QUESTIONS - ANGELA - QUADRATIC EQUATIONS

QUESTIONS / ANSWERS - ANGELA - QUADRATIC EQUATIONS
On 10/30/06, Angela wrote:
> Angela has left a new comment on your post "MATHS – 6.EQUATIONS":
>
> For this problem 23x^2 + 17x = -11 I get 23x^2 + 17x + 11=0 as no solution
> because this can't be factored. Is that correct? Angela
> ----------------------------------------------------------------
ya.you are correct.there are no real factors.test is .........discriminant =
(coefficient of x)^2-4*(coefficient of x^2)*(constant) should be greater than or equal to zero.
17^2-4(23)(11) = -723
if this is negative , there no real factors for the function..that is there are no real values of x which satisfy this equation.hence we say there are no real solutions.

QUESTIONS - ANGELA - COMPLEX NUMBERS

QUESTIONS/ANSWERS - ANGELA
COMPLEX NUMBERS

On 10/30/06, Angela - noreply-comment@blogger.com - wrote:
Angela has left a new comment on your post PANCHATANTRA - 5.COMMUNICATION

Your time is greatly appreciated. You have helped me tremendously. Here are
some problems I need help with... Divide 10-9i by 3-2i and make the denominator
a real number.

=(10-9i)(3+2i)/(3-2i)(3+2i)
=(30+20i-27i-18i*i)/(9-4i*i)
=(48-7i)/13

QUESTIONS - ANGELA - CONICS

QUESTIONS/ANSWERS
CONICS

On 10/30/06, Angela <noreply-comment@blogger.com> wrote:

> Angela has left a new comment on your post "PANCHATANTRA - 5.COMMUNICATION":
>
> Determine the vertex of the function f(x)= x^2+2x-3

F(X)=[X^2+2*X*1+1]-1-3=(X+1)^2-4
HENCE VERTEX IS AT X+1=0...OR...X=-1...THEN ..F(X) =-4
SINCE (X+1)^2 IS ALWAYS GREATER THAN OR EQUAL TO ZERO,WE HAVE A MINIMUM
VALUE OF F(X)=-4 AT THE VERTEX.(1,-4)

Does the graph of the
> function open up or down?

UP
Why?
BECAUSE THE VERTEX IS A MINIMUM.

Determine the range of the function above.

at x=-1 ,the function has a minimum of -4 as x tends to minus or plus
infinity , the function tends to plus infinity.
hence range is -4 to infinity.....(-4,infinity)

On
> what intervals is the function above increasing or decreasing? Use set
> notation.

the function has a minimum at x =-1
> hence from x=-infinity to -1 the function decreases ..
x =(-infinity,-1)
and
increases from x= -1 to plus infinity...(-1,infinity)

MATHS - 7.INDUCTION

Mathematical Induction :

This is a well taught and useful method at intermediate or even school level to prove certain formulae for sum of a series etc. The method is as follows.

Generally,we are given a series with n terms and given a formula to prove that its sum is given by that formula.

The method of proof is ,

1. we test the formula for applicability in the case of n=1 OR 2 OR WHATEVER IS RELEVANT TO THE PROBLEM .
2. After verifying that it is applicable , we assume that the formula is correct for n = k say , that is for any particular value of “n”.
3. Next we go on to prove that , if it is true for n = k, then , it is also true for n = k + 1 .
4. Now we reason out that since we proved its validity at n = 1 at the beginning ,then it should be true for n = 1 + 1 = 2 as per our later verification that if the formula applies for n = k then , it is also true for n = k + 1. Hence it must be true for n = 2 , 3 , 4 ..etc…that is for all integral values of n .

It is very important to follow both the steps mentioned above in that order , The first step ensures that the given formula is correct by checking at the lowest level of n = 1 or 2 or whatever is relevant.

Then we should check its applicability as per the third step to prove its validity at any integral value of n.

This method has certain limitations by its very nature .

1.It can be applied only for an integral value of “n” the number of terms.

2.We should know the formula or answer ahead , to prove that it is applicable., that is we are not deriving a formula here but proving a given formula.

There are several standard examples for this at the school and college levels and hence we shall confine ourselves with a few important points to remember in adopting this method.

We should adopt a good nomenclature for the problem ,defining clearly all the terms. For example when dealing with problems on summation of series, put

sum to ‘n’ terms as s(n) and ‘n’ th. term as t(n) and note that

s(n) = t(1) +t(2) +……..t(n-1) +t(n) and

s(n+1)=s(n) + t(n+1).

After initial verification for n=1,use the given formula to be proved for s(n) in

s(n+1)=s(n)+t(n+1)

As we have assumed that the given formula is true for ‘n’. Substitute (n+1) for n in L.H.S. to get at what we have to obtain from R.H.S. Now take L.H.S. and substituting for t(n+1) try to maneuver the right hand side to make it equal to left hand side .

In dealing with problems to prove divisibility , proceed similarly and try to convert the R.H.S. in to multiples of the required divisor by splitting / combining them , using the assumed fact for a value of ‘n’ and other general known factors.

Here , now we shall include an example from determinants asked by a reader, which uses this concept in a different context and is quite illustrative of the power of this method.

PROBLEM:

To show that determinant of A = determinant of its transpose

TERMINOLOGY:
let |A| be determinant of order n.
let its transpose be |T|
let elements of |A| be represented by aij...that is the |A|=
a11,a12......a1n
a21,a22......a2n
.........................
an1,an2......ann
similarly let elements of transpose |T| be tij…that is |T|=
t11,t12……..t1n		since |T| is transpose		|T|=	a11,a21,……….an1
t21,t22……..t2n		of |A|,we have tij=aji			a12,a22,………..an2
…………………..					……………………….
tn1,tn2……..tnn					a1n,a2n,…………ann
let cofactor of element aij be |Pij|...and cofactor of element tij be |Qij|
hence expanding |A| by elements of its column no.1
|A| = sigma [ai1*|Pi1|]...from i=1 to n......................................1
similarly expanding |T| by elements of its row.no.1
|T| = sigma[t1i*|Q1i] from i=1 to n................................................2
INDUCTION:
1.let us test it for n=2
|A| = a11,a12
a21,a22
=A11*A22 - A12*A21
|T|= a11,a21
a12,a22
=a11*a22-a12*a21 = |A|
hence the theorem is correct for n=2
2.let us assume that the theorem is true for n th order determinant
that is |A| = |T| for order of n
3.let us prove that then it is true for n+1
expanding |A| and |T| of order n+1 by the elements in their first column and first row respectively,we get
|A| = sigma[ai1*|Pi1|] where |Pi1| is as we know a determinant of order n.
=sigma[t1i*|Pi1|].............since…. ai1=t1i
=sigma[t1i*|Q1i].............since for determinants of order n ,it is already assumed under 2, that the theorem is valid,hence |Pi1|=|its transpose|=|Q1i|
|A|=sigma[t1i*|Q1i|]=|T|

Questions - Differential Equations

Questions / Answers
Differential Equations

On 10/23/06, Anonymous wrote:
> Anonymous has left a new comment on your post "Questions - Differential
> Equations":
>
> Can someone help me with this? For a second order differential equation of
> the form y" = f(x,y'), the substitutions v = y', v' = y" lead to a first
> order differential equations of the form v' = f(x,v). Provided that this is
> differential equation for v can be solved, y can be obtained by integrating
> dy/dx = v(x) Notice that one arbitary constant is introduced in solving the
> differential equation for v and that another is introduced when integrating
> v(x) to obtain y. Solve the following differential equation x^2y" + 2xy' - 1
> = 0
>
>
>
> Posted by Anonymous to Maths,management,counselling at 9:26 AM
--------------------------------------------------------------------
x^2y"+2xy'-1=0
put
x=e^z....dx/dz=e^z=x.....dz/dx=1/x
y' = dy/dx = (dy/dz)(dz/dx)=Dy/x..where Dy represents dy/dz
y"= d^2y/dx^2 = (d/dx){dy/dx}=(d/dx){Dy/x}
=(1/x^2)[(d^2y/dz^2)-(dy/dz)]=(1/x^2)[D^2y-Dy]
=(1/x^2)[D(D-1)]y
so the given eqn. becomes
D(D-1)y+2Dy-1=0
[D^2-D+2D]y = 1
[D^2+D]y=1
D(D+1)y=1
c.f
y=c1e^(0z)+c2e^(-1z)......
p.i........
[1/D(D+1)]1...etc
this is a standard differential eqn. in z which you can solve for y.then substitute e^z =x
or z=ln(x)

Questions - Differential equations

Questions / Answers

On 10/23/06, Anonymous wrote:
> Anonymous has left a new comment on your post "QUESTIONS -
> Lillian.Blackwell":
>
> Need some help!! Consider a second order differential equation of the form
> y" = f(y,y'). If we let v equal y', we obtain v' = f(y,v). This equation
> uses the variables x, y and v and hence is not the form of the first order
> equations. However, it is possible to eliminate the variable x by thinking
> of y as an independent variable; the chain rule gives dv/dx = (dv/dy)(dy/dx)
> = (dv/dy)v, and hence the original differential equation can be written as v
> (dv/dy) = f(y,v) Provided this equation can be solved, we obtain v as a
> function of y. A relation between y and x results from solving dy/dx = v(y)
> Again there are two arbitary constants in the final result. Use this
> technique to solve the following differential equation. y" + y(y')^3 = 0
> thanks
Posted by Anonymous to Maths,management,counselling at 9:29 AM
----------------------------------------------------------------------------------
Differential Equations
y" +y(y')^3 = 0
put
y'=dy/dx = u
y" = du/dx = (dy/dx)/(dy/du)=udu/dy
hence
udu/dy+yu^3=0
du/dy = -yu^2
-du/u^2=ydy
integrating
1/u = y^2/2 +k
u=dy/dx=1/(k+y^2/2)....you can integrate to get arc tan function ...

PANCHATANTRA – 7 - 360 DEGREE ANALYSIS

360 DEGREE ANALYSIS

The students assembled around Vishnu Sarma in the evening and posed their query of the day….

Sir! Now a days we are hearing that students are being asked to report on their teachers. We know of teachers examining students and reporting. But this new trend appears strange!

Vishnu Sarma smiled and said -

Good! You can test and report on me!

Let me explain you with - again a story from an old classic journal.

There was a king. His washer-man who also serves as his body masseur, was massaging the king one day. He was always wondering about the hours of hard manual work he had to put in every day to serve the Palace and the disproportionately less compensation he got both monetarily and otherwise. On the other hand, he felt that the Minister hardly lifts a finger, only commanding others but gets a high compensation and respect from all. He could not control himself any more and told the king about his assessment and the purported injustice meted out to him. At that moment , there was a commotion out side the Palace and the king asked the washer man to find out about it.

The washer man went out and came back reporting that it was a procession going on.

The king enquired “what procession?”

The washer man went out again , came back reporting that it was a wedding procession.

The king enquired ”who is getting married ?”

The washer man went out and came back reporting that the son of the reputed merchant of the kingdom was getting married.

The king enquired “to whom?”

The washer man went out again , came back reporting that the bride was the daughter of another famous merchant.

By this time, the Minister came for his daily briefing with the king. The king then told him that he heard a commotion outside the palace a little while ago and wanted to know what it was about?

The Minister went out and came back in a few minutes and narrated all the details about the procession and suggested that it is a good development to promote trade and relations with the neighboring kingdom as the bride’s father is a very famous merchant in the neighboring kingdom and had a good access to the king there. If the king permits, the Minister concluded, he will take up the matter. The king gave his ascent and bade him good bye.

Then the king turned towards the washer-man to give his response to his submission. When the washer man exclaimed “pardon me Sir! I understood the difference” and left.

Well my dear that is the 360 degree reporting .

So it is important to consider all possibilities before jumping to one conclusion. We can call this as 360 degree out-look which is very much essential in every walk of life and learning 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”.

It was getting dark and the students got up to go to their homes pondering over

the moral.

Questions - Differential Equations

Questions/Answers
Differential equations.
Anonymous has left a new comment on your post "Questions - Continuous
Compounding":

Can someone help me with the question below? Seems long, but they are
straightforward answers. Question) Classify the following equations -
linear, nonlinear, seperable, exact, homogeneous, or one that requires an
integrating factors as follows: . If an integrating factor is required,
specifiy it, BUT DO NOT SUBMIT SOLUTIONS TO ANY OF THE EQUATIONS BELOW.

YOU GAVE 20 PROBLEMS !!!.I AM GIVING THE METHODS TO FOLLOW WITH ONE
EXAMPLE EACH. YOU MAY PROCEED ACCORDINGLY AND TRY.

LEGEND:- Y' = DY/DX..Y"=D^2Y/DX^2 ETC..
L..LINEAR...NL..NON LINEAR...
S...VARIABLE SEPERABLE
H...HOMOGENEOUS..
NH..NON HOMOGENEOUS
E....EXACT....NE.....NOT EXACT
IF...INTEGRATING FACTOR

LINEAR: ALL THE DERIVATIVES SHALL BE IN FIRST DEGREE.
EX. 5Y""+4Y"' +2Y" +Y'=5X

NONLINEAR:ONE OR MORE DERIVATIVES WILL BE OF DEGREE 2 OR MORE
Y'^2+Y=5X

SEPERABLE: ALL X TERMS CAN BE SEPERATED OUT FROM ALL Y TERMS AND PUT
ON EITHER SIDE OF =
EX. (Y^2+Y+1)DY=(X^2+1)DX

HOMOGENEOUS:WHEN Y = VX IS PUT IN THE EQN.ALL 'X' TERMS CANCEL OUT
LEAVING ONLY V AND Y
EX. DY/DX = (3X+4Y)/(2X+Y)...PUT Y=VX TO GET.....DY/DX =
X(3+4V)/X(2+V)=(3+4V)/(2+V)

EXACT:THE EQUATION CAN BE PUT IN THE FORM OF
DERIVATIVE OF F(X) = CONSTANT ,THEN IT IS CALLED EXACT.THE TEST FOR
EXACT EQN.IS IF......
MDX+NDY=0
THEN
DOM/DOY = DON/DOX..WHERE DOM/DOY
REPRESENTS PARTIAL DERIVATIVE OF M WRT X.
EX. XDY+YDX = C
LHS = D(XY)

INTEGRATING FACTOR:IF AN EQUATION IS NOT EXACT BUT CAN BE MADE INTO AN
EXACT EQN.BY MULTIPLYING WITH SOME FUNCTION OF X AND Y THEN THAT
FUNCTION IS CALLED 'INTEGRATING FACTOR'
EX. YDX-XDY = 0 IS NOT EXACT SINCE
DOM/DOY=1 AND DON/DOX = -1
BUT THIS CAN BE MADE BY MULTIPLICATION WITH 1/Y^2
(YDX-XDY)/Y^2 = D(X/Y)
---------------------------------------------------
(a) dy/dx = (x^3 -2y) / x

L.....NH.....NE .....IF=X

(b) (x + y)dx - (x -y)dy = 0

L.....H.......PUT Y=VX AND IT BECOMES S

(c) dy/dx = (2x + y)/ (3+ 3y^2 - x)

L.....NH......E

(d) (x + e^y)dy - dx = 0

L......NE.........IF = E^(-Y)

Questions - Continuous Compounding

Questions/Answers....Continuous Compounding.

Anonymous has left a new comment on your post "COMMENTS/RESPONSE MATHS -
> 6.EQUATIONS":
>
> Can someone help me? Determine the sum accumulated in 20years by each of
> the following investment programs. Suppose that the interest is 9.5%
> compounded continuously and that the deposits are also made continuously. a)
> Nothing initially and $1500 per year for 20 years b) $10,000 initially and
> $1000 per year for 20 years c) $20,000 initially and $500 per year for
> 20years d) $30,000 initially and no additional deposits Note that in each
> case the total amount deposited is $30,000 Thanks
> ------------------------------------------------------------------------
ANSWER
---------------------------------------------------------------------------
formula for continuous compounding is
amount = p*e^(nr).....where p=principal.....n= number of
years....r=interest rate in fraction.
in case of periodic investment the formula for continuous compounding is
IMPORTANT:-YOUR WRITE UP GIVES THE IMPRESSION THAT THE INITIAL
INVESTMENT IS AT THE BEGINING OF THE 20 YEAR PERIOD,WHERE AS THE
PERIODIC INVESTMENTS START AT THE BEGINING OF NEXT YEAR.THIS IS VERY
IMPORTANT AND MAKES A DIFFERENCE.NORMALLY,THE PERIODIC INVESTMENTS
ALSO START AT THE BEGINING OF EACH YEAR.PLEASE CHECK AND INFORM IF THE
ASSUMPTION IS CORRECT OR NOT.THE FOLLOWING SOLUTION IS BASED ON THAT
ASSUMPTION.
amount = y*[e^(n-1)r + e^(n-2)r + e^(n-3)r +...............+ e^2r +
e^1r + e^0r]...
this is g.p.with first term = 1 , common ratio = e^r and number of
terms = 20.its sum is
amount = y*[e^(nr) - 1 ] /[e^(r) -1]
IF INSTALMENTS ALSO START FROM DAY ONE OF THE 20 YEAR PERIOD THEN THE FORMULA IS
AMOUNT = Y*[ E^NR+ E^(N-1)R + E^(N-2)R +...........+ E^2R + E^1R ]
THIS IS A G.P. WITH FIRST TERM = E COMMON RATIO = E^R
AND NUMBER OF TERMS =20
AMOUNT = Y*E[E^(NR) - 1 ] / [E^(R) - 1 ]
hence for the four cases given the answer is as follows
X=initial= 0 10000 20000 30000
periodicY=1500 1000 500 0
amount=(X*(e^(nr)))+Y*((e^(nr)

)-1)/((e^r)-1)….n=20….r=0.095
amount= 85580.4 123912.5 162244.7 200576.8
col.1 is number of years for the investment to earn interest.col.2,3,4,5 are the 4 alternatives.
20 0 66858.9 133717.9 200576.8
19 9119.957173 6079.971449 3039.985724 0
18 8293.442216 5528.961478 2764.480739 0
17 7541.831885 5027.887923 2513.943962 0
16 6858.337793 4572.225195 2286.112598 0
15 6236.786764 4157.857843 2078.928921 0
14 5671.565081 3781.043388 1890.521694 0
13 5157.567781 3438.378521 1719.18926 0
12 4690.152548 3126.768365 1563.384183 0
11 4265.097785 2843.398524 1421.699262 0
10 3878.564489 2585.709659 1292.85483 0
9 3527.061571 2351.374381 1175.68719 0
8 3207.414331 2138.27622 1069.13811 0
7 2916.735782 1944.490521 972.2452607 0
6 2652.400577 1768.267051 884.1335257 0
5 2412.021296 1608.014197 804.0070987 0
4 2193.426884 1462.284589 731.1422947 0
3 1994.643042 1329.762028 664.8810141 0
2 1813.874396 1209.249598 604.6247988 0
1 1649.488283 1099.658855 549.8294276 0
0 1500 1000 500 0
TOTAL AMOUNT AT THE END OF 20 YRS.
85580.36968 123912.5242 162244.6787 200576.8333

QUESTIONS - Lillian.Blackwell

Questions/Answers..

Lillian.Blackwell to me

Could you please help me with finding all the points of inflection with the

below problem:

F(x) =x^4-18x^2+4

-----------------------------------------------

For points of inflection the following criteria must be satisfied……

AT x = a

1.F”(a)=0 or does not exist

2. But F’(a) exists and

3.F”(x) changes sign when passing through the point x=a

Then the point [x=a and y=F(a)]

is the point of inflection of the curve.

F(x) =x^4-18x^2+4

F’(x) = 4x^3-36x

F”(x) =12x^2-36

F”(x)=0 at 12(x^2-3)=0…or…x = =sqrt(3)….or…..- sqrt(3)

F’(x) exists at x=sqrt(3)…..&……x= -sqrt(3)

Now make a table as follows to study change of sign in F(x)

x…x is less than - sqrt(3)………-sqrt(3)sqrt(3).....x is more than sqrt(3)

sign of F”………+ ve……………………- ve …...........………………+ve

since F”(x) changes sign while passing through x=-sqrt(3)….and ….x=sqrt(3)

There are 2 points of inflection..they are

[-sqrt(3),9-54+4]……and ……[sqrt(3),9-54+4]

that is [-sqrt(3),-41] and [sqrt(3),-41]

Questions - Nadia Beninigton

Questions/Answers..

Nadia Beninigton

to me

please refer your diagram.

a) Name the point where POB is meeting the circle as Q.

PQB is a secant.

Then as per the theorem on tangent length and secant.we have

PT^2 = PQ*PB

3^2 = PQ*10

PQ=9/10

QB = diameter = 2*radius = PB-PQ=10-9/10 = 91/10

Radius = 91/20 = 4.55

b) Name one secant through center as PQOR where Q and R are the points where the secant meets the circle. Name the other secant as PST, where S and T are the points where the secant meets the circle.

Then as per the above theorem we get

PQ*PR = PS*PT

2*(PQ+QR)=PS(PS+ST)

2(2+diameter)=3(3+4)=21

4+2*2*radius=21

4R=21-4=17

Radius = 17/4 = 4.25

COMMENTS/RESPONSE MATHS - 6.EQUATIONS

I MAILED THE RESPONSE BUT IT BOUNCED.HENCE POSTING THE RESPONSE BELOW

Anonymous has left a new comment on your post "MATHS – 6.EQUATIONS":

Can you help me with this problem? A tank with a capacity of 500gallons originally containing 200 gallons of water with 50 pounds of salt in the solution. Water containing 2 pounds of salt per gallon is entering at a rate of 3 gallons per minute, and the mixture is allowed to flow out of the tank at a rate of 2 gallons per minute. Find the amount of salt in the tank at anytime t prior to the moment when the solution begins to overflow from the tank. Find the concentration (in pounds per gallon) of salt at the moment the tank begins to overflow. Compare this concentration with the theoretical limiting concentration if the tank had infinite capacity. Thanks

LET US FORMULATE THE PROBLEM
INITIAL SOLUTION IN TANK = 200 G
INITIAL SALT IN TANK = 50 LB.
INFLOW = 3 GPM..
WITH SALT CONTENT = 2 LB/G
OUT FLOW = 2 GPM.
NET INFLOW = 3-2 = 1 GPM

LET AT ANY TIME ....T......,MINUTES
THE SALT IN TANK BE S LB.
SOLUTION IN TANK 200 + T*1= (200+T) G
SALT CONCN. = S / (200+T) LB/G
AFTER DT M.
INFLOW = 3DT G
SALT IN = 3DT*2 = 6DT LB.
OUT FLOW = 2DT G
SALT OUT = 2DT*S / (200+T) LB.
CHANGE IN SALT CONTENT DURING DT IS
DS = SALT IN - SALT OUT
DS = 6DT - 2SDT / (200+T)
DS/DT = [1200+6T-2S] / (200+T)
THIS IS A STANDARD NON HOMOGENIOUS EQN.WHICH YOU CAN CONVERT INTO
HOMOGENEOUS EQN. BY PUTTING
200+T = U ...

SO THAT YOU GET A HOMOGENEOUS
EQN.....WHICH YOU CAN SOLVE BY SUBSTITUTION
S = V U
THE CONSTANT OF INTEGRATION CAN BE OBTAINED FROM S = 50 WHEN T=0
THIS WILL GIVE YOU A GENERAL EQN FOR SALT CONTENT TILL THE TANK GETS
FULL WHICH NEEDS (500-200) / 1 = 300 MTS.AS
1 GPM IS THE NET INFLOW
HENCE YOU CAN GET SALT CONTENT
AT T = 300 TO FIND THE ANSWER WHEN TANK IS ABOUT TO OVER FLOW.
TO GET CONCN.DIVIDE S BY TOTAL SOLUTION.

= S / (200+T)
OBVIOUSLY IF THE TANK HAS INFINITE CAPACITY, IT WILL EQUAL THE
CONCENTRATION OF INCOMING SOLUTION
THAT IS 2 LB/GALLON.

PANCHATANTRA – 6.PROJECT MANAGEMENT

The students assembled around Vishnu Sarma in the evening and posed their query of the day….

Sir! . You gave us some examples on communication which were illuminating. We realize that there could be much more to learn on that, but we have a nagging doubt on another aspect. We very often find senior administrators shifted and put in charge of large projects or assignments, once in petroleum, then in steel, then in power and so on and so forth. Many a times they are not even aware of any technical aspects of the task at hand.

How are they appointed like that and how do they Manage?

Vishnu Sarma replied –

Well, this once again is a large canvas to cover, but let me give the principal ideas suitable for your understanding to handle such cases of Project Management. Any job, big or small, which has a specific content, can be called a Project.

There are three things-one should try to watch and control,

1.Time schedule for completion,

2.Cost allocation for the project and

3.Specified quality of the product to be delivered.

Let us briefly dwell on each of these important aspects with simple examples.

1. Time schedule

This is often not given its due importance or is under-estimated, particularly in developing countries, though it is the most important feature. In-fact it can out weigh all other considerations. But, due to a horde of reasons including, lack of planning & accountability, bureaucratic hurdles & other indecision zones, they have a poor record on this score. However the situation is improving and there is a lot more awareness to day on this aspect than earlier.

To understand its importance, let us say you wanted a dress to be delivered for a wedding. Suppose it is delayed and delivered after the wedding. Then it becomes irrelevant.

Imagine an examination or a sports competition without time limit. They lose their relevance.

When a project is delayed, its fruits cannot be realized in time, the money spent will idle and a host of other losses may occur.

Our teacher used to convey it in a very telling manner in this way.

When a student failed to answer a question in time, he used to ask about his younger brothers or sisters and retort ‘why your younger sister can do this problem!’

When the student protested saying she cant do ,the teacher used to add coolly
'in 3 years!!!’, as the sister is 3 years younger. This in short conveys the message effectively. There is no use if the job is completed late after the due date.

2. Cost

By the same reasoning, if the cost of a Project escalates, it may not yield the desired profits and may even be borne sick. This aspect has a better appreciation, but failures in practice are innumerous, largely due to lack of accountability and planning.

3. Quality

Once again, though this is understood, it is usually the most often sacrificed resort by many as the easiest expedient. But it will have a telling effect in the end. You might have noted our use of the word ‘specified quality’. This is important and there is no necessity to go for superior quality if it is going to effect time and cost as they are generally interrelated. If a tiled flooring is specified, there is no necessity to go for marble flooring.

Hence, if the Project Manager takes care of these 3 aspects, with the aid of suitable experts, and facilitates the work towards that end, he can fulfill the requirements.

It was getting dark and the students got up to go to their homes pondering over the
moral.

MATHS – 6.EQUATIONS

Simultaneous equations involving 2 unknowns - x and y are often referred by school students for different methods of solution and more importantly to find whether they are consistent, independent or inconsistent. Let us see 3 simple examples and elaborate on this particular aspect of existence and uniqueness of solution.

1. x+y =2 and x-y =0
2. x+y =2 and 2x+2y = 4
3. x+y =2 and 2x+2y =3

We adopt 2 methods to determine this aspect.

1.Algebraic method

2.Graphic method

I.Independent equations

Test 1 . Algebraic method.

Find the ratios of coefficients of unknowns in the 2 given equations

In ex.1…it is

1/1 =1 for x

and

1/-1 =-1 for y.

the ratios are not equal

IF THE RATIOS ARE DIFFERENT,THEY ARE INDEPENDENT EQUATIONS AND WE GET UNIQUE SOLUTION.

Here we find that x = y = 1 is the one and only one solution.

In ex.2….it is

1 / 2 for x

and

1 / 2 for y

the ratios are equal.

IF THE RATIOS ARE EQUAL THEN THEY ARE DEPENDENT EQUATIONS AND WE CAN HAVE TWO POSSIBILITIES.

A. We may have infinite solutions

We test for this possibility by finding the ratio of constant terms. Here the ratio is 2 / 4 = 1 / 2 which is same as the ratio of coefficients of unknowns.

IF RATIO OF COEFFICIENTS OF UNKNOWNS IS SAME AS THE RATIO OF CONSTANT TERMS THEN THEY ARE DEPENDENT / CONSISTENT AND WE HAVE INFINITE SOLUTIONS.

Here we have x = y = 1 or x = 0 and y =2 or x=2 and y=0 etc..are all possible solutions. In-fact any set of values x = x and y = 2 – x is a solution.

PLEASE NOTE THE COMMENTS UNDER EQUATIONS AND IDENTITIES IN THIS CONTEXT

B. We may have no solution.

IF RATIO OF COEFFICIENTS OF UNKNOWNS IS NOT SAME AS THE RATIO OF CONSTANT TERMS THEN THEY ARE CALLED INCONSISTENT AND WE HAVE NO SOLUTION.

In ex.3..it is

1 / 2 for x

1 / 2 for y

and

2 / 3 for constants which is not equal to the above ratio.

Hence the equations are in-consistent and we have no solution.

Test 2…Graphic method.

Draw graphs of the 2 equations on the same scale on the same paper.

If the 2 graphs intersect , then the point of intersection is the solution.

In Ex.1 we find that the 2 graphs…straight lines in fact intersect
at x = y = 1.Hence they are independent equations and the point of intersection viz.x = y = 1 is the solution.

In Ex.2 …we find that the 2 graphs ….straight lines in fact are coincident. Hence they are dependent/consistent and any point on the line is a solution. Say x = y = 1 or…x=2 and y =0 etc..

In Ex.3...we find that th 2 graphs...straight lines in-fact are parallel and do not intersect.hence, they are inconsistent equations and there is no solution.