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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 .

Partial Derivatives

Consider   z=f(x,y)

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

df/dx

= dz/dx  or  fx

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

df/dy

=dz/dy   or  fy

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

we get the second order partial derivatives

(d/dx)(df/dx)

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

we get another second order partial derivatives

(d/dy)(df/dx)

Similarly two more derivatives

(d/dx)(df/dy)

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

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 .

Simple Pendulum

A particle of mass   m   is attached by a light inextensible string of length    l   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   O   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  T  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)

Arcs and Chords

Let us take the arc of a curve and a fixed point   P  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    P   on the curve, and this point  P   has some arcual length   s    from   A.

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

y=f(x)

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

r=f(theta)
------------Polar Equation

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

f(x,y,c)=0

where    c    is a parameter .

Let ,  phai(x,y,dy/dx)=0

be the differential equation of the family of curves given by

f(x,y,c)=0

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

mm'=-1

therefore ; m'=-1/m=-dx/dy

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

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;

f(x,y,c')=0

Which touches the envelope at    (x,y).   The values of    x,y,dy/dx    for the curve at    P   satisfies the given differential equations. Also the value of   x,y,dy/dx   at   P   for the envelope are the same as for t…

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   x   and   y   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   y    .

Working Rule for Inverse Circular Functions

Simplify the given expression , For example ;
If   (1/tan) z   it is to be differentiated then put   z   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)…

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 …

De Moivre's Theorem

We know that the trigonometrical form of a complex number   z   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  n  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   n   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…

Newtonian Law of Attraction

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    m2gm   be the masses of the two particles placed at a distance of the    r cms   .  Then the force of attraction    F    between them is such that

F   varies   m1 , m2    when r  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   P   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  …