UPSC Exam > UPSC Notes > Mathematics Optional Notes for UPSC > Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative
Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative | Mathematics Optional Notes for UPSC PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Edurev123
3. Gradient, Divergence and Cur in
Cartesian and Cylindrical Coordinate and
Directional Derivative
3.1 Find the directional derivative of :
(i) ?? ?? ?? ?? -?? ?? ?? ?? ?? ?? ?? at (?? ,-?? ,?? ) along ?? -axis
(ii) ?? ?? ???? +?? ?? ?? ?? at (?? ,-?? ,?? ) in the direction of ?? ??ˆ-??ˆ-?? ??ˆ
(???????? :?? +?? =???? Marks)
Solution:
Approach : The directional derivative in any direction is the dot product of the gradient
with that direction.
(i)
?? (?? ,?? ,?? ) =4?? ?? 3
-3?? 2
?? 2
?? 2
??? =
??? ??? ??ˆ+
??? ??? ??ˆ+
??? ??? ??ˆ
=(4?? 3
-6?? ?? 2
?? 2
)??ˆ-6?? 2
?? ?? 2
??ˆ+(12?? ?? 2
-6?? 2
?? 2
?? )??ˆ
Directional derivative along ?? -axis
=??? ·??ˆ
=12?? ?? 2
-6?? 2
?? 2
??
Directional derivative along ?? -axis at (2,-1,2)
=(??? ·??ˆ
)|
(2,-1,2)
=12·2·2
2
-6·2
2
·(-1)
2
·2=48
(ii)
?? (?? ,?? ,?? ) =?? 2
???? +4?? ?? 2
??? =(2?????? +4?? 2
)??ˆ+(?? 2
?? )??ˆ+(?? 2
?? +8???? )??ˆ
??? |
(1,-2,1)
=??ˆ+6??ˆ
? Directional derivative along (2??ˆ-??ˆ-2??ˆ
)
=(??ˆ+6??ˆ
)·(
2??ˆ-??ˆ-2??ˆ
3
)
=(
-1-12
3
)=-13
Page 2
Edurev123
3. Gradient, Divergence and Cur in
Cartesian and Cylindrical Coordinate and
Directional Derivative
3.1 Find the directional derivative of :
(i) ?? ?? ?? ?? -?? ?? ?? ?? ?? ?? ?? at (?? ,-?? ,?? ) along ?? -axis
(ii) ?? ?? ???? +?? ?? ?? ?? at (?? ,-?? ,?? ) in the direction of ?? ??ˆ-??ˆ-?? ??ˆ
(???????? :?? +?? =???? Marks)
Solution:
Approach : The directional derivative in any direction is the dot product of the gradient
with that direction.
(i)
?? (?? ,?? ,?? ) =4?? ?? 3
-3?? 2
?? 2
?? 2
??? =
??? ??? ??ˆ+
??? ??? ??ˆ+
??? ??? ??ˆ
=(4?? 3
-6?? ?? 2
?? 2
)??ˆ-6?? 2
?? ?? 2
??ˆ+(12?? ?? 2
-6?? 2
?? 2
?? )??ˆ
Directional derivative along ?? -axis
=??? ·??ˆ
=12?? ?? 2
-6?? 2
?? 2
??
Directional derivative along ?? -axis at (2,-1,2)
=(??? ·??ˆ
)|
(2,-1,2)
=12·2·2
2
-6·2
2
·(-1)
2
·2=48
(ii)
?? (?? ,?? ,?? ) =?? 2
???? +4?? ?? 2
??? =(2?????? +4?? 2
)??ˆ+(?? 2
?? )??ˆ+(?? 2
?? +8???? )??ˆ
??? |
(1,-2,1)
=??ˆ+6??ˆ
? Directional derivative along (2??ˆ-??ˆ-2??ˆ
)
=(??ˆ+6??ˆ
)·(
2??ˆ-??ˆ-2??ˆ
3
)
=(
-1-12
3
)=-13
3.2 Examine whether the vectors ?? ?? ,?? ?? and ?? ?? are copla, 1 ar, where ?? ,?? and ??
are the scalar functions defined by:
and
?? =?? +?? +?? ?? =?? ?? +?? ?? +?? ?? ?? =???? +???? +????
(2011: 15 Marks)
Solution:
??? =(
?
??? ?? +
?
??? ?? +
?
??? ???
)(?? +?? +?? )
=?? +?? +???
??? =2?? ?? +2?? ?? +2?? ???
Similarly , ??? =(?? +?? )?? +(?? +?? )?? +(?? +?? )???
Now, ??? ,??? ,??? would be co-planar if their scalar triple product is zero.
??? ×??? =|
?? ?? ???
2?? 2?? 2?? ?? +?? ?? +?? ?? +?? |
=?? (2???? +2?? 2
-2???? -2?? 2
)+?? (2???? +2?? 2
-2?? 2
-2???? )+
???
(2?? 2
+2???? -2?? 2
-2???? ).
? ??? (??? ×??? )= (?? +?? +???
)·[(2???? +2?? 2
-2???? -2?? 2
)??
+(2?? ?? +2?? 2
-2?? 2
-2???? )?? +(2?? 2
+2???? -2?? 2
-2???? )???
]
=2???? +2?? 2
-2?? 2
-2?? 2
+2???? +2?? 2
-2?? 2
-2???? +2?? 2
+2???? -2?? 2
-2????
=0
? The vector ??? ,??? ,? ware co-planar.
3.3 If ??? be the position vector of a point, find the value(s) of ?? for which the vector
?? ?? ??? is (i) irrotational, (ii) solenoidal.
(2011: 15 Marks)
Solution:
A vector ???
is said to be solenoidal if divergence of ???
=0.
i.e.,
?·???
=0
Also, div (?? ???
)=(grad ?? )·???
+?? div ???
Page 3
Edurev123
3. Gradient, Divergence and Cur in
Cartesian and Cylindrical Coordinate and
Directional Derivative
3.1 Find the directional derivative of :
(i) ?? ?? ?? ?? -?? ?? ?? ?? ?? ?? ?? at (?? ,-?? ,?? ) along ?? -axis
(ii) ?? ?? ???? +?? ?? ?? ?? at (?? ,-?? ,?? ) in the direction of ?? ??ˆ-??ˆ-?? ??ˆ
(???????? :?? +?? =???? Marks)
Solution:
Approach : The directional derivative in any direction is the dot product of the gradient
with that direction.
(i)
?? (?? ,?? ,?? ) =4?? ?? 3
-3?? 2
?? 2
?? 2
??? =
??? ??? ??ˆ+
??? ??? ??ˆ+
??? ??? ??ˆ
=(4?? 3
-6?? ?? 2
?? 2
)??ˆ-6?? 2
?? ?? 2
??ˆ+(12?? ?? 2
-6?? 2
?? 2
?? )??ˆ
Directional derivative along ?? -axis
=??? ·??ˆ
=12?? ?? 2
-6?? 2
?? 2
??
Directional derivative along ?? -axis at (2,-1,2)
=(??? ·??ˆ
)|
(2,-1,2)
=12·2·2
2
-6·2
2
·(-1)
2
·2=48
(ii)
?? (?? ,?? ,?? ) =?? 2
???? +4?? ?? 2
??? =(2?????? +4?? 2
)??ˆ+(?? 2
?? )??ˆ+(?? 2
?? +8???? )??ˆ
??? |
(1,-2,1)
=??ˆ+6??ˆ
? Directional derivative along (2??ˆ-??ˆ-2??ˆ
)
=(??ˆ+6??ˆ
)·(
2??ˆ-??ˆ-2??ˆ
3
)
=(
-1-12
3
)=-13
3.2 Examine whether the vectors ?? ?? ,?? ?? and ?? ?? are copla, 1 ar, where ?? ,?? and ??
are the scalar functions defined by:
and
?? =?? +?? +?? ?? =?? ?? +?? ?? +?? ?? ?? =???? +???? +????
(2011: 15 Marks)
Solution:
??? =(
?
??? ?? +
?
??? ?? +
?
??? ???
)(?? +?? +?? )
=?? +?? +???
??? =2?? ?? +2?? ?? +2?? ???
Similarly , ??? =(?? +?? )?? +(?? +?? )?? +(?? +?? )???
Now, ??? ,??? ,??? would be co-planar if their scalar triple product is zero.
??? ×??? =|
?? ?? ???
2?? 2?? 2?? ?? +?? ?? +?? ?? +?? |
=?? (2???? +2?? 2
-2???? -2?? 2
)+?? (2???? +2?? 2
-2?? 2
-2???? )+
???
(2?? 2
+2???? -2?? 2
-2???? ).
? ??? (??? ×??? )= (?? +?? +???
)·[(2???? +2?? 2
-2???? -2?? 2
)??
+(2?? ?? +2?? 2
-2?? 2
-2???? )?? +(2?? 2
+2???? -2?? 2
-2???? )???
]
=2???? +2?? 2
-2?? 2
-2?? 2
+2???? +2?? 2
-2?? 2
-2???? +2?? 2
+2???? -2?? 2
-2????
=0
? The vector ??? ,??? ,? ware co-planar.
3.3 If ??? be the position vector of a point, find the value(s) of ?? for which the vector
?? ?? ??? is (i) irrotational, (ii) solenoidal.
(2011: 15 Marks)
Solution:
A vector ???
is said to be solenoidal if divergence of ???
=0.
i.e.,
?·???
=0
Also, div (?? ???
)=(grad ?? )·???
+?? div ???
??? ?? ?? will be solenoidal if
div (?? ?? ?? ) =0
? (grad ?? ?? )·?? +?? ?? div (?? ) =0
? (?? ?? ?? -1
grad ?? )·?? +?? ?? ·3 =0
? div ?? =(??ˆ
?
??? +??ˆ
?
??? +??ˆ
?
??? )(?? ??ˆ+?? ??ˆ+?? ??ˆ
)
=1+1+1=3
and grad ?? (4)=?? '
(4) grad ?? ? (?? ?? ?? -1
·
??
?? )·?? +3?? ?? =0
? ?? ?? ?? -2
(?? ·?? )+3?? ?? =0
? ?? ?? ?? -2
·?? 2
+3?? ?? =0
? ?? ?? (?? +3)=0??? =-3
A vector ?? is said to be irrotational if
?×???
=0
Also, ?×(?? ???
) =(grad ?? )×???
+?? (?×???
)
??? ?? ?? will be irrotational if
?×(?? ?? ?? ) =0
? (grad ?? ?? )×?? +?? ?? (?×?? ) =0
? (?? ?? ?? -1
·
??
?? )·?? +?? ?? ·0=0
Hence, ?? ?? ?? is irrotatlonal for all the real values of ?? .
3.4 A vector field is given by
????
=(?? ?? +?? ?? ?? )?? +(?? ?? +?? ?? ?? )??
Verify that the field ????
is irrotational or not. Find the scalar potential.
(2015 : 12 Marks)
Solution:
A vector field ??
is said to be irrotational if curl ??
=0, i.e.,
?×??
=0
?
??
×??
=
|
|
??ˆ ??ˆ ??ˆ
?
??? ?
??? ?
??? ?? 2
+?? ?? 2
?? 2
+?? 2
?? 0
|
|
Page 4
Edurev123
3. Gradient, Divergence and Cur in
Cartesian and Cylindrical Coordinate and
Directional Derivative
3.1 Find the directional derivative of :
(i) ?? ?? ?? ?? -?? ?? ?? ?? ?? ?? ?? at (?? ,-?? ,?? ) along ?? -axis
(ii) ?? ?? ???? +?? ?? ?? ?? at (?? ,-?? ,?? ) in the direction of ?? ??ˆ-??ˆ-?? ??ˆ
(???????? :?? +?? =???? Marks)
Solution:
Approach : The directional derivative in any direction is the dot product of the gradient
with that direction.
(i)
?? (?? ,?? ,?? ) =4?? ?? 3
-3?? 2
?? 2
?? 2
??? =
??? ??? ??ˆ+
??? ??? ??ˆ+
??? ??? ??ˆ
=(4?? 3
-6?? ?? 2
?? 2
)??ˆ-6?? 2
?? ?? 2
??ˆ+(12?? ?? 2
-6?? 2
?? 2
?? )??ˆ
Directional derivative along ?? -axis
=??? ·??ˆ
=12?? ?? 2
-6?? 2
?? 2
??
Directional derivative along ?? -axis at (2,-1,2)
=(??? ·??ˆ
)|
(2,-1,2)
=12·2·2
2
-6·2
2
·(-1)
2
·2=48
(ii)
?? (?? ,?? ,?? ) =?? 2
???? +4?? ?? 2
??? =(2?????? +4?? 2
)??ˆ+(?? 2
?? )??ˆ+(?? 2
?? +8???? )??ˆ
??? |
(1,-2,1)
=??ˆ+6??ˆ
? Directional derivative along (2??ˆ-??ˆ-2??ˆ
)
=(??ˆ+6??ˆ
)·(
2??ˆ-??ˆ-2??ˆ
3
)
=(
-1-12
3
)=-13
3.2 Examine whether the vectors ?? ?? ,?? ?? and ?? ?? are copla, 1 ar, where ?? ,?? and ??
are the scalar functions defined by:
and
?? =?? +?? +?? ?? =?? ?? +?? ?? +?? ?? ?? =???? +???? +????
(2011: 15 Marks)
Solution:
??? =(
?
??? ?? +
?
??? ?? +
?
??? ???
)(?? +?? +?? )
=?? +?? +???
??? =2?? ?? +2?? ?? +2?? ???
Similarly , ??? =(?? +?? )?? +(?? +?? )?? +(?? +?? )???
Now, ??? ,??? ,??? would be co-planar if their scalar triple product is zero.
??? ×??? =|
?? ?? ???
2?? 2?? 2?? ?? +?? ?? +?? ?? +?? |
=?? (2???? +2?? 2
-2???? -2?? 2
)+?? (2???? +2?? 2
-2?? 2
-2???? )+
???
(2?? 2
+2???? -2?? 2
-2???? ).
? ??? (??? ×??? )= (?? +?? +???
)·[(2???? +2?? 2
-2???? -2?? 2
)??
+(2?? ?? +2?? 2
-2?? 2
-2???? )?? +(2?? 2
+2???? -2?? 2
-2???? )???
]
=2???? +2?? 2
-2?? 2
-2?? 2
+2???? +2?? 2
-2?? 2
-2???? +2?? 2
+2???? -2?? 2
-2????
=0
? The vector ??? ,??? ,? ware co-planar.
3.3 If ??? be the position vector of a point, find the value(s) of ?? for which the vector
?? ?? ??? is (i) irrotational, (ii) solenoidal.
(2011: 15 Marks)
Solution:
A vector ???
is said to be solenoidal if divergence of ???
=0.
i.e.,
?·???
=0
Also, div (?? ???
)=(grad ?? )·???
+?? div ???
??? ?? ?? will be solenoidal if
div (?? ?? ?? ) =0
? (grad ?? ?? )·?? +?? ?? div (?? ) =0
? (?? ?? ?? -1
grad ?? )·?? +?? ?? ·3 =0
? div ?? =(??ˆ
?
??? +??ˆ
?
??? +??ˆ
?
??? )(?? ??ˆ+?? ??ˆ+?? ??ˆ
)
=1+1+1=3
and grad ?? (4)=?? '
(4) grad ?? ? (?? ?? ?? -1
·
??
?? )·?? +3?? ?? =0
? ?? ?? ?? -2
(?? ·?? )+3?? ?? =0
? ?? ?? ?? -2
·?? 2
+3?? ?? =0
? ?? ?? (?? +3)=0??? =-3
A vector ?? is said to be irrotational if
?×???
=0
Also, ?×(?? ???
) =(grad ?? )×???
+?? (?×???
)
??? ?? ?? will be irrotational if
?×(?? ?? ?? ) =0
? (grad ?? ?? )×?? +?? ?? (?×?? ) =0
? (?? ?? ?? -1
·
??
?? )·?? +?? ?? ·0=0
Hence, ?? ?? ?? is irrotatlonal for all the real values of ?? .
3.4 A vector field is given by
????
=(?? ?? +?? ?? ?? )?? +(?? ?? +?? ?? ?? )??
Verify that the field ????
is irrotational or not. Find the scalar potential.
(2015 : 12 Marks)
Solution:
A vector field ??
is said to be irrotational if curl ??
=0, i.e.,
?×??
=0
?
??
×??
=
|
|
??ˆ ??ˆ ??ˆ
?
??? ?
??? ?
??? ?? 2
+?? ?? 2
?? 2
+?? 2
?? 0
|
|
=??ˆ(0-0)-??ˆ(0-0)+??ˆ
(2???? -2???? )
=0
?
???
is irrotational.
Now, it can be written as grad of a scalar field, i.e., to find ?? so that
??? =??
i.e., ??ˆ
??? ??? +??ˆ
??? ??? =(?? 2
+?? ?? 2
)??ˆ+(?? 2
+?? 2
?? )??ˆ
?
??? ??? =?? 2
+?? ?? 2
;
??? ??? =?? 2
+?? 2
??
? ?? =
?? 3
3
+
?? 2
?? 2
2
+?? (?? ) (*)
Differentiating w.r.t. ?? and comparing with (*)
??? ??? =?? 2
?? +?? '
(?? )
? ?? '
(?? )=?? 2
?? (?? )=
?? 3
3
+?? ? ?? (?? ,?? )=
?? 3
3
+
?? 3
3
+
?? 2
?? 2
2
+??
3.5 For what values of the constants ?? ,?? and ?? the vector
????
=(?? +?? +???? )??ˆ+(???? +?? ?? -?? )??ˆ+(-?? +???? +?? ?? )??ˆ
is irrational. Find the divergence in cylindrical coordinates of this vector with
these values.
(2017: 10 Marks)
Solution:
Irrational ? Curl ???
=0
Page 5
Edurev123
3. Gradient, Divergence and Cur in
Cartesian and Cylindrical Coordinate and
Directional Derivative
3.1 Find the directional derivative of :
(i) ?? ?? ?? ?? -?? ?? ?? ?? ?? ?? ?? at (?? ,-?? ,?? ) along ?? -axis
(ii) ?? ?? ???? +?? ?? ?? ?? at (?? ,-?? ,?? ) in the direction of ?? ??ˆ-??ˆ-?? ??ˆ
(???????? :?? +?? =???? Marks)
Solution:
Approach : The directional derivative in any direction is the dot product of the gradient
with that direction.
(i)
?? (?? ,?? ,?? ) =4?? ?? 3
-3?? 2
?? 2
?? 2
??? =
??? ??? ??ˆ+
??? ??? ??ˆ+
??? ??? ??ˆ
=(4?? 3
-6?? ?? 2
?? 2
)??ˆ-6?? 2
?? ?? 2
??ˆ+(12?? ?? 2
-6?? 2
?? 2
?? )??ˆ
Directional derivative along ?? -axis
=??? ·??ˆ
=12?? ?? 2
-6?? 2
?? 2
??
Directional derivative along ?? -axis at (2,-1,2)
=(??? ·??ˆ
)|
(2,-1,2)
=12·2·2
2
-6·2
2
·(-1)
2
·2=48
(ii)
?? (?? ,?? ,?? ) =?? 2
???? +4?? ?? 2
??? =(2?????? +4?? 2
)??ˆ+(?? 2
?? )??ˆ+(?? 2
?? +8???? )??ˆ
??? |
(1,-2,1)
=??ˆ+6??ˆ
? Directional derivative along (2??ˆ-??ˆ-2??ˆ
)
=(??ˆ+6??ˆ
)·(
2??ˆ-??ˆ-2??ˆ
3
)
=(
-1-12
3
)=-13
3.2 Examine whether the vectors ?? ?? ,?? ?? and ?? ?? are copla, 1 ar, where ?? ,?? and ??
are the scalar functions defined by:
and
?? =?? +?? +?? ?? =?? ?? +?? ?? +?? ?? ?? =???? +???? +????
(2011: 15 Marks)
Solution:
??? =(
?
??? ?? +
?
??? ?? +
?
??? ???
)(?? +?? +?? )
=?? +?? +???
??? =2?? ?? +2?? ?? +2?? ???
Similarly , ??? =(?? +?? )?? +(?? +?? )?? +(?? +?? )???
Now, ??? ,??? ,??? would be co-planar if their scalar triple product is zero.
??? ×??? =|
?? ?? ???
2?? 2?? 2?? ?? +?? ?? +?? ?? +?? |
=?? (2???? +2?? 2
-2???? -2?? 2
)+?? (2???? +2?? 2
-2?? 2
-2???? )+
???
(2?? 2
+2???? -2?? 2
-2???? ).
? ??? (??? ×??? )= (?? +?? +???
)·[(2???? +2?? 2
-2???? -2?? 2
)??
+(2?? ?? +2?? 2
-2?? 2
-2???? )?? +(2?? 2
+2???? -2?? 2
-2???? )???
]
=2???? +2?? 2
-2?? 2
-2?? 2
+2???? +2?? 2
-2?? 2
-2???? +2?? 2
+2???? -2?? 2
-2????
=0
? The vector ??? ,??? ,? ware co-planar.
3.3 If ??? be the position vector of a point, find the value(s) of ?? for which the vector
?? ?? ??? is (i) irrotational, (ii) solenoidal.
(2011: 15 Marks)
Solution:
A vector ???
is said to be solenoidal if divergence of ???
=0.
i.e.,
?·???
=0
Also, div (?? ???
)=(grad ?? )·???
+?? div ???
??? ?? ?? will be solenoidal if
div (?? ?? ?? ) =0
? (grad ?? ?? )·?? +?? ?? div (?? ) =0
? (?? ?? ?? -1
grad ?? )·?? +?? ?? ·3 =0
? div ?? =(??ˆ
?
??? +??ˆ
?
??? +??ˆ
?
??? )(?? ??ˆ+?? ??ˆ+?? ??ˆ
)
=1+1+1=3
and grad ?? (4)=?? '
(4) grad ?? ? (?? ?? ?? -1
·
??
?? )·?? +3?? ?? =0
? ?? ?? ?? -2
(?? ·?? )+3?? ?? =0
? ?? ?? ?? -2
·?? 2
+3?? ?? =0
? ?? ?? (?? +3)=0??? =-3
A vector ?? is said to be irrotational if
?×???
=0
Also, ?×(?? ???
) =(grad ?? )×???
+?? (?×???
)
??? ?? ?? will be irrotational if
?×(?? ?? ?? ) =0
? (grad ?? ?? )×?? +?? ?? (?×?? ) =0
? (?? ?? ?? -1
·
??
?? )·?? +?? ?? ·0=0
Hence, ?? ?? ?? is irrotatlonal for all the real values of ?? .
3.4 A vector field is given by
????
=(?? ?? +?? ?? ?? )?? +(?? ?? +?? ?? ?? )??
Verify that the field ????
is irrotational or not. Find the scalar potential.
(2015 : 12 Marks)
Solution:
A vector field ??
is said to be irrotational if curl ??
=0, i.e.,
?×??
=0
?
??
×??
=
|
|
??ˆ ??ˆ ??ˆ
?
??? ?
??? ?
??? ?? 2
+?? ?? 2
?? 2
+?? 2
?? 0
|
|
=??ˆ(0-0)-??ˆ(0-0)+??ˆ
(2???? -2???? )
=0
?
???
is irrotational.
Now, it can be written as grad of a scalar field, i.e., to find ?? so that
??? =??
i.e., ??ˆ
??? ??? +??ˆ
??? ??? =(?? 2
+?? ?? 2
)??ˆ+(?? 2
+?? 2
?? )??ˆ
?
??? ??? =?? 2
+?? ?? 2
;
??? ??? =?? 2
+?? 2
??
? ?? =
?? 3
3
+
?? 2
?? 2
2
+?? (?? ) (*)
Differentiating w.r.t. ?? and comparing with (*)
??? ??? =?? 2
?? +?? '
(?? )
? ?? '
(?? )=?? 2
?? (?? )=
?? 3
3
+?? ? ?? (?? ,?? )=
?? 3
3
+
?? 3
3
+
?? 2
?? 2
2
+??
3.5 For what values of the constants ?? ,?? and ?? the vector
????
=(?? +?? +???? )??ˆ+(???? +?? ?? -?? )??ˆ+(-?? +???? +?? ?? )??ˆ
is irrational. Find the divergence in cylindrical coordinates of this vector with
these values.
(2017: 10 Marks)
Solution:
Irrational ? Curl ???
=0
Curl ?
??
=?×???
[
?? ?? ?? ?
??? ?
??? ?
??? ?? ?? h]
( if ?? =???? +???? +h?? )
=|
|
?? ?? ?? ?
??? ?
??? ?
??? ?? +?? +???? ???? +2?? -?? -?? +???? +2?? |
|
=??(?? -(-1)-?? (-1-?? )+?? (?? -1)
=(?? +1)?? +(?? +1)?? +(?? -1)?? =0
? ?? =-1,?? =1,?? =-1
? ???
=(?? +?? -?? )?? +(?? +2?? -?? )?? +(-?? -?? +2?? )??
We find div ???
and express it in cylindrical coordinates.
???
=(?? +?? -?? )?? +(?? +2?? -?? )?? +(-?? -?? +2?? )?? div ???
=?·???
=
??? ?? ??? +
??? ?? ??? +
??? ?? ??? =1+2+2=5 (constant)
Hence, divergence in cylindrical co-ordinates =5.
3.6 Let ???? =?? ?? ??ˆ+?? ?? ??ˆ+?? ?? ??ˆ
. Show that curl (curl ???? )=?? ?????? (?????? ???? )-?? ?? ???? .
(2018: 12 Marks)
Solution:
Curl ( Curl ?? )=?×(?×?? ) .
Now, we know that
??
×(???
×??
)=(??
·??
)???
-(??
·???
)??
Given:
?? 1
??ˆ+?? 2
??ˆ+?? 3
??ˆ
=??
?×?? =|
|
??ˆ ??ˆ ??ˆ
?
??? ?
??? ?
??? ?? 1
?? 2
?? 3
|
|
=??ˆ(
??? 3
??? -
??? 2
??? )-??ˆ(
??? 3
??? -
??? 1
??? )+??ˆ
(
??? 2
??? -
??? 1
??? )
Read More
|
387 videos|203 docs
|
Related Exams
About this Document
|
Oct 31, 2024 Last updated |
Document Description: Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation.
The notes and questions for Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative have been prepared according to the UPSC exam syllabus. Information about Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative covers topics
like and Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative.
Introduction of Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative in English is available as part of our Mathematics Optional Notes for UPSC
for UPSC & Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative in Hindi for Mathematics Optional Notes for UPSC course.
Download more important topics related with notes, lectures and mock test series for UPSC
Exam by signing up for free. UPSC: Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative | Mathematics Optional Notes for UPSC
Description
Full syllabus notes, lecture & questions for Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative | Mathematics Optional Notes for UPSC - UPSC | Plus excerises question with solution to help you revise complete syllabus for Mathematics Optional Notes for UPSC | Best notes, free PDF download
Information about Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative
In this doc you can find the meaning of Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative defined & explained in the simplest way possible. Besides explaining types of
Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative theory, EduRev gives you an ample number of questions to practice Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative tests, examples and also practice UPSC
tests
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
practice quizzes
,Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative | Mathematics Optional Notes for UPSC
,video lectures
,Summary
,Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative | Mathematics Optional Notes for UPSC
,Viva Questions
,shortcuts and tricks
,study material
,Objective type Questions
,Important questions
,Gradient
,Previous Year Questions with Solutions
,Exam
,Semester Notes
,Extra Questions
,past year papers
,Sample Paper
,Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative | Mathematics Optional Notes for UPSC
,mock tests for examination
,Free
,Gradient
,Gradient
,ppt
,MCQs
;
Additional Information about Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative for UPSC Preparation
Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative Free PDF Download
The Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative is an invaluable resource that delves deep into the core of the UPSC exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative now and kickstart your journey towards success in the UPSC exam.
Importance of Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative
The importance of Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative cannot be overstated, especially for UPSC aspirants.
This document holds the key to success in the UPSC exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative Notes
Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative Notes on EduRev are your ultimate resource for success.
Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative UPSC Questions
The "Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative UPSC Questions" guide is a valuable resource for all aspiring students preparing for the
UPSC exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative on the App
Students of UPSC can study Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Gradient, Divergence and Cur in Cartesian and Cylindrical Coordinate and Directional Derivative is prepared as per the latest UPSC syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup