integral calculus differentiation under the integral … · 2013-10-18 · integral calculus...

INTEGRAL CALCULUS

DIFFERENTIATION UNDER THE INTEGRAL SIGN:

Consider an integral involving one parameter � and denote it as

�� , ��

where a and b may be constants or functions of �. To find the derivative of �� when it exists it is not possible to first evaluate this integral and then to

find the derivative, such problems are solved by using the following rules.

(A) Leibnitz’s Rule for Constant limits of Integration:

Let ��, �� , �� be continuous functions of x and � then �� , �� , ��,

where a, b are constants and independent of parameter �

(B) Leibnitz’s Rule for Variable Limits of Integration:

If in the integral � �� , �� , �� satisfies the same conditions, and are functions of the

parameter�, then

�� , �� , �� ,�� , ��

***********************************************************

**

Example 1: Evaluate � �� and hence show that � �� !" by using differentiation under integral sign.

Solution: Let ��#� � � ��

Differentiate w.r.t � by Leibnitz’s Rule under integral sign. $ # %��#�& � � ''#� ()�� sin ��

� � ()�� sin � �

� � -()�� ). �� )/01�.234 �5 � then ��#� � �6��)4# � 7…………… . . �1� ;< �=� 6>( ?<�@6��6 7 , AB6 # � ∞ =� �=�

$ ��∞� � �6��)4∞� 7 � �D2 � 7 $ 7 � D2 $ ��#� � !" � 6��)4# AB6 # � 0 G( H(6 � �� !"�

*****************************************************

Example 2: using differentiation under the integral sign, evaluate � ��)4IJK� �4 , # L 0

Solution: Let ��#� � � ��)4IJK� � ………… . . �1�4 6>(� ��. %��#�& � � �� )4IJK� � � � �� IJK�IJK � � � 4 4 � �� 4 443. Integrating both sides w.r.t # $ ��#� � log�1 � #� � Y …………… . . �2� From �1� when # � 0 ��0� � 0 From �2� when I�0��log �1��C $ Y � 0 The solution is The solution is The solution is The solution is _�`� � abc�d � `� ***************************************************** Example 3: Example 3: Example 3: Example 3: Evaluate � j�k�l ��43�2� �� using the method of differentiation

under integral sign.

Solution: Let I = � j�k�l ��43�2� �� Differentiate w.r.t a by Leibnitz’s rule under integral sign $ � %��& � � ''� �6��)4 �� 1��1 � �"� �

�

� � ��1 � �"�"� . 1��1 � �"� � �

Let 4�43�2�2��43�2� � s�3t�43�2�2� � u�3v�43�2� by partial fractions Solving, z � 0, { � �2�2)4 , Y � 0, | � )4�2)4 $ � %��& � 1�" � 1� } �"�1 � �"�"� � 1�1 � �"� ~ � � D2�� 1�

�

Integrating w.r.t � $ �� !" � 4�34�� !" log�� 1� � C Then C �0 by putting ��0

***************************************************** Example 4: Example 4: Example 4: Example 4: Evaluate � �<H�1 � # ?<@��! using the method of differentiation under integral sign **************************************************************************************************************************************************************************************************************************************************************** Reduction fReduction fReduction fReduction formulaeormulaeormulaeormulae:::: I� � Sin�θ d θ II� � cos�θ d � III� � sin�θcos�θ � And to evaluate I� � sin�θ d�2 � II� � cos�θ d�2 � III� � sin�θcos�θ d�/" �

With � = x

Or

Similarly, (b)

Thus (1) and (2) are the required reduction formulae

∫∫−

= dxxxdxxann sin.sinsin)( 1

dxxxxnxxnn )cos(cos.sin)1()cos.(sin 21

−−−−=−−

∫

dxxxnxxnn )sin1.(sin)1(cos.sin 221

−−+−=−−

∫

dxxnxdxnxxnnn

∫∫ −−−+−=−− sin)1(sin)1(cos.sin 21

dxxnxxdxxnnnn

∫∫−−

−+−=21 sin)1(cos.sinsin

)1(...........sin1cos.sin

sin 21

dxxn

n

n

xxdxx

n

n

n

∫∫−

−

−+−=

)2(...........cos1cos.sin

cos 21

dxxn

n

n

xxdxx

n

n

n

∫∫−

−

−+−=

TO Evaluate

Then

i)

ii)

=

=

etc

=

( put x = a sinθ, so that dx = a cosθ dθ

Also when x = 0, θ = 0, when x = a, θ =

=

θ

θ = π/2)

iii) Evaluate � ��%��3 ��&�� ( put x = a tan θ, so that dx = a sec2θ dθ

Also when x = 0, θ = 0, when x = ∞, θ =

π/2)

= � � �� /�� =

d��d � cos"�)"θ dθ�2

= d��d . ��)��)��………�.d��)��)��………�.� . �� .

iv) Evaluate � sin�x cos"x dx

= � sin"x �sinx cosx�" dx

= � 4)�J�"�" ��"�" �" dx

= 4��sin"2x � sin"2x cos2x� dx

=

4� -� 4)�J��" dx � 4"� sin"2x � cos2x. 2�dx5 =

44� -� dx � � cos 4x dx � ��"� dx5

= dd¡ -¢ � d� £¤¥�¢ � d� £¤¥��¢5

v) Evaluate � cos�θ sin 6θ d�/� �

= � cos�θ �2sin3θcos3θ� d�/� �

= §� sin θ cos¨3θ d�/� �

= §�� sin x cos¨x d�/" x

= §� . �©¡.�.�d�.§.¡.�.� � dd�

(Put 3� = x ; so that 3d� = dx. Also when � = 0, x = 0

When � = �/6, x = �/2

vi) Evaluate � ��1 � �"� /"�4

= � @=�� 6�?<@"� /"?<@ 6 6!/" =

= � @=��6 ?<@� 6 6!/"

= �§ . d © �.d¡. �. � . �� ¡��¡

(put x = sint so that dx = cost dt, when x = 0, t = 0 when x = 1, t = �/2 )

************************************************************************ Tracing of Curves:Tracing of Curves:Tracing of Curves:Tracing of Curves: For the evaluation of Mathematical quantities such as Area, Length, Volume and Surface area we need the rough graph of the equation in either Cartesian or parametric or polar form depending on the statement of the problem. We use the following theoretical steps to draw the rough graph. A) Cartesian Curves: y = f (x)

a) Symmetry:

i) If the power of y in the equation is even, the curve is symmetric

about x- axis

ii) If the power of x in the equation is even, the curve is symmetric

about y- axis

iii) If both the powers x and y are even then the curve is symmetric

about both the axis.

iv) If the interchange of x and y leaves the equation unaltered then the

curve is symmetric about the line y = x

v) Replacing x by – x and y by – y leaves the equation unchanged the

curve has a symmetry in opposite quadrants.

b) Curve through the origin:

The curve passes through the origin, if the equation does not contain

constant term.

c) Find the origin, is on the curve. If it is, find the tangents at 0, by equating

the lowest degree terms to zero.

i) Find the points of intersections with the coordinate axes and the tangents

at these points.

For, put x = 0 find y; and put y = 0,

find x. At these points, find �®��. �� ®�� ∞, then the tangent is parallel to y axis.

If �®�� 0, then the tangent is parallel to x axis.

d) Asymptotes: express the equation of the curve in the form y = f (x). Equate

the denominator to zero. If the denominator contains x, then there is an

asymptote.

e) Find the region in which the curve lies.

f) Find the interval in which the curve is increasing or decreasing.

B) Parametric Form: x=f(t), y=g(t)

In this case we try to convert the parametric form into Cartesian form by

eliminating the parameter (if possible). Otherwise we observe the following

I) Find dy/dt and dx/dt and hence dy/dx.

II) Assign a few values for t and find the corresponding value for x, y ,y’.

III) Mark the corresponding points, observing the slope at these points.

C) Polar curves: r = f(�)

a) Symmetry: 1. If the substitution of - � for � in the equation , leaves the

equation unaltered, the curve symmetrical about the initial line.

2. If the power of r are even, the curve is symmetrical

about the pole.

b) Form the table, the value of r, for both positive and negative values of �

and hence note how r varies with � . Find in particular the value of � which gives

r = 0 and r = ∞ .

c) Find tan ¯ . This will indicate the direction of the tangent.

d) Sometimes by the nature of the equation it is possible to ascertain the value of r and � that are contained between certain limits.

e) Transform into Cartesian, if necessary and adopt the method given before.

f) Sketch the figure.

PROBLEMS FOR TRACING THE CURVES

1. Astroid : � � � �°��± , ² � � �³��± �°´ �� ²�� ) It is symmetrical about the x-axis

Limits |�| ¶ �

and |·| ¶ �

The curve lies entirely within the square bounded by the lines � � ¸� , · � ¸ �

Points: we have

when t = 0 or

, when t � As t increases x

From 0 to

+ve and

decreases from

a to 0

From to π

-ve and

increases

numerically

from 0 to

As t increases from π to 2π, we get the reflection of the curve ABC in the x

The values of t > 2π give no new points.

Hence the shape of the curve is as shown in the fig.

Here ox = oy = a

Y

-Y

-X O

when t = 0 or

Y Portion traced

+ve and

decreases from

+ve and

increases from

0 to a

From 0 to ∞ A to B

ve and

increases

numerically

from 0 to -a

+ ve and

decreases

from a to 0

From ∞to 0 B to C

π, we get the reflection of the curve ABC in the x

give no new points.

Hence the shape of the curve is as shown in the fig.

X

Portion traced

A to B

B to C

, we get the reflection of the curve ABC in the x - axis.

It is symmetrical about the y axis. As such we may consider the curve only for

positive values of x or .

Limits: The greatest value of y is 2a and the least value is zero. Therefore the curve

lies entirely between the lines y = 0 and y = 2a.

Points: We have

As increases x

From 0 to π

increases from

0 to aπ

From π to 2π

increases from

a π to 2aπ

As decreases from 0 to

axis. Hence the shape of the curve is as in the fig.

2. Cardioid: r

Fig.,

A cardioids is symmetrical about the initial line and lies entirely within the circle r

= 2a. Its name has been derived from the Latin word ‘ Kardia’

Because it is a heart shaped curve.

*********************************************************************

***

Y


The greatest value of y is 2a and the least value is zero. Therefore the curve

lies entirely between the lines y = 0 and y = 2a.

Y Portion traced

increases from increases from

0 to 2π From ∞ to 0 0 to A

increases from

to 2aπ

decreases

from 2a to 0 From 0 to ∞ A to B

decreases from 0 to - 2π, we get the reflection of the arch OAB in the y

axis. Hence the shape of the curve is as in the fig.

X

is symmetrical about the initial line and lies entirely within the circle r

= 2a. Its name has been derived from the Latin word ‘ Kardia’- meaning heart.

Because it is a heart shaped curve.

*********************************************************************

Initial Line


The greatest value of y is 2a and the least value is zero. Therefore the curve

Portion traced

0 to A

A to B

, we get the reflection of the arch OAB in the y-

is symmetrical about the initial line and lies entirely within the circle r

meaning heart.

*********************************************************************

APPLICATIONS OF CURVE TRACING

I) Length

II) Area

III) Volume

IV) Surface area

Table to find the values: Area, Length ,VolumeThe surface area

Quantity Coordinate

system

Area (A) Length (S) By revolving about the axis of

rotation to form solid

Volume (V) Surface area (SA)

Cartesian

form y = f (x) � �� ¹1 � ·4" ��

� � D·"�� 2D· @� ��

�

Where

�1�� º1 � ·4"

Parametric

form x= x(t)

y= y(t)

� �� 6 6�� º��4�" � �·4�"�

� 6 � D·" �6 6�� 2D· @6 6�

�

Where

�1�j � º��4�" � �·4�"

Polar form

r = f (�) � 12�� »"� � ¹»" � »4"�� D·" ��

� � 2D· @� ��

Where �1�¼ �º»" � »4"

3. Find the entire length of the cardioid » � � �1 � ?<@ ��, Also show that the upper

half is bisected by � � ! The cardioid is symmetrical about the initial line and for its

upper half, increases from 0 to π Also z�@<, �¾�¼ � -a sin θ. $ Length of the curve � 2 � ¹À»" � -�¾�¼5"Á! d� = 2� ºÂ� �1 � ?<@ ��" � �� @=��"Ã�!

= 2 � º�2�1 � ?<@��!

= 4a � º?<@�/2�!

= 4a Ä��Å/"4/" Ä �

= 8a (sin π/2 - sin 0) = 8a

$ Length of upper half of the curve of the cruve is 4a. Also length of the arch AP

from 0 to π/3

= 2 � º�2�1 � ?<@��Æ� = 2 � cos ¼" . �Æ�

= 4�|sin �/2| !/

= 4� -4"� 05 = 2a ( half the length of upper half of the cardioids )

Fig.

1. Find the entire length of the curve �� ²�� =��

Solution:

The equation to the curve is �2� � · 2� =�2�), the curve is symmetrical about the axis and

it meets the ‘x’ – axis at x = a

Fig.

If S1 = the length of the curve AB

Then required length is 4S1

S = 4S1 = � ¹1 � ��®��" ��

Now, �2� � · 2� =�2� $ �®�� ®��4/

S = 4 � ¹1 � �®��"/ ��

= 4 � ¹�2/� 3 ®2/��2/� ��

= 4 � ¹�2/��2/� ��

= 4 � �4/ �)4/ � � 4� �4/ Ç�2/�"/ È �

s = 6a units

Fig.

2. Find the perimeter of cardioid r = a (1+cosθ).

Solution:

The equation to the curve is symmetrical about the initial line.

Fig.

$ The required length of the curve is twice the length of the curve OPA

At O, θ = π and at A θ = 0

Now, r = a(1+cosθ) $ �¾�¼ � ��@=��

s = � � ¹»" � ��¾�¼�" �!

s = � � ºa"�1 � cosθ�" � �" @=�"� �!

s = 2 � 2acos Å" dθ�

s = �� É �³� ��d� Ê��

= 8a units

3. Find the area of the �2� � · 2� =�2� Solution: The parametric equation to the curve �2� � · 2� =�2� is given by :

x� � ?<@ � , · � � @=� �

Area = 4 � · �� Put x� � ?<@ � , · � � @=� �

$ dx = -3a cos2θ sinθ dθ when x = 0, θ = π/2 ; when x = a, θ = 0

A = 4 � �@=� ��!/" -3a cos2θ sinθ dθ

= 12�" � @=��!/" cos2θ dθ

= 12�" � �3" . 4"3"� �4" !"�

= !�2� Sq. units

4. Find the area of the cardioid r = a (1+cosθ).

Solution: The curve is symmetrical about the initial line. $ Total area = 2 © area above the line θ = 0

A = 2� 4"! »"� is the formula for area in polar curves

A = 2� 4"! a" �1 � cosθ�"�

A = � �"! �2?<@" ¼"�" �

A = 4�" � ?<@� ¼"! �

Put θ/2 = t $ � � 26 = 4�" � ?<@�6!/" 26 = §�� © d� © ��

= !�2" Sq. units

5. Find the area bounded by an arch of the cycloid

x� � �6 � @=�6�, · � � �1 � ?<@6 �, 0 ¶ 6 ¶ 2D and its base.

Solution: x� � �6 � @=�6�, · � � �1 � ?<@6 � for this arch ‘t’ varies from 0 to 2π.

$ z»(� � � · �"!jË

= � � �1 � ?<@6 � � �1 � ?<@6 �"!jË 6 since dx = � �1 � ?<@6 �6 = �� 1 � ?<@6 �" 6"!jË

=4�" � @=�� j" 6"!jË

= 8�" � @=��6 6!jË

= 16�" � @=��6 6!/"jË

= d¡��. �� . d� . ��

= 3D�" ÌÍ. B�=6@

6. Find the volume generated by revolving the cardiod r = a (1+cosθ) about the

initial line.

Solution: For the curve, varies from 0 to

____________________________________________________________________

Find the volume of the solid obtained by revolving the Astroid x2/3

+ y2/3

= a2/3

Solution: the equation of the asteroid is x2/3

+ y2/3

= a2/3

Volume is obtained by revolving the curve from x = 0 to x = a about x-axis and taking

two times the result.

Î � 2� D·"��

� 2� D��@=� ��"��3� ?<@2�@=�� !" �

� ��d�� ÏÐ³� Ï�³±�

_____________________________________________________________________

___

Problems for practice:

1. Find the surface area of r = a (1 - cosθ)

2. Find the volume of the solid obtained by revolving the cissoid ·"�2� � �� about its asymptote.

3. Find the length between [0,2D ] of the curve � � �� sin ��, · � ��1 �cos ��. ____________________________________________________________________

Find the surface area of solid generated by revolving the astroid �2� � · 2� =�2� about the axis.

Solution: The required surface area is equal to twice the surface area generated by

revolving the part of the astroid in the first quadrant about the axis.

Taking x� � ?<@ 6 , · � � @=� 6 we have,

Surface area = 2 � 2D·� � 4 � · �1�j 6!/" !/"

=4 � · ¹��j� � ��®�j�" 6!/"

= 4π� �asin t� Ñ ��3acos"t sint�"�/" � �3acos"t sint�"Ò1/2dt = 12a " � sin�t cos t dt�/" Put z = sint

= 4"Ó D�" ÌÍ. B�=6@

7. Find the surface area of the solid generated when the cardioid r = a (1+cosθ)

revolves about the initial line.

Solution: The equation to the curve is r = a (1+cosθ). For the upper part of the

curve, θ varies from 0 to π

Put x = r cosθ, y = rsinθ

$ Surface area = � 2D ·@!

= 2D � �» @=�� 1�¼ �!

= 2D � �» @=��¹»" � ��¾�¼�" �!

= 2D � ��1 � ?<@��. @=�� º�"�1 � cos ��" � �"@=�"� �!

= 16ÔÔÔÔa2

UNIT UNIT UNIT UNIT IV:IV:IV:IV: VECTOR CALCULUSVECTOR CALCULUSVECTOR CALCULUSVECTOR CALCULUS

Scalar and Vector point functions:

(I) If to each point p(R) of a region E in space there corresponds a definite

scalar denoted by f(R), then f (R) is called a ‘scalar point function’ in E. The

region E so defined is called a scalar field.

Ex: a) The temperature at any instant

b)The density of a body and potential due to gravitational matter.

(II) If to each point p(R) of a region E in space there corresponds a definite

vector denoted by F(R), then it is called the vector point function in E. the

region E so defined is called a vector field.

Ex: a) The velocity of a moving fluid at any instant

b) The gravitational intensity of force.

Note: Differentiation of vector point functions follows the same rules as those

of ordinary calculus.

If F (x,y,z) be a vector point function then

Ø6 � 'Ø'� '�'6 � 'Ø'· '·'6 � 'Ø'Ù 'Ù'6

Ø6 � 'Ø'� �6 � 'Ø'· ·6 � 'Ø'Ù Ù6

Ø � 'Ø'� � � 'Ø'· · � 'Ø'Ù Ù

Ø � Ú ''� � � ''· · � ''Ù ÙÛØ

( 1)

Vector operator del ( ∇∇∇∇ )

The operator is of the form

∇∇∇∇ �� = �� Ý ��® � 7 ��Þ GRADIENT, DIVERGENCE, CURL (G D C)

Gradient of the scalar point function:

It is the vector point function f defined as the gradient of the scalar point

function f and is written as grad f, then

grad f = ∇∇∇∇f f f f = �= �� Ý ��® � 7 ��Þ� � � = '�'� � Ý '�'· � 7 '�'Ù

DIVERGENCE OF A VECTOR POINT FUNCTION

The divergence of a continuously differentiable vector point function F(div F)is

defined by the equation

=ß Ø � ∇. Ø � = 'Ø'� � Ý 'Ø'· � 7 'Ø'Ù �� Ø � � = � à Ý � ¯ 7 6>(�

=ß Ø � ∇. Ø � Ú= ''� � Ý ''· � 7 ''ÙÛ · �� = � à Ý � ¯ 7� ∇ · Ø � '�'� � 'à'· � ''̄Ù

CURL OF A VECTOR POINT FUNCTION

The curl of a continuously differentiable vector point function F is defined by

the equation

curl F � ∇ © Ø � = © 'Ø'� � Ý © 'Ø'· � 7 © 'Ø'Ù

curl F � ∇ © Ø � Ú= ''� � Ý ''· � 7 ''ÙÛ © ��= � àÝ � ¯7� curl F � ∇ © Ø � ââ = Ý 7''� ''· ''Ù� à ¯ ââ

∇ © Ø � = Ú'¯'· � 'à'ÙÛ � Ý Ú'¯'� � '�'ÙÛ � 7 Ú'¯'� � '�'ÙÛ

∇� �� ∇ © � ∇ · Ø

DEL APPLIED TWICE TO POINT FUNCTIONS

Le being vector point functions, we can form their divergence and curl,

whereas being a scalar point function, we can have its gradient only. Then we

have Five formulas:

div grad f � ∇"� � '"�'�" � '"�'·" � '"�'Ù" ?B»� H»� � � ∇ © ∇� � 0 3. =ß ?B»� Ø � ∇ · ∇ © Ø � 0 =ß ?B»� Ø � ∇ · ∇ © Ø � 0 ?B»� ?B»� Ø � H»� =ß Ø � ∇"F

∇ © �∇ © Ø� � ∇�∇ · Ø� � ∇"F H»� =ß Ø � ?B»� ?B»� Ø � ∇"F ∇�∇ · Ø� � ∇ © �∇ © Ø� � ∇"F

PROOF:

(I) To prove that ã�ä � ã · ãä � ã · �³ åå� � æ åå² � ç ååè� ä � ∇ · Ú= '�'� � Ý '�'· � 7 '�'ÙÛ

� curl grad f � ∇ © ∇f � Ú= ''� � Ý ''· � 7 ''ÙÛ © Ú= '�'� � Ý '�'· � 7 '�'ÙÛ

� '"�'�" � '"�'·" � '"�'Ù"

� é '"'�" � '"'·" � '"'Ù"ê�

� ∇"f G>(»( ∇"� '"'�" � '"'·" � '"'Ù" =@ ?��( ë�A��?=�� <A»�6<»

�� ∇"f � 0 =@ ?��( ë�A��?( ′@ (ÍB�6=<�

(II) To prove that �Ï´ì í´�� î � �

curl grad f � Ü © Üf � Ú= ''� � Ý ''· � 7 ''ÙÛ © Ú= '�'� � Ý '�'· � 7 '�'ÙÛ

� ââ= Ý 7''� ''· ''Ù'�'� '�'· '�'Ùâ

â

� i é '"�'· 'Ù � '"�'· 'Ùê � j é '"�'� 'Ù � '"�'� 'Ùê � ké '"�'� '· � '"�'� '·ê

Ü © Üf � 0

(III) To prove thatã · ã © ñ � �∑ ³ åå�� · �³ © åñå� � æ © åñå² � ç © åñåè�

� ∑ = · �= © �2ó��2 � Ý © �2ó�® �� 7 © �2ó�Þ ��

� ∑�= © = · �2ó��2 � = © Ý · �2ó�� ® � = © 7 · �2ó�� Þ�

� ∑�7 · �2ó�� ® � Ý · �2ó�� Þ�

� 0

(IV) To prove that ã © ã © ñ � �∑ ³ åå�� © �³ © åñå� � æ © åñå² � ç © åñåè�

�ô= © é= © '"Ø'�" � Ý © '"Ø'� '· � 7 © '"Ø'� 'Ùê

�ôõé= · '"Ø'�"ê = � �= · =� '"Ø'�"ö � õé= · '"Ø'� '·ê Ý � �= · Ý� '"Ø'� '·ö� õé= · '"Ø'� 'Ùê 7 � �= · 7� '"Ø'� 'Ùö

�ô= · '"Ø'�" = � = · '"Ø'� '· Ý � = · '"Ø'� 'Ù 7 �ô'"Ø'�"

�ô= ''� Ú= · 'Ø'� � Ý · 'Ø'· � 7 · 'Ø'ÙÛ �ô'"Ø'�"

� Ü�Ü · F� � Ü"F curl curl F � Ü�Ü · F� � Ü"F

(V) To prove that

We have by (IV) which implies Ü�Ü · F� � ?B»� ?B»� Ø � Ü"F Ü�Ü · F� � Ü © �Ü © Ø� � Ü"F

� � 6 � 1, · � 6", Ù � 6 � 5 = � 3 Ý � 2 7

. ø´°Ðì�ù: Ì><G 6>�6 ã"r� � n�n � 1�r�)" úbaûü¤b¥: Ü"�r�� Ü · �Ür�� Ü · Ún r�)4 RrÛ � nÜ · �r�)"R� � nÂ�Ür�)"�. R � r�)"�Ü · R�Ã � n À�n � 2�r�) Rr · R � r�)"�3�Á � nÂ�n � 2�r�)��r"� � 3r�)"Ã � n�n � 1�r�)"

Otherwise: Ü"�r�� ý2�þ��ý�2 � ý2�þ��ý�2 � ý2�þ��ý�2 Now ý�þ��ý� � nr�)4 ýþý� � nr�)4 �þ � nr�)"x �∂^2 �r^n ��/�∂x^2 � � nÂr^�n � 2� � �n � 2� r^�n � 3� ∂r/∂x xÃ � nÂr^�n � 2� � �n � 2� r^�n � 3� x/r xÃ � nÂr�)" � �n � 2�r�)�x"Ã …………… . �1�

SImilarly, ý2�þ��ý�2 � nÂr�)" � �n � 2�r�)�y"Ã………… . . �2� ý2�þ��ý�2 � nÂr�)" � �n � 2�r�)�z"Ã ……………�3� Adding equations �1�, �2�and �3�, gives Ü"�r�� nÂ3r�)" � �n � 2�r�)��x" � y" � z"�Ã � nÂ3r�)" � �n � 2�r�)�r"Ã � n�n � 1�r�)" In particular Ü" �4þ� � 0

Ex: A particle moves along the curve � � 6 � 1, · � 6", Ù � 6 � 5 find the

components of velocity and acceleration at t=2 in the direction of = � 3 Ý � 2 7

Solution: ;>( A<@=6=<� ß(?6<» <� 6>( A�»6=?�( �6 6=�( 6 =@,

�� = � · Ý � Ù 7 �� 6 � 1�= � 6" Ý � �6 � 5� 7

;>( ß(�<?=6· Î� � ��j =@ H=ß(� �·

Î� � ��j � 36" = � 2 6 Ý � 7

;>( �??(�(»�6=<� � � ��j � �2��j2 =@ H=ß(� �·

� � �2��j2 � 6 6 = � 2 Ý ;>(� �6 6=�( 6 � 2, Î� �� »( H=ß(� �·

Î� � ��j � 12 = � 4 Ý � 7

� � �2��j2 � 12 = � 2 Ý Î(�<?=6· ß(?6<» =� 6>( =»(?6=<� <� = � 3Ý � 27 =@ � Î� · ��3 3" �√4� � � �12 = � 4 Ý � 7� · ��3 3" �

√4� � � "�√4� z??(�(»�6=<� =� 6>( =»(?6=<� <� = � 3Ý � 27 =@ � � · ��3 3" �

√4� � � �12 = � 2 Ý� · ��3 3" �√4� � � 4�

√4� ø´°Ðì�ù: Ø=� 6>( B�=6 ß(?6<» �<»�� 6< 6>( @B»��?( �· Ù" �4 �6 6>( A<=�6 ��1,�1,2� �°ìÏ±³°�:Î(?6<» �<»�� 6< 6>( H=ß(� @B»��?( =@ Ü��· Ù"� Ü��· Ù"� � �= �� Ý ��® � 7 ��Þ� ��· Ù"� � · Ù" = � 3�·"Ù" Ý � 2�· Ù 7 z6 ��1,�1,2� � 4 = � 12 Ý � 4 7

�(�?( 6>( (@=»( B�=6 �<»�� 6< 6>( @B»��?( =@ � � �)4" 3��√4�34��34� � 4

√4¨� �4= � 12Ý � 47�

ø´°Ðì�ù: Ø=� 6>( =»(?6=<�� (»=ß�6=ß( <� ��, ·, Ù� � �·" �·Ù �6 6>( A<=�6 �2, �1,1� =� 6>( =»(?6=<� <� = � 2Ý � 27 �°ìÏ±³°�:Î(?6<» �<»�� 6< 6>( @B»��?( ��, ·, Ù� =@ Ü� i. e. Ü� � �= �� Ý ��® � 7 ��Þ� ��·" � ·Ù � Ü� � ·" = � �2�· � Ù � Ý � �3·Ù"� 7 z6 �2, �1,1� Ü� � = � 3 Ý � 3 7 |=»(?6=<�� (»=ß�6=ß( =� 6>( =»(?6=<� <� = � 2Ý � 27 =@ � �Ü��",)4,4� · �3" 3"�√43�3� � �= � 3Ý � 37� · �3" 3"�

√�

� 4)�)� � )44 Ü"�r�� n�n � 1�r�)4 Ü"�r�� Ü�Ür�� ø´°Ðì�ù:Ø=� 6>( ?<�@6��6@ �, �, ? @B?> 6>�6 6>( ß(?6<» Ø � �� · � �Ù��̂� �� ? · � 2Ù�7� � �� 2· � Ù��̂ =@ =»»<6�6=<�� °ìÏ±³°�:

�=ß(� Ø � �� · � �Ù��̂� �� ? · � 2Ù�7� � �� 2· � Ù��̂ =@ =»»<6�6=<�� Ì=�?( 6>( ß(?6<» �=(� =@ =»»<6�6=<�� ;>(»(�<»( Ü © Ø � 0

YB»� Ø � â �̂ �̂ 7�� ® ��Þ � � · � �Ù �� 2· � Ù � � ?· � 2Ù â � �? � 1��̂ � �1 � ��̂� �� 1�7� =. (. , �? � 1��̂� �1 � ��̂� �� 1�7 � 0� ;>=@ =@ A<@@=��( <��· G>(�, ? � 1 � 0, 1 � � � 0, � � 1 � 0 G>=?> =�A�=(@ � � 1, � � 1, ? � 1

Orthogonal curvilinear co-ordinates

Let the rectangular co

function of (u,v,w),

So that x = x(u,v,w),y = y(u,v,w),z = z(u,v,w) …..(1)

Suppose that (1) can be solved for u,v,w in terms of x,y,z

i,e u = (x,y,z), v = v(x,y,z),w = w(x,y,z) ….(2)

We assume that the functions in (1) and (2) are single valued functions

and have continuous partial derivatives so that the

(x,y,z) and (u,v,w) is unique. Then (u,v,w) are called curviline

of (x,y,z). Each of u,v,w has a level of surface t

The surface

surface through

Each pair of these co

ordinate curves. The curve

only w changes along this curve. Similarly we define u and v

In vector notation, (1) can be written as

R = x(u,v,w)I + y(u,v,w)J + z(u,v,w)K

Let the rectangular co-ordinates (x,y,z) of Any point be expressed as

So that x = x(u,v,w),y = y(u,v,w),z = z(u,v,w) …..(1)

Suppose that (1) can be solved for u,v,w in terms of x,y,z

(x,y,z), v = v(x,y,z),w = w(x,y,z) ….(2)


partial derivatives so that the correspondence

(x,y,z) and (u,v,w) is unique. Then (u,v,w) are called curvilinear co

of (x,y,z). Each of u,v,w has a level of surface through an arbitrary point .

are called co

Each pair of these co-ordinate surface intersects In curves called the co

The curve of intersection of will be called the w

only w changes along this curve. Similarly we define u and v-curves.

(1) can be written as

R = x(u,v,w)I + y(u,v,w)J + z(u,v,w)K

ordinates (x,y,z) of Any point be expressed as


correspondence between

ar co-ordinates

bitrary point .

called co-ordinate

ordinate surface intersects In curves called the co-

of intersection of will be called the w-curve, for

curves.

The co-ordinate curves for ρ

horizontal circles with centers on the Z

x =

ρ are rays perpendicular to the Z-axis; for

horizontal circles with centers on the Z-axis; for z lines parallels to the Z

x = ρ cos ф,

axis; for ф

axis; for z lines parallels to the Z-axis.

y =

So that scale factors are h1

2) Spherical polar co-

Let p(x,y,z) be any point whose projection

Then the Spherical polar co-

.

The level surfaces

about O, cones about the Z-axis with vertex at O and planes through the Z

The co-ordinate curves for r are rays from

centre at O (called meridians); for

axis.

x = OQ cosф

= OP cos(90-θ)cos

= r sinθ cosф,

y = OQ sinф = r sinθ

z = r cosθ

So that the scale factors are

Also the volume element

y = ρ sin ф, z=z

1=1, h2 = ρ, h3= 1. Also the volume element

dV=ρ dρ dф dz.

-ordinates:

Let p(x,y,z) be any point whose projection on the xy-plane is Q(x,y) .

-ordinates of p are such that

are respectively spheres

axis with vertex at O and planes through the Z

ordinate curves for r are rays from the origin; for θ, vertical circles with

centre at O (called meridians); for ф, horizontal circles with centres on the Z

)cosф

sinф

So that the scale factors are



plane is Q(x,y) .

r = op,

are respectively spheres

axis with vertex at O and planes through the Z-axis.

, vertical circles with

, horizontal circles with centres on the Z-

integral calculus differentiation under the integral … · 2013-10-18 · integral calculus...

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