Statement:-
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) then there exists at least one c in (a,b) such that f’(c)/g'(c) = f(b)-f(a)/g(b)-g(a).
Proof:-
Let us define a function
F(x) = f(x){g(b)-g(a)} - g(x){f(b)-f(a)}
This function is continious in [a,b] and differentiable in (a,b), since f(x) and x are continious in [a,b] and differentiable in (a,b).
Here F(a) = f(a){g(b)-g(a)} - g(a){f(b)-f(a)}
= f(a)g(b)-g(a)f(b)
F(b) = f(b){g(b)-g(a)} - g(b){f(b)-f(a)} = F(a)
The by Rolle's Theorem there exist at least one point c in (a,b) such that F'(c) = 0.
f'(c){g(b)-g(a)} = g'(c){f(b)-f(a)}
f’(c)/g'(c) = f(b)-f(a)/g(b)-g(a) g'(x) not equal to 0 in (a,b) and g(b) not equal to g(a)
Monday, June 21, 2010
Lagrange's Theorem
Statement:-
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) then there exists at least one c in (a,b) such that f’(c) = f(b)-f(a)/b-a.
Proof:-
Let us define a function
F(x) = f(x)(b-a)-[f(b)-f(a)]x........(i)
which is continious in [a,b] and differentiable in (a,b), since f(x) and x are continious in [a,b] and differentiable in (a,b).
Here F(a) = f(a)(b-a)-[f(b)-f(a)]a
= bf(a)-af(b)-af(b)+af(a)=bf(a) - af(b)
F(b) = f(b)(b-a)-[f(b)-f(a)]b
= bf(b)-af(b)-bf(b)+bf(a)
=bf(a)-af(b) = F(a)
The by Rolle's Theorem there exist at least one point c in (a,b) such that F'(c) = 0.
But F'(c) = f'(c)(b-a)-[f(b)-f(a)].
f'(c)(b-a)-[f(b)-f(a)] = 0
f'(c) = f(b)-f(a)/b-a
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) then there exists at least one c in (a,b) such that f’(c) = f(b)-f(a)/b-a.
Proof:-
Let us define a function
F(x) = f(x)(b-a)-[f(b)-f(a)]x........(i)
which is continious in [a,b] and differentiable in (a,b), since f(x) and x are continious in [a,b] and differentiable in (a,b).
Here F(a) = f(a)(b-a)-[f(b)-f(a)]a
= bf(a)-af(b)-af(b)+af(a)=bf(a) - af(b)
F(b) = f(b)(b-a)-[f(b)-f(a)]b
= bf(b)-af(b)-bf(b)+bf(a)
=bf(a)-af(b) = F(a)
The by Rolle's Theorem there exist at least one point c in (a,b) such that F'(c) = 0.
But F'(c) = f'(c)(b-a)-[f(b)-f(a)].
f'(c)(b-a)-[f(b)-f(a)] = 0
f'(c) = f(b)-f(a)/b-a
Rolle’s Theorem
Statement:-
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) and f(a) = f(b), there exists at least one c in (a,b) such that f’(c) = 0.
Proof:-
Since f(x) is continuous in closed interval [a,b], there exist c,d in [a,b] such that
f(c) = M(the maximum value of f(x))
f(d) = m the minimum value of f(x)
CASE I
If M = m, then f(x) is constant in [a,b] and in this case f’(x) = 0 for all x in (a,b). That is theorem is true in this case.
CASE II
If M = m, then at least one of them is different from f(a) and f(b). Let M = f(a). We shall show that f’(c) = 0 with c in (a,b)
At x = c, with h>0
RHD = lim h 0 f(c+h)-f(c)/h
= lim h 0 f(c+h)-M/h = -ve/+ve <= 0 ……….i
LHD = lim h 0 f(c-h)-f(c)/-h
= lim h 0 f(c-h)-M/-h = -ve/-ve >= 0 ……….ii
Since f’(x) exist in (a,b), I and ii must be equal and this will be possible only when both i and ii are equal to zero.
Hence f’(c) = 0 with c in (a,b).
CASE III
If m is different from f(a) and f(b) we can similarly show that f’(d) = 0 with d in (a,b). This completes the proof.
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) and f(a) = f(b), there exists at least one c in (a,b) such that f’(c) = 0.
Proof:-
Since f(x) is continuous in closed interval [a,b], there exist c,d in [a,b] such that
f(c) = M(the maximum value of f(x))
f(d) = m the minimum value of f(x)
CASE I
If M = m, then f(x) is constant in [a,b] and in this case f’(x) = 0 for all x in (a,b). That is theorem is true in this case.
CASE II
If M = m, then at least one of them is different from f(a) and f(b). Let M = f(a). We shall show that f’(c) = 0 with c in (a,b)
At x = c, with h>0
RHD = lim h 0 f(c+h)-f(c)/h
= lim h 0 f(c+h)-M/h = -ve/+ve <= 0 ……….i
LHD = lim h 0 f(c-h)-f(c)/-h
= lim h 0 f(c-h)-M/-h = -ve/-ve >= 0 ……….ii
Since f’(x) exist in (a,b), I and ii must be equal and this will be possible only when both i and ii are equal to zero.
Hence f’(c) = 0 with c in (a,b).
CASE III
If m is different from f(a) and f(b) we can similarly show that f’(d) = 0 with d in (a,b). This completes the proof.
Sunday, June 13, 2010
Self Induction
When there is change in current through a solonoid then there will be induced emf in opposite direction of cause which produces it. This process is called as self induction
Now flux linked with the coil of solonoid
φ α I
φ = LI----------1 where L is called inductance of coil
Now we have
e = -dφ/dt
e = -d(LI)/dt
= -LdI/dt--------2
If dI/dT = 1amp/sec
Then e=L
Hence inductance of a solonoid can be defined as the back emf induced when the rate of change of current is 1 unit
Self inductance of a solonoid
Let a solonoid of length L having total no, of turns of coil N
Let Current I passes through it
then B = UonI
= UoN/l *I
φ = BAN
= UoAN^2I/L
we have
L = φ/I = UoAN^2/L
Now flux linked with the coil of solonoid
φ α I
φ = LI----------1 where L is called inductance of coil
Now we have
e = -dφ/dt
e = -d(LI)/dt
= -LdI/dt--------2
If dI/dT = 1amp/sec
Then e=L
Hence inductance of a solonoid can be defined as the back emf induced when the rate of change of current is 1 unit
Self inductance of a solonoid
Let a solonoid of length L having total no, of turns of coil N
Let Current I passes through it
then B = UonI
= UoN/l *I
φ = BAN
= UoAN^2I/L
we have
L = φ/I = UoAN^2/L
Brewsters Law
We can get polarized light from reflection according to Brewsters law which states that when a light incident on a glass slab at certain angle then, we can get the polarized light in reflected part polarized in direction perpendicular to plane of incidence. This particular angle is called polarizing angle and in this case the sum of angle of refraction and angle of polarization will be 90 i.e angle between reflected and refracted rays will be 90
p+r = 90
i.e r = 90
Now according to snell's lae
refractive index = sini/sinr
= sinp/sin(90-p)
= sinp/cosp
=tanp
Therefore refractive index of material is equal to tangent of polarizing angle.
Malus law
The intensity of the light coming out from analyzer will be directly proportional to the square of cosine of angle between polarizer and analyzer.
Now, if a be the amplitude of the light used and (theta) be the angle between analyzer and polarizer then acos(theta) will be the component of a along analyzer and asin(theta) will be that along perpendicular direction. Therefore intensity of light which pass through analyzer is
I = (acos(theta))^2
since a is constant
I α cos^2(theta)
Calcite crystal
It is a cuboid made up of six 119m having angle 109 and 71. There will be two corners in which all parallepgram will meet at obtuse angle are called blunt corners. The axis of crystal. The light incident on optical axis will pass without splitting. In actual practice light incident parallel to potical axis will pass without splitting. SO that optical axis is not just a line, it is a direction
Principal section
Principal section is a plane which contains optical axis and perpendicular to opposite faces
Quarter wave plate
A special type of crystal which introduces a path difference of λ/4(i.e quarter of wavelength) between two emergent light is called quarter wave plate
optical path for o-ray = Uot
optical path for e-ray = Uet
In case of clacite crystal Ue
In case of quarts crystal Ue>Uo
Uet - Uot = λ/4
Half wave plate
A special type of crystal which introduces path diference of λ/2(i.e) half of wave length) between two emergent light is called half wave plate
In case of clacite crystal Ue
In case of quarts crystal Ue>Uo
Uet - Uot = λ/2
p+r = 90
i.e r = 90
Now according to snell's lae
refractive index = sini/sinr
= sinp/sin(90-p)
= sinp/cosp
=tanp
Therefore refractive index of material is equal to tangent of polarizing angle.
Malus law
The intensity of the light coming out from analyzer will be directly proportional to the square of cosine of angle between polarizer and analyzer.
Now, if a be the amplitude of the light used and (theta) be the angle between analyzer and polarizer then acos(theta) will be the component of a along analyzer and asin(theta) will be that along perpendicular direction. Therefore intensity of light which pass through analyzer is
I = (acos(theta))^2
since a is constant
I α cos^2(theta)
Calcite crystal
It is a cuboid made up of six 119m having angle 109 and 71. There will be two corners in which all parallepgram will meet at obtuse angle are called blunt corners. The axis of crystal. The light incident on optical axis will pass without splitting. In actual practice light incident parallel to potical axis will pass without splitting. SO that optical axis is not just a line, it is a direction
Principal section
Principal section is a plane which contains optical axis and perpendicular to opposite faces
Quarter wave plate
A special type of crystal which introduces a path difference of λ/4(i.e quarter of wavelength) between two emergent light is called quarter wave plate
optical path for o-ray = Uot
optical path for e-ray = Uet
In case of clacite crystal Ue
In case of quarts crystal Ue>Uo
Uet - Uot = λ/4
Half wave plate
A special type of crystal which introduces path diference of λ/2(i.e) half of wave length) between two emergent light is called half wave plate
In case of clacite crystal Ue
In case of quarts crystal Ue>Uo
Uet - Uot = λ/2
Dopplar's effect
When source of sound and observer in motion either or both (or relative motion) then their will be apparent change in frequency. This phenomenon is known as Doppler's effect.
CaseI : Sound in motion observer in rest
Let us consider a source of sound which generates sound of frequency f wavelength λ and velocity V be moving toward a stationary observer O with velocity Us. Then observer feels apparent change in wave length as
λ = (V-Us)/f--------------1
Now relative motion of wave with respect to observer will be V - 0 = V
So the apparent frequency observered by the observer will be
f1 = v /λ = V/(V-Us) * f-----------2
Here V-Uo < U so clearly f1 > f SO loud sound is heared when the source moves towards stationary observer.
CASEII: When source is moving away the observer then Us will be negative and hence apparent freq.
f1 = V/(V+Us) * f--------------3
CASEIII : WHen observer in motion and source in rest
Let an observer O be in motion and source is in rest. Let the observer wave toward the source with velocity Vo, let the velocity of sound wave be V. Wave length λ and frequency f
In this case the wave length of sound remains unchanged and be given as V/P - λ
Here since the observer is moving towards source relative velocity of wave with respect to observer is V+Vo
Therefore the apparent change in frequency will be
f1 = relativevelocity/wave length = V+Vo\λ
.-.f1 = (V+Uo)/V * f------------4
Since V+Vo is greater than V the pitch of sound increases and sound will be heard louder
CASE IV:
If the observer is moving away from source relative velocity will be V-Vo apparent freq.
f1 = (V-Vo)/V * f --------------5
CaseI : Sound in motion observer in rest
Let us consider a source of sound which generates sound of frequency f wavelength λ and velocity V be moving toward a stationary observer O with velocity Us. Then observer feels apparent change in wave length as
λ = (V-Us)/f--------------1
Now relative motion of wave with respect to observer will be V - 0 = V
So the apparent frequency observered by the observer will be
f1 = v /λ = V/(V-Us) * f-----------2
Here V-Uo < U so clearly f1 > f SO loud sound is heared when the source moves towards stationary observer.
CASEII: When source is moving away the observer then Us will be negative and hence apparent freq.
f1 = V/(V+Us) * f--------------3
CASEIII : WHen observer in motion and source in rest
Let an observer O be in motion and source is in rest. Let the observer wave toward the source with velocity Vo, let the velocity of sound wave be V. Wave length λ and frequency f
In this case the wave length of sound remains unchanged and be given as V/P - λ
Here since the observer is moving towards source relative velocity of wave with respect to observer is V+Vo
Therefore the apparent change in frequency will be
f1 = relativevelocity/wave length = V+Vo\λ
.-.f1 = (V+Uo)/V * f------------4
Since V+Vo is greater than V the pitch of sound increases and sound will be heard louder
CASE IV:
If the observer is moving away from source relative velocity will be V-Vo apparent freq.
f1 = (V-Vo)/V * f --------------5
Faradays Lenz
Faraday's Law of Electromagnetic induction states that
1) When there is change of flux linked iwth a conductor then emf will be induced.
2) Induced emf will continue till the change in flux continues.
3) The induced emf is directly proportional to change of flux
Lenz law
It states that the induced emf is in the direction so as to oppose the cause producing it
Lenz law is in accordance with conservation of energu. If we are taking a magneti with north pole pointing towards the coil, then the magnetic field of coil due to induced emf will be in near end of magnet which repel magnet. So some work must be done against this force of repulsion which will be converted into electrical energy.
1) When there is change of flux linked iwth a conductor then emf will be induced.
2) Induced emf will continue till the change in flux continues.
3) The induced emf is directly proportional to change of flux
Lenz law
It states that the induced emf is in the direction so as to oppose the cause producing it
Lenz law is in accordance with conservation of energu. If we are taking a magneti with north pole pointing towards the coil, then the magnetic field of coil due to induced emf will be in near end of magnet which repel magnet. So some work must be done against this force of repulsion which will be converted into electrical energy.
Ampere's Circutial law:
It states that the line integral of magnetic field over a closed surface is equal to Uo times the total current enclosed by the surface.
i.e {B.dl = UoI
If we consider a straight conductor I, then it will produce a circular magnetic field. Now magnetic field at a point at a distance r from conductor can be give as
B = UoI/2πr
Now {B.dl = {Bdlcosϴ
={Bdl
= {UoI/2πr dl
=UoI/2πr 0{2πr dl
{B.dl = UoI
Applicatiojn of Ampere's law
1) Magnetic field due to straight conductor
Let us consider a straight conductor carrying current I. Consider a point P at distance R from conductor where magnetic field is to be determined. Since this is straight conductor magnetic lines of force are circular. Let us draw a circle of radius r passing through P.
then from Ampere's Law,
{B.dl = UoI
{Bdlcosϴ = UoI
B = UoI/2πr
2) Magnetic field due to solenoid
Let us consider a solonoid having number of turns per unit length n carry current I through it. It produces magnetic field along its axis Now to find magnetic field draw a rectangle of length PQ = l
By ampere's law
{B.dl = UoI
{B.dl = P{Q B.dl + R{Q B.dl + S{R B.dl + P{S B.dl
{B.dl = B P{Q dl =Bl----------1
Now
{B.dl = UoI
{B.dl = UonlI--------2
From I and 2
B = UonI
i.e {B.dl = UoI
If we consider a straight conductor I, then it will produce a circular magnetic field. Now magnetic field at a point at a distance r from conductor can be give as
B = UoI/2πr
Now {B.dl = {Bdlcosϴ
={Bdl
= {UoI/2πr dl
=UoI/2πr 0{2πr dl
{B.dl = UoI
Applicatiojn of Ampere's law
1) Magnetic field due to straight conductor
Let us consider a straight conductor carrying current I. Consider a point P at distance R from conductor where magnetic field is to be determined. Since this is straight conductor magnetic lines of force are circular. Let us draw a circle of radius r passing through P.
then from Ampere's Law,
{B.dl = UoI
{Bdlcosϴ = UoI
B = UoI/2πr
2) Magnetic field due to solenoid
Let us consider a solonoid having number of turns per unit length n carry current I through it. It produces magnetic field along its axis Now to find magnetic field draw a rectangle of length PQ = l
By ampere's law
{B.dl = UoI
{B.dl = P{Q B.dl + R{Q B.dl + S{R B.dl + P{S B.dl
{B.dl = B P{Q dl =Bl----------1
Now
{B.dl = UoI
{B.dl = UonlI--------2
From I and 2
B = UonI
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