## Thursday, 26 December 2013

### Motion Of A Particle

Principles in  the formation of the Equation of Motion of a Particle :-

For Motion in Straight Line ;

Let the mass of the body (particle) be    m    and let the distance of the particle measured from a suitable origin be   x   at the time  t   .

Then acceleration is      (d/dt)(dx/dt)

therefore by Second Law of Motion

m[(d/dt)(dx/dt)] = Forces in the direction of    x   increasing

The forces usually are ;-

1)---   g    vertically downwards in a gravitational field like earth .
2)---  tension in a string
3)---  reactions or stress at point where the point may be in contact with other particles .
4)---  forces of attraction
5)---  forces of resistance to motion by any means may be by atmosphere , friction, winds or by any other means .

In  taking the forces we must fix the sign properly if the forces acts along the line .

If a force  F   acts at an angle    A     to the straight line , then the resolved part of the force along the line is    FcosA  . If   acts on on a particle and if the particle is moving along the horizontal line , then   g    , being vertical , has no resolved part horizontally  and    does not effects the Motion .

Also note that velocity,    v=dx/dt     , and acceleration

(d/dt)(dx/dt) = dv/dt = (dx/dt)(dv/dx) = v(dv/dx)

For Motions in two dimension;

Let two perpendicular axes are   X      and    Y     at the plane of motion . The forces are resolved in these two directions and respectively

m(d/dt)(dx/dt)       and       m(d/dt)(dy/dt)

## Monday, 23 December 2013

### Equations of First Order

In this post we will discuss about the equation of first order but not first degree .It is usually denoted

dy/dx    by   p .

Thee are three types of such equations

1) Equations solvable for .
2)Equation solvable for    y  .
3)Equation solvable for    .

Equation Solvable for   p   :-

examples like this    p.p +2py  cot x = y.y   and its solution is

[y-(c/1+cos x)][y-(c/1-cos x)] = 0

is the equation solvable for   p   .

Equation Solvable for    y   :-

Let in the given differential equation , on solving for   y   , given that ;

y=f(x,p)  ---------------(1)

Differentiating with respect to    , we obtain ;

p=dy/dx=A(x,p,dp/dx)

so that we obtain a new differential equation with variables     and     .

Suppose that it is possible to solve the equation

Let the solution be
F(x,p,c)=0  ----------(2)
where     is the arbitrary constant .

The equation of  (1)  may be exhibited in either of the two forms . We may either eliminate   p   between  (1)  and  (2)  and obtain  A(x,y,c)   as the required solution or we may solve   (1)  and  (2)  for  x , y   and obtain .

x=f'(p,c)   and   y=f"(p,c)

as required solution where   p  is the parameter .

Equations Solvable for    x   :-

Let the given differential equation , on solving for  x   , gives

x=f(p,y)  -----------------------(1)

differentiating with respect to   y  we obtain

1/p=dy/dx=A(y,p,dp/dx)  ; say

So that we obtain a new differential equation in variables    y      and   p   , Suppose that it is possible to solve the equation .

Let the solution be
F(p,y,c)=0   -------------------(2)

After the elimination    p   between    (1)   and   (2)     will give the solution .  Express   x  and    in terms    of    and   c   where    p   is to be regarded as parameter .

### Equations of Tangent and Normal

Explicit Cartesian Equations :-

If      be the angle which the tangent at any point   (x, y)   on the curve     y = f (x)     makes with    x   axis then ;
tan A = dy/dx = f' (x)

Therefore , the equation of the tangent at any point     (x , y)     on the curve    y = f (x)      is

Y - y = f' (x) (X - x)  -------------(1)

where   X , Y   are the current co-ordinates of any point on the tangent .

The normal to the curve    y = f (x)    at any point    (x , y)     is the straight line which  passes through that point ans is perpendicular to the tangent to the curve at the point so that its slope is ;

-1/f (x)

Hence the equation of the normal at   (x , y)    to the curve    y= f (x)    is ;

(X - x) + f' (x) (Y - y) = 0

Implicit Cartesian Equations :-

If any point    (x , y) , then the curve   f (x, y) = 0

Where   Dy/Dx    is not equivalent to  0   .

dy/dx =  - (Df/Dx) / (Df/Dy)

Hence the equations of the tangent and the normal at any point
(x , y)   on the curve    f (x , y) = 0  are ;

(X - x)(Df / Dx) + (Y - y) (Df / Dy) = 0      and

(X - x) (Df / Dy) - (Y - y)(Df / Dx) = 0

Parametric Cartesian Equations  :-

At the pont  of the curve    x = f (t) , y = F(t)  ;
where we have  f'(t)   is not equivalent to   0    ;
we have ;

dy/dx = (dy/dt) (dt/dx) = F' (t)/f' (t)

Hence the equations of the tangents and the normal at any point    t    of the curve    x=f(t) , y=F(t)   are ;

[X-f(t)]F'(t)-[Y-F(t)]f'(t)=0
[X-f(t)]f'(t)+[Y-F(t)]F'(t)=0

respectively .

## Sunday, 22 December 2013

### Maxima and Minima

Greatest  and Least Value :-

In this section we shall be concerned with the application of Calculus to determining the values of a function which are greatest or least in their immediate neighborhood technically known as  Maximum and Minimum Values .

It will be assumed that    f (x)   possesses continuous derivatives of every order that come in equation .

Maximum Value of a Function :-

Let     be any interior points of the interval of definition of a function   f (x)    , if it is the greatest of all its value for values of   x   lying in some neighborhood of   c   . To b more definite and to avoid the vague words   "Some Neighborhood" we say that    f (c)    is a maximum value of function if there exists some interval     ( c - D , c + D )       around     c   such that   .

f (c) > f (x)

for all values of     other then      c   lying in the interval

So that ,      f (c)    is  maximum value of      f (x)     if .

f (c) > f ( c+ h )     i.e.       f ( c + h ) - f (c) < 0

where value of    h   lying between   -D   and   D  .

Minimum Value of a Function :-

f (c)   is said to be a minimum value of    f (x)    , if it is the least of all its values for values of   x    lying in some neighborhood of     c    .

This is equivalent to saying that     f (c)    is a minimum value of    f (x)   ,if there exist a positive     D     such that ;

f (c) < f ( c + h ) ,   i.e.      f ( c + h ) - f (c) > 0

for value of    h    lying between    -D   and   .

For values of   h     sufficiently small in numerical value  .

The term Extreme Value is used both for a maximum as well as for a minimum value , so that    f (c)   is an extreme value if    f ( c+ h ) - f (c)     keeps on invariable sign for value of     h    sufficiently small numerically .

## Friday, 20 December 2013

### Velocity and Acceleration

Velocity and Acceleration of a Moving Particles :-

If a particle is moving along a straight line , and if at any instant     t     the position        of the particle be given by the distance   s    measured along the path from a suitable fixed point    on it , then      denoting the velocity and   f   the acceleration of the particle at the instant ,

We have;

v = rate of displacement

= rate of change of   s   with respect to time

= ds/dt ;

and ,           f = rate of change of velocity with respect to time

= dv/dt

= ( d/dt ) ( ds/dt )

If instead of moving in a straight line , the particle be moving in any manner in a plane , the position of a particle at any instant   t   being given by the Cartesian Co-ordinates    x , y     referred to a fixed  set of axes the components of velocity and acceleration parallel to those axes will similarly be given by

vx = rate of displacement parallel to    x-axis  = dx/dt

vy = rate of displacement parallel to    y-axis   = dy/dt

fx = rate of chang of     vx = ( d/dt ) ( dx/dt )

fy = rate of change of    vy = ( d/dt ) ( dy/dt )

## Thursday, 19 December 2013

### Differential Equations

In this post we will discuss about "Formation of Ordinary Differential Equations"

Let ;     f ( x , y , c1 ) = 0  ---------------------(1)

be an equation containing   x ,  y  an  on arbitrary constant   c1   .

Differentiating  (1)  we get ;

( Df/Dx ) + ( Df/Dx ) (dy/dx ) = 0 ------(2)

equation  (2)  will in general contains   c1   . If   c1  be eliminated between  (1)  and  (2)  , we shall get a relation involving   x,  y  and   dy/dx  which will evidently be a differential equation  of the first order .

Similarly , if we have an equation

f ( x , y , c1 , c2 ) = 0 ----------------(3)

containing two arbitrary constant  c1  and  c2  , then by differentiating this twice , we shall get two equations . Now between these two equations and given equations , in all three equations , if the two arbitrary constant   c1  and   c2  be eliminated , we shall evidently get a differential equation  of the second order .

in general , if we have an equation ;

f ( x , y , c1 , c2 ,  .......cn ) = 0 ----------(4)

containing  n  arbitrary constants  c1 , c2 , .....cn      then by differentiating this   n  times , we shall get    equations . Now between these    n   equations  and the given equation in all   ( n+1 )   equations , if the   n   arbitrary constants   c1 , c2 , ...cn   be eliminated , we shall evidently get a differential equation for  n th    order , for there being    n   differentiations , the resulting equation must contain a derivative of the   n th order .

### Homogeneous Equations

We have to understand about Homogeneous Equation in Calculus .

If  and  N  of the equation   Mdx + Ndy = 0   are both of the same degree in  x  and  y   , and are homogeneous , the equation is said to be homogeneous  . Such an equation can be put in the form

dy/dx = f ( y/x )

Every homogeneous equation of the above type can be easily solved by putting     y = vx
where   v  is a function of , and consequently

dy/dx = v + x ( dv/dx )

whereby it reduced to the form     v + xdv/dx = f ( v )

i.e.        dx/x = dv/ [ f(v) - v ]

in which the variables are separated

A Special Form :-

The equation of the form

dy/dx = [ ( a1x +b1y +c1 ) / ( a2x + b2y +c2 ) ]

where ,  [ a1/a2 is not equal to b1/b2 ] --------------(1)

can be easily solved by putting
x = x' + h
and              y = y' + k
where   and are constant

So that ,              dx = dx'    and      dy = dy'
and choosing   h  and  k  in such a way that

a1h + b1k + c1 = 0    and    a2h + b2k + c2 =0        -------(2)

For now the equation reduces to the form

dy'/dx' = ( a1x' +b1y' ) / ( a2x' + b2y' )

which is homogeneous in   x'  and  y'  and hence solved .

## Monday, 16 December 2013

### Clairaut's Equation

An equation of the form

y = px + f (p) , where p = dy/dx

is called Clairaut's Equation .

Differentiating both sides of the equation with respect to   x   , we have  ,

p = p + xdp/dx + f' (p) dp/dx  or   dp/dx {x + f' (p)} = 0

therefore , either ,  dp/dx = 0 ------------ (1)

or ,     x + f' (p) =0 ------------(2)

from (1) ,    p = C --------------(3)

Now if    p    be eliminated between  (3)  and the original equation , we get   y = Cx + f(C)  as the general or complete solution of the equation .

Again , if  p  be eliminated between  (2)  an the original equation , we shall obtain a relation between   x  and  y  which also satisfies the differential equation , and as such can be called a solution  of the given equation . Since this solution does not contain any arbitrary constant nor can it be derived from the complete solution by giving any particular value to the arbitrary constant , it is called the Singular Solution of the differential equation .

Thus we see that the Equation of Clairaut's form has two kinds of solution .

a)   The complete solution ( linear in  x  and  y  ) containing one arbitrary constant .

b)   The singular solution containing no arbitrary constant .

Now , to eliminate  p  between

y = px + f(p)  and   0 = x + f'(x)

is the same as to eliminate   between ,

y = Cx + f(C)   and   0 = x + f'(C)

i.e. , the same as the process of finding the envelop of the line  y = Cx + f(C)  for different values of  .

Thus , the singular solution represents the envelope of the family of straight lines represented by the complete solution .

## Thursday, 12 December 2013

### Method of Isoclines

It is only in the limited number of cases that a differential equation may be solve analytically be the preceding methods , and in many practical cases where the solution of a differential equation is needed under given initial conditions and the above methods  fail , a graphical method , the method of isoclines is sometimes adopted . We proceed to explain below this method in case of simple differential equation of the first order .

Let us consider an equation of the type

dy/dx = f (x ,y ) ------------(1)

As already explained before , the general solution of this equation involves one arbitrary constant of integration . and hence represents a family of curves and in general , one member of the family passes through a given point  ( x , y ) .

Now if in  (1)  we replace   dy/dx  by   m   we get an equation   f( x , y) = m    , which for any particular numerical value of   m  represents a curve , at every point of which the value of  dy/dx  i.e.  the slope of the tangent line to the family of curves represented by the general solution of   (1)  is the same as the numerical value of   . This curve   f(x , y) = m   is called an Isoclinal or Isocline .

Which may be graphically constructed on a graph paper

Through different points on anyone isocline , short parallel lines are drawn having their common slope equal to the particular value of  m  for that Isocline .

## Tuesday, 10 December 2013

### Simpson's Rule

Approximate evaluation of a definite integral :
Simpson's Rule .

In many cases , a definite integral can not be obtained either because the quantity to be integrated can not be expressed as a mathematical function , or because the indefinite integral of the unction itself can not be determined directly . In such cases formula of approximation are used . One such important formula is Simpson's Rule . By this rule the definite integral of any function is expressed in terms of the individual values of any number of ordinates within the interval , by assuming that the function within each of the small ranges into which the whole interval may be divided can be represented to a sufficient degree of approximation by a parabolic function .

In numerical analysis, Simpson's rule is a method for numerical integration, the numerical approximation of definite integrals. Specifically, it is the following approximation:
$\int_{a}^{b} f(x) \, dx \approx \frac{b-a}{6}\left[f(a) + 4f\left(\frac{a+b}{2}\right)+f(b)\right].$
Simpson's rule also corresponds to the three-point Newton-Cotes quadrature rule.

Suppose that the interval $[a, b]$ is split up in $n$ subintervals, with $n$ an even number. Then, the composite Simpson's rule is given by
$\int_a^b f(x) \, dx\approx \frac{h}{3}\bigg[f(x_0)+2\sum_{j=1}^{n/2-1}f(x_{2j})+ 4\sum_{j=1}^{n/2}f(x_{2j-1})+f(x_n) \bigg],$
where $x_j=a+jh$ for $j=0, 1, ..., n-1, n$ with $h=(b-a)/n$; in particular, $x_0=a$ and $x_n=b$. The above formula can also be written as
$\int_a^b f(x) \, dx\approx\frac{h}{3}\bigg[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+\cdots+4f(x_{n-1})].$

The error committed by the composite Simpson's rule is bounded (in absolute value) by
$\frac{h^4}{180}(b-a) \max_{\xi\in[a,b]} |f^{(4)}(\xi)|,$
where $h$ is the "step length", given by $h=(b-a)/n.$

In other words this Simpson's Rule can be written as  :

h/3 [sum of the extreme ordinates  +  2.sum of the remaining odd ordinates   +  4.sum of the even ordinates]

## Saturday, 7 December 2013

### Areas of Plane Curves

Areas in Cartesian Co - ordinate :-

Suppose we want to determine the area A1 bounded by the curve   y = f(x)   , the x - axis an two fixed ordinates   x = a   and   x = b    . The function   f(x)  , is supposed to be single - valued , finite and continuous in the interval  (a , b ) .

The process of finding the area , bounded by any defined contour line is called Quadrature , "the term meaning the investigation of the size of a square which shall have the same area as that  of the region under consideration" .

Here are the example of some Plain Curves :-

Name Implicit equation Parametric equation As a function graph
Straight line $a x+b y=c$ $(x_0 + \alpha t,y_0+\beta t)$ $y=m x+c$
Circle $x^2+y^2=r^2$ $(r \cos t, r \sin t)$
Parabola $y-x^2=0$ $(t,t^2)$ $y=x^2$
Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a \cos t, b \sin t)$
Hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $(a \cosh t, b \sinh t)$

## Thursday, 5 December 2013

### Standared Methods of Integration

The different methods of Integration will aim at reducing a given Integral to one of the Fundamental or known Integrals . As a matter of facts , there are two principal processes :

1)   The method of substitution , i.e. the change of the independent variable .

Integration by substitution can be derived from the fundamental theorem of calculus as follows. Let ƒ and ϕ be two functions satisfying the above hypothesis that ƒ is continuous on I and ϕ is continuous on the closed interval [a,b]. Then the function ƒ(ϕ(t))ϕ(t) is also continuous on [a,b]. Hence the integrals
$\int_{\phi(a)}^{\phi(b)} f(x)\,dx$
and
$\int_a^b f(\phi(t))\phi'(t)\,dt$
in fact exist, and it remains to show that they are equal.
Since ƒ is continuous, it possesses an antiderivative F. The composite function Fϕ is then defined. Since F and ϕ are differentiable, the chain rule gives
$(F \circ \phi)'(t) = F'(\phi(t))\phi'(t) = f(\phi(t))\phi'(t).$
Applying the fundamental theorem of calculus twice gives
\begin{align} \int_a^b f(\phi(t))\phi'(t)\,dt & {} = (F \circ \phi)(b) - (F \circ \phi)(a) \\ & {} = F(\phi(b)) - F(\phi(a)) \\ & {} = \int_{\phi(a)}^{\phi(b)} f(x)\,dx, \end{align}
which is the substitution rule.

2) Integration by Parts ;

Integrating the product rule for three multiplied functions, u(x), v(x), w(x), gives a similar result:
$\int_a^b u v \, dw = u v w - \int_a^b u w \, dv - \int_a^b v w \, du.$
In general for n factors
$\frac{d}{dx} \left(\prod_{i=1}^n u_i(x) \right)= \sum_{j=1}^n \prod_{i\neq j}^n u_i(x) \frac{du_j(x)}{dx},$
$\Bigl[ \prod_{i=1}^n u_i(x) \Bigr]_a^b = \sum_{j=1}^n \int_a^b \prod_{i\neq j}^n u_i(x) \, du_j(x),$