Wednesday, 26 February 2014

Criteria of Maximum and Minimum


We have these two criteria for judging whether a function has a maximum or minimum at a particular point .

For a maximum at   x=c  :- 

---  Criterion A :-  (1) dy/dx=0 and
       (2) dy/dx  is possitive at  c-h  ;

           dy/dx  is negative at   c+h

---  Criterion B :-  (1) dy/dx=0  and

        (2)  (d/dx)(dy/dx)  is negative 

For  a minimum at   x=d  :-

---  Criterion A :-  (1)  dy/dx=0  and

      (2)  dy/dx  is negative at  d-h  ;

           dy/dx  is positive at  d+h  

---  Criterion B :-  (1)  dy/dx=0  and

     (2)  (d/dx)(dy/dx)  is positive . 

Monday, 24 February 2014

Partial Derivatives


Consider   z=f(x,y) 

If we differentiate  w.r.t.  x   considering   y   as constant , we get the partial derivatives as


= dz/dx  or  fx  

Similarly keeping  constant we can differentiate  w.r.t.  y  we get ;


=dz/dy   or  f 

If we differentiate   df/dx   again w.r.t.   x  keeping    y   as constant  

we get the second order partial derivatives 


If we differentiate   df/dx  w.r.t.   y ,  keeping    x  as constant ,

we get another second order partial derivatives


Similarly two more derivatives


and (d/dy)(df/dy) .

Saturday, 22 February 2014

Determination of Singular Solutions


Let ;       phai(x,y,dy/dx)=0

be a given differential equation and

let ;    f(x,y,c)=0

be its general solution .

Now it is known that the envelope of any family of curves

        f(x,y,c)=0  ----------  (1)

is contained in the locus obtained on eliminating ,    c      between  (1) and

[delta f(x,y,c)]/[delta c]=0  ---  (2)

Let this eliminate be ,

   sai (x,y)=0  ----------   (3)

As the elimination given by  (3)  may represent loci other than the envelope .

It is necessary to verify if any part of locus represented by  (3)  is or is not a solution of the given differential equation .

Friday, 21 February 2014

Simple Pendulum


A particle of mass   m   is attached by a light inextensible string of length     to a fixed point   O.   It oscillates in a vertical plane under the force of gravity through a small angle .

We have to fined the period of oscillation .

Let    be the point of suspension, 
OA   the vertical .
At any time  t,   let   P  be the position of the particle,
Angle AOP   being   theta redians    and
arc AP   being   s.

The force that act on the particle are its weight   mg  vertically downwards and the tension of the string along   PO.

Resolving along the   tangent PQ  to the circle

m(d/dt)(ds/dt)=-mg sin theta   ------ (1)

The amplitude of the oscillation is small and

therefore  sin theta=theta

since ;  s=theta l ,

equation  (1)  reduces to

(d/dt)(ds/dt)=gs/l      -------------- (2)

Thus the motion is simple harmonic and the period of a Complete Oscillation is

2 pai multiplied by under root of (l/g)

Thursday, 20 February 2014

Arcs and Chords


Let us take the arc of a curve and a fixed point   on it . Now take a variable point   Q   on the curve and let    Q--->P.     Then it is definitely ;

lim Q-->P(arcPQ/chordPQ)=1

Length of arc as a function :- 

Let    y=f(x)   be the equation of a curve on which we take a fixed point    A   .

To any given value of   x   corresponds a value of

  y, viz.,f(x) ;

To that pair of numbers   x    and   f(x)   corresponds a point     on the curve, and this point   has some arcual length   s    from   A.  

 Thus   "s"   is a function of   x   for the curve


Similarrly , we can see that   "s"   is a function of parameter   "t"   for the curve

x=f(t),   y=F(t)
         ------------Parametric Equation

and in function of   theta   for the curve

        ------------Polar Equation

Wednesday, 19 February 2014

Orthogonal Trajectories


Definition :-  A curve which cuts every member of given family of curves according to a given law is called a trajectory of the given family .

----   We shall consider only the case when each trajectory cuts every member of a given family at a constant angle . The trajectory will be called orthogonal , if the constant angle is a right angle . For example , every line through the origin of co-ordinates is an orthogonal trajectory of the family of concentric circle with center at the origin .

How to fined the orthogonal trajectories of the family of curves  

where    c    is a parameter .

Let ,  phai(x,y,dy/dx)=0 

be the differential equation of the family of curves given by


 If     (dy/dx)=m   at a point   (x,y)  on one of the curves of the system and if another curve cuts that curve at right angle , then    m'   its slope must be given by the equation


therefore ; m'=-1/m=-dx/dy 

Hence at  (x,y)  on the orthogonal trajectory , these equation must be satisfied


Hence this is the differential equation of the orthogonal system .

Monday, 17 February 2014

Singular Solutions


In addition to the General Solutions and Particular Solutions, obtained by giving particular values to the arbitrary constant in general solution a differential equation , may also posses other solutions . The solutions of differential equations , other then the general and particular , are known as Singular Solutions . In this connection , we have the following result .

Whenever the family of curves

f(x,y,c)=0     --------------(1)

represented by the differential equation

phai(x,y,dy/dx)=0    -------(2)

posses an envelope , the equation of the envelope is the singular solution of the differential equation (2) .

Suppose that the family of curves possesses an envelope. Take any point    P(x,y)      on the envelope . They exists a curve of the family , say;


Which touches the envelope at    (x,y).   The values of    x,y,dy/dx    for the curve at     satisfies the given differential equations. Also the value of   x,y,dy/dx   at    for the envelope are the same as for the curve. Thus we see that the values of the   x,y,dy/dx   at every point of the envelope satisfy the given differential equation. Hence the equation of the envelope is a solution of differential equation.

This solution does not contain any arbitrary constant and in general can not be obtained from the general solution by giving particular values to the arbitrary constant .

Thursday, 13 February 2014

Some Working Rules in Calculus


Here are some working rules used in Differential Calculus for solving any problem :-

Working Rule for Differentiation of Implicit Function

1)   Differentiate the given relation between    and    with respect to    x   .

2)   Bring the terms containing     dy/dx     on one side .

3)   Divide both side by co-efficient of    dy/dx    , this will give      dy/dx   .

4)   In order to simplify the value of    dy/dx   , use the relation between    x   and     .

Working Rule for Inverse Circular Functions

Simplify the given expression , For example ;
If   (1/tan) z   it is to be differentiated then put    in the form of    tan(theta)   ;
So that ; [1/tan z] = [1/tan] [tan(theta)] = theta .

For this certain substitutions are helpful , they are ;

If   square of   a  -  square of   x  occurs
put    x=a sine(theta)  or  a cos(theta)

If   square of   a  +  square of   x  occurs
put    x=a tan(theta)   or  a cot(theta)

If   square of   x  -  square of   a  occurs
put   x=a sec(theta)   or   a cosec(theta)

If  (a+x)/(a-x)  or  (a-x)/(a+x)   occurs

put   x=a cos(2theta) .

If you follow this working rules you will see that the solution of problem in differential calculus will be easy . If anyone need that working rules are help them to solve the problem then from my next blog we will discuss many working rules like this .

Sunday, 9 February 2014

Pedal Equation


The   (p , r) or Pedal Equation of Curve :-

We have been acquainted with two types of equation of any curve ; one Cartesian Equation    (x,y)   and the other the Polar Equation containing    (r,theta) . When the equation of any curve is given in terms of   (p,r)   where     p   is the length of the perpendicular from the pole on the     tangent   and   r    is the radius vector , then that form of the curve  is called the Pedal Equation .

Find the pedal equation of a curve from its polar form :-

Let the polar equation of any curve be ,

      f(r,theta)=0  --------------------(1)

Let the coordinates of any point on the curve be    (r,theta)    and let the length of perpendicular from the pole on the tangent at    (r,theta)    be    p    .

If     phai    be the angle between the tangent and the radius vector ,

then we know that ,

tan (phai) = r .d theta /dr  ----------(2)

and   p=r . sin(phai)  ---------------(3)

Now , if we eliminate    theta     between the equations  (1) , (2) and (3) then we shall get an equation in terms of   p   and  r   and thus will be required an equation of the curve .

Friday, 7 February 2014

De Moivre's Theorem


We know that the trigonometrical form of a complex number    is given by  

          z = r(cos theta + i sin theta)

 where ,    r =  /z/ , theta = arg z 

The product of two complex number is a complex number , As such 

 n  th  power of (a+ib)  is also a complex number .

But the methods of ordinary Algebra do not provide us with any precious method for computing

 n  th  power of (a+ib)   where  may be an integer or fraction .

De Moivre's Theorem  helps us to compute the value of

 n  th  power of (a+ib)  by changing it in a trigonometrical form .

The general enunciation of  De Moivre's Theorem :-  

For all values of    and   theta   , real or complex ;

cos n theta + i sin n theta      is a value of

n th power of    cos theta + i sin theta 

The theorem holds for real and non real complex values of    theta    and     n 

The expression     cos theta + i sin theta   is some times abbreviated to

cos theta

So , De Moivre's Theorem  is ;

n th power of    cos theta + i sin theta  

= cos n theta + i sin n theta 

Sunday, 2 February 2014

Newtonian Law of Attraction

 Diagram of two masses attracting one another

Newtonian Law of Attraction :-

Every particle of matter attracts every other particle of matter with a force which varies directly as the product of the masses of the particles and inversely as the square of the distance between them . 

Let   m1 gm   and    m2 gm   be the masses of the two particles placed at a distance of the    r cms   .  Then the force of attraction      between them is such that

                             F   varies   m1 , m2    when is constant .

                            F    varies   1/(r.r)    when  m1 , m2    are constant

Therefore by theorem of joint variation   F   is given by

            F  varies   (m1m2) /(r.r)

i.e.   F= gama  (m1m2)/(r.r)

         where   gama   is called constant of attraction  .

So , the Newtonian law of attraction is 
            Attraction = gama (product of mass)/(square of distance)

Important :- 

  --------  The attraction of a rod   AB   at an external point    is the same as that of the arc of a circle , of like material with center   P   and of radius equal to perpendicular from   P   on   AB   , which is intercepted by the lines   PA   and    PB   .

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