB & BC

* University / College Calculus

* other syllabuses that involve Applications of Integration

* any precocious or independent learner who loves to learn

[AP_Calculus_IGSB] Integrating an Exponential with Square Root

Problem

Introduction

     Here is an integration problem that has no clues as to what to do with it.  Hmmm ... the integrand (3 to the power of square root something) does not look like it can be simplified.  [H09, H10]  How about a substitution?  [H12]  But what substitution?  Usually, we substitute the “ugliest” part of the integrand.  What constitutes the “ugliest” requires experience and observation.  In this case, the square root expression looks pretty nasty.  But how do we even integrate  3  to the power of something?

How to Integrate the Exponential

Solution

Comment

     In this problem, the original variable of integration is  x.  When doing substitutions, it is usually easier to make  x  the subject, and then replace the  “dx” with its equivalent.  After the substitution [H11], we integrate by parts and then substitute back to express everything in terms of  x.

Heuristics Used

H04. Look for pattern(s)        [look for the “ugliest” part]

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H12* Think of a related problem

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics (challenge)

* IB Mathematics HL (challenge)

* Advanced Placement (AP) Calculus AB & BC

* University / College calculus

* other syllabuses that involve integration

* any precocious or independent learner who is interested

[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

[H2_20151125DEST] Differential Equation via Substitution

Question

Introduction

     This question was taken from a Facebook group.  The suggested substitution was added in to make it accessible to students taking H2 Mathematics.  In the original question, no suggested substitution was given.

Strategy

     Observe that the given equation has  x  and  y.   After substitution, we should get an equation with  v  and  x only.  How to make  y  “disappear”?  One way is recognise patches that can be substituted for  v.  Another way is to make  y  the subject and differentiate that with respect to  x.  Replace the derivative ^dy/_dx  with your new expression.  Here, I do both at once!

Solution

Remarks

     Usually we try to express  y  in terms of  x,  but here, it is more convenient to express  x  in terms of  y.

     Since  A  is an arbitrary constant,  A+1  is still an arbitrary constant.  Also, whenever there is a “ln” appearing in the solution, it is good to introduce “ln” with an arbitrary constant.  So we can lump  A  and  1  together and call it  ln B.  Nothing is lost in this process.  ln B  is able to achieve all possible numbers.   ln B  is negative if  B  is a fraction between  0  and  1,  and is positive if  B  is more than  1  and is zero if  B = 1.

     This is not an applied differential equation, so there are no exogenic reasons or hints for us to suppose that  x + y + 2  is a positive quantity.  As such, we still need to leave it as | x + y + 2|.  To get rid of the absolute or modulus sign properly, we can replace  ±B  with  C.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13. Use Equation / write a Mathematical Sentence

Heuristic that cannot be used here

H08. Make suppositions

Suitable Levels

*  GCE ‘A’ Level H2 Mathematics

*  IB HL Mathematics Calculus Option

*  Advanced Placement (AP) Calculus BC

*  university / college calculus

*  other syllabuses that involve differential equations

*  anyone who is game for a challenge

[U_Calculus_STKC58] Exploiting symmetry for a Complicated Integral

Question

     This problem appears as problem 58 from a Facebook group and it is set by Kunihiko Chikaya.  Ordinary integration problems are already challenging, but this one is tough on steroids. 

The Standard Approach

Solution 1

With this result, I realised that there is a short cut.  We can make use of symmetry.  Note that  sin(p – x) = sin x   and   cos(p – x) = - cos x.

Solution 2

Heuristics Used

H04. Look for pattern(s):         “onions”, exploit symmetry

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* University / College Mathematics

* challenge for ‘A’ Level H2 Mathematics

* challenge for IB HL Mathematics

* other syllabuses that involve trigonometry and integration 

[S2_20151029QFCS] Chunking and Piggy-back

Question 

Introduction

     We have seen here, here and here that recognising chunks is useful in mathematics.  It is an example of a having what Carolyn Kieran calls a structural view of algebra, which is a type of pattern recognition.  In this article, I show how, using chunking and substitution, we can piggy-back or ride on solutions of a simpler equation to obtain solutions of a more complicated equation.

     The solution for part (a) of is easy enough.  Just factorise (or AmE. “factor”)

                            (y + 2)(y – 7) = 0

Hence                 y + 2 = 0   or   y – 7 = 0

                                 y = -2   or   y = 7

This is Standard Operating Procedure.  But part (b) seems like a monster of an equation.  Oh dear!  What shall we do?

Observation

     Can you observe anything appearing more than once?

     Now do you notice any chunks that are repeated?  Once you can see the connection, you can make a substitution  y = x³ – 1  and then make use of the answer in part (a).  We  copy and paste the chunk (shown in green) into the previous solution and proceed from there.

Solution to part (b)

                                 x³ – 1 = -2     or     x³ – 1 = 7

                                       x³ = -1     or           x³ = 8

                                        x = -1     or            x = 2

Solved!

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (Sec 2 ~ Grade 8-9)

* any syllabus that includes algebraic factorisation (factoring) and substitution

* anyone who is interested in and ready for algebra

[AP_Calculus_20150516LIM] Large abyss with no Limit?

Question

Introduction

     This looks like a scary limit to calculate.  Indeed, it is quite tricky, as  x  goes towards negative infinity and there is a square root.  However, we can often work out the correct answer informally first.

Informal Solution (intuitive approach)

Formal Solution

Final Remark

     Having the informal solution (which can be worked out on rough paper or even mentally) gives us an idea of the answer.  We can counter-check our formal solution with the informal solution for confirmation.

Suitable Levels

* Advanced Placement (AP) Calculus AB & BC

* University / college level calculus

* other syllabuses that involve limits

[AM_20150510PLRF27] Looking for a Polynomial's Missing Link

Question

Introduction

     The first part of this question is rather standard.  The second part is quite challenging, especially if you are trying to connect with the earlier part.

Reminder

     To solve polynomial equations of degree  3  or higher (that are tested in school tests and exams), we often need to guess a rational (fraction or integer) root.  By the way, whole numbers are rational numbers because we can always put them into fractions upon  1  as denominator.  So how do we guess the roots?  The following is a very important theorem that guides us as to what numbers to try.

So we consider all the possible factors of the constant term  a₀  for the numerator and

all the possible factors of the coefficient  a_n  of the highest power for the denominator and consider the + and the – of all the possible fractions formed.  Usually, we try those with denominator 1 i.e. the integers first.

Solution

Remarks

     Note that the solution consists of only the part in blue.  Black is used for explanations, which are lengthy because of the dense interplay of ideas and subtleties involved.

     For the second part, if you are not able to see the connection, then use the standard method to solve the equation.  Here we realise that when the  x  is replaced by  ^v/₂,  the coefficients are reduced to the original coefficients.  However, these are in reverse order.  This indicates that one needs to use the reciprocal, so you divide throughout by  v³,  so that the highest power becomes just a constant.  I know you would not have thought of this if you have not seen this kind of question before, but this is the trick to use.

     Do not be discouraged by difficult question.  Have a growth mindset.  Every time you encounter a difficult question, learn how the trick ticks.  Your brain muscles will get stronger.  Try to apply the same trick when you see a similar question next time.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* GCE ‘A’ Level H2 Mathematics (revision)

* IB Mathematics (revision)

* other syllabuses that involve polynomials, Remainder and Factor Theorem

[S2_20150501AXC] 2011: Chunking and Substitution with Algebra

Question

Introduction

     This lower secondary algebra question seems complicated, doesn’t it?  Can you spot any chunk that is repeated, or almost the same?  This is one of the keys to solving the problem.  Another key that you need is the relevant algebraic identities and tricks.

Reminders

     First, let us review some of these useful formulas.                                        

This is the square-of-difference identity, illustrated here.

This swapping trick is illustrated here. 

     Looking back at the question, do you notice anything that is repeated?  Can you see any chunks that are the same or almost the same?  (n – 2011)  is almost the same as  (2012 – n)  isn’t it?  Whenever you see a repeated chunk, it is a good idea to substitute that chunk with another variable that you invent.  To name this new variable, you can use any letter that is not used before, so as not to conflict with existing letter(s).

Solution

Summary

     To solve the given problem, we have used the following:-

     (1)  square-of-difference identity

     (2)  swapping technique

     (3)  observation of repeated chunks

     (4)  using substitution with the chunks

Heuristics Used

H04. Look for pattern(s)  e.g. chunking, observation

H09. Restate the problem in another way  e.g. swapping, identities

H10. Simplify the problem e.g. substitution for chunks

H11. Solve part of the problem

H12* Think of a related problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

· Lower Secondary Mathematics

· other syllabuses that involve whole numbers and ratios

[JCH2BXQNSR_20150429] Binomial Expansion for a Quotient

Question

Solution

Comment

     Note in the above working that  3(x + kx²)² = 3(x² + 2kx³ + k²x⁴),  but since we do not need the  x³  and  x⁴  terms, we omit them and just write “3(x² + ... )”.

Heuristics Used

H12* Think of a related problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* revision for GCE ‘A’ Level H2 Mathematics

* AP Calculus

* other syllabuses infinite binomial series