SHARING.... -experiences of teaching maths as a hobby for over 50 years to school and college students. -challenges in project management and the common sense approach to it from old classics.

Sunday, March 01, 2015




MATHS FOR ALL.......REPOSTED FROM EARLIER BLOG 
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Let us continue our journey. We have seen four methods of problem solving in the previous article and took a break with a game and reference to Boolean algebra. Before we continue with other methods , let us deal with few more examples with relevant tips required in applying these methods and some common pitfalls that need to be avoided.
We considered a couple of examples already concerning locus and rational numbers. The one issue we have to keep in mind in taking up specific examples is our boundaries, start and finish lines in taking up this exercise. Hence, we shall try to see that these examples are generally understandable to high school students as well , or serve as a broad concept study to them, even though they may relate to higher standard of under-graduate level. For this purpose wherever needed we shall give the relevant formulae, which the high-school students may accept as taught to understand the further working. Similarly, when we take up examples of high school level, we shall try to see that they include a broader concept usefull for application to under-graduate students.
Once, we were going on a walk from our house to a park about 4 km. away. Our grand son accompanied us on his cycle. As he goes faster on his cycle , I just wanted to give him a little more exercise to his body and a small problem to think about. I told him, we walk at about 4 km.per hr speed , where as you may go at about 8 km.,per hour on your cycle. We are both starting at the same time and the park is about 4 km. away. So ,what you do is go on your cycle ,reach the park, turn back ,meet us wherever we are on the way , turn back again to go to the park continuing like this till we reach the park , every time turning back at the park to meet us wherever we are and then again turning back to go to the park. He did like wise enjoying the experience of meeting us to say that he has already made so many rounds etc.. So finally when we reached the park , I asked him O.K. you did a good job , but can you tell me how long a distance you travelled in this fashion. He went scilent for some time and finally replied , I feel it should be 8 km., though I cant tell you how I got it or whether it is correct or not.
That is it .He is a middle school student and he replied by what we call as “Intution” The surprising fact is that it is the correct answer. When prodded further , finally , he came out with the theory that on his cycle ,he goes at 8 km.per hour , where as , we are walking at 4 km.per hour and hence he should have covered double the distance of our walk.
Well that is the advantage of keeping things simple. Though , he could still not explain it to the satisfaction of examiners ,but the only missing link in his explanation is the fact that he cycled for same time as we were walking whatever direction he was travelling . Now imagine what could have happened, if the same problem is given for IIT entrance with some high sounding words and the students over awed by the level of the examination start thinking big using time and distance formulae , formulating a series ,trying to find their sum ,all by seeing some imaginary hidden devils in the problem as it is given in IIT exam. Wondering how ? Let us see how ..

Speed of walk = 4 kmph
Speed of cycle = 8 kmph
Time of start = 0 hrs say.
Distance to park = 4 km.
Jouney No.1 for cycle :
Time taken to reach park by cycle = 4/8 = 0.5 hrs .
Relative speed between cycle and walk (in same direction)=8 - 4=4 kmph.
Hence distance of seperation after journey No.1 = 0.5 hr x 4 kmph = 2 km.
Total distance travelled by cycle up to Journey No.1 = 8 x 0.5 = 4 km.
Journey No. 2 for cycle :
Relative speed between cycle and walk (in opposite direction) =8 + 4=12 kmph
Time required to meet = 2 / 12 = 0.166666..
Distance travelled by cycle = 0.16666x 8 = 1.3333 km…..
Total distance travelled by cycle up to Journey No.2 = 4 +1.33333..= 5.3333..
Like this we can continue to a series whose sum will finally turn out to be 8 km. as we deduced earlier by keeping things simple or rather what they deserve to be !!
We may recall what our dashing opening batsman in cricket , Sehwag said about his play ….” I will not think of the bowler’s reputation..I will see the ball ..if it can be hit. it will be hit “
That is the point So our tip is to look at the problem for what it deserves and not given by whom ?
Few lines on the aspect of intution brought in at the beginning of this problem. Some may say it is linked with IQ ..”GOD” given etc. May be, but then what do we mean by saying some one is intelligent? Is he good at every thing in this world, No. .one may be good at maths. another may be good in English etc. Now what do we mean by good in maths or English ? How this can be brought out in a way that is understandable. Well, one way is to see “ given a page or so of relevant material in a particular topic in which we say the person is intelligent , he should be able to read ,understand ,explain and apply it in a shorter time than others”. Then we can say that he is good in that topic. O.K., Leaving out the “God” given part from it, cant we improve our performance in this test by taking interest in the topic of our choice and putting extra hours of hard work to make up for the difference in comprehension. The answer is that it is largely possible particularly when we have time at our disposal and a strong determination to improve coupled with interest ofcourse. So let us take inspiration from the hare and the tortoise story and move on .

Now to cement the moral of giving a problem its due , let us try another
example. What happens to  x ^ n ....[ x to the power of n ] 
as x becomes larger and larger that is as x tends to infinity. The immediate response is that it also tends to infinity without asking any question.. But think what happens if n is negative .It tends to zero. Similarly if we consider the same problem , this time ,making n tend to infinity with x having different values. Once again there are several answers depending on the value of x …whether it is negative less than –1 ,or –1 or between -1 and 0 or zero or between 0 and +1 or +1 or greater than +1 , as you can easily see .
So it is important to consider all possibilities before jumping to one conclusion. We can call this as 360 degree look which is very much essential in normal life too. It is quite the thing done , when evaluating marriage proposals , we enquire from all around. .that is elders , youngsters, colleagues etc.. This is also the modern thought in performance appraisal system of employees to write their confidential reports. Reports are sought from superiors, colleagues, juniors and the employee too. For example you can give an evaluation report on your teacher apart from that normally written by their head or principal . This along with the opinions of your teacher’s colleagues , all taken together is expected to give a true picture to improve matters. The key word here to be noted is “to improve matters ..that is we should be really interested in “improving matters”.
Now on to another example to give another dimension to this 360 degree thinking .Let us now call it thinking with an open mind ..if we are not getting ideas in conventional approach ,let us consider ,analogous situations else where.
This is called “analogy method ” and it is the most natural way of learning and a powerfull tool too. We all learn from analogies as we grow up from child hood. The problem we are going to take up for this purpose came up in IIT entrance exam about 25 years ago when ,we were on a holiday in vizag. Our neighbour’s daughter who appeared for the exam came up with the paper to us saying that she could not know how to proceed with this problem in trigonometry.

If in a triangle ,
Cos 2A + Cos 2B +Cos 2C = - 1.5
Show that the triangle is equilateral..
She said , there are only 2 equations ,one the above and the second that ABC being a triangle A + B + C = 180, with this ,how can we prove that A= B= C= 60, that too using trigonometry ? Exactly ..when she is not able to make head way in trigonometry ,why cant she use algebra ? In fact she was using an algebraic idea that she needs 3 equations to find 3 unknowns, but was inhibited to take it forward ,probably due to some mind block. These are the shackles that need to be broken. This is what we call open mind ,try to use all your knowledge be it algebra or trigonometry or calculus ..let us see how algebra of high school level helps to solve this problem..
Her reasoning is correct ..there are only 2 equations so we cannot have more than 2 unknowns to find out…so let us convert it to only a two variable equation using trigonometric formulae..( at this point the high school students may accept the following trigonometric formulae as taught )
COS[2A]=2COS^2[A]-1........CO[A]+COS[B]=2COS[0.5(A+B)]COS[0.5(A-B)
USING THE ABOVE WE GET ….
Cos 2A + Cos 2B +Cos 2C + 1.5 =0
2COS(A+B)COS(A-B)+2COS^2(C)-1+1.5=0......PUTTING C=180-A-B
2COS(A+B)COS(A-B)+2COS^2(A+B)+0.5=0
Y^2+YCOS(A-B)+0.25=0....WHERE ...Y=COS(A+B)
Y=COS(A+B)=0.5[-COS(A-B)-{COS^2(A-B)-1}^0.5]
NOW Y=COS(A+B) IS REAL AS THE TRIANGLE EXISTS..BUT COS^2(A-B)<=1 
AND  SO FOR REAL ROOTS...CO^2(A-B)=1..OR..COS(A-B)=1..OR..A-B=0..OR..A=B
SO COA(A+B)= - 0.5 ...OR...A+B=120....SO A=B=60 & SO C=60
We can see how use of elementary formula of quadratic equation helped us to solve this problem easily. As mentioned earlier this was given about 25 yrs. ago. Some variations of this problem have come up subsequently ,like …..
Show that in any triangle, Cos 2A + Cos 2B +Cos 2C is greater than or equal to –3/2.
Just to reinforce our suggestion of open mind, let us note that this can be easily solved using vector analysis taught in first year intermediate.
For this purpose consider triangle ABC, with O as cicum-centre . Then obviously angles BOC, COA, AOB will be equal to 2A, 2B and 2C respectively as
angle at center is double the angle at the circumference of a circle. Now taking square of the vector sum , (OA+OB+OC) being equal to dot product of (OA+OB+OC) with (OA+OB+OC) and noting that this being a perfect square
It is always greater than or equal to zero., we can easily prove the result using the standard formulae for dot product …this is left as an exercise to the readers.


Tail Piece : The reference to GOD in this article brings to memory some of the media highlight and criticism of our SHAR Scientists going to Tirupati with a replica of the launch vehicle to invoke divine blessings before the actual launch, recently. Well some may not like to call by name “GOD”, but we talk of few things like above mentioned IQ etc as “GOD” given or ordained/destined. Some may not accept this reference , but like in maths , they can call some “x” or some unknown destiny which guides several things in this nature. It is a lot better to proceed as in maths assuming some “x” than coming to a stop. Finally as the saying goes “GOD” is like “PROFIT” to a Marvari business man. It is there if you believe it to be there or else it is not there!!! ( The reference to a community here is purely as the saying goes as in the case of many Sardarji incidents etc. and is not meant to cause any ill feeling or disrespect)




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