Wednesday, October 28, 2015

Question

Introduction
This question was most probably taken from an Integrated Programme (IP) school in Singapore.  For your information, students in Integrated Programme schools do not take the GCE ‘O’ Levels, and each IP school is free to design its own individual curriculum.  In practice, they incorporate the mainstream GCE ‘O’ Level topics, as well as additional topics, and/or teach topics in advance, and may call their syllabuses by different names.  They tend to set more challenging questions than the mainstream schools, which are already targetting their internal examination standards above the ‘O’ Levels.  In other words, they tend to cram in more, but never less.  Part (d) tests approximate change, which had been taken out of the mainstream syllabus at this time of writing.
Note that the IP schools tend to be schools that traditionally attract the academically best students from each cohort.  Even before the IP programme was introduced, these schools were already setting harder questions.  Anyway, this blog welcomes everybody from all around the world who is willing to learn, regardless of the type of school they are from, even home-schoolers and independent learners!  Let us see how we can employ re-usable tactics to tackle this question.

Get to the root of the matter, fast!
The question is on the applications of differential calculus on a quadratic curve (a parabola).  Observe that the equation of the curve is given in completed square form.  From the equation, can you spot the line of symmetry (centre line) and the  y-intercept immediately (like within 5 seconds)?  [You need to know all the basic facts at your finger tips and make observations.]
The line of symmetry always passes through the maximum or (in this case) minimum point.  We know that the minimum is when the squared term  (x – 3)2  is zero.  So
the line of symmetry is  x = 3.
To get the  y-intercept, put  x = 0.  This gives  y = (-3)2 = 9.  So
C = (0, 9)  and equation of  CD is  y = 9
Since  PQ = 2k,  distance from  P  to the centre line = k.  All the above are basic observations that should be carried out mentally within one minute and you should be able to mark the diagram with pencil notes (shown above in blue).  Once this is done, let us get on to the real business.

Solution

Final Remarks
For part (c), they have already told you it’s maximum, so you do not need to prove that it is maximum.  However, intuitively it is obvious there is a maximum: imagine if  k = 0  or  k = 3, then we get very thin rectangles with area zero.  As  k  increases from  0, the area gets bigger and after that shrinks towards zero again.
Actually, was this problem really so difficult?  What did we do to solve it?  Let’s review
· Know all basic facts and skills thoroughly at immediate recall (e.g. extremum of parabola lies on line of symmetry, how to spot that from completed square form, how to find y-intercept)
· Use heuristics: e.g. make observations.  Use simple facts you already know (e.g. subtract lengths, substitute values of  x  to find  y  which is “height” have x-axis etc,).  Practice positive psychology: instead of worrying, write down everything you can deduce.  Then try to find connections.
· Apply the formulas
· For part (d), if it is not in your syllabus, do not worry about it.  But if you are curious or feel the itch to learn more, it is also not too difficult.  See the boxed formulas in the solution above.  Just remember that the ratio of small changes  DA/Dk  is approximately equal to the derivative  dA/dk.  You can detach the  Dk,  bring it to the other side of the equation, and that allows you to approximate  DA.

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
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
· GCE ‘O’ Level Additional Mathematics
· other syllabuses that involve applications of differentiation

· anyone who is interested in calculus!