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Page 1
Edurev123
2. Functions of Two and Three Variables
2.1 If ?? =?? ±?? .???? and ?? =?? ±?? .???? with approximately what accuracy can you
calculate the polar coordinates ?? and ?? of the point ?? (?? ,?? ) ? Express your
estimates as percentage changes of the value that ?? and ?? have at point (?? ,?? ) .
(2009 : 20 Marks)
Solution:
Approach : ?? and ?? are functions of two variables ?? and ?? . We write the total derivative of
each to calculate error from the error in the variables.
We have ?? =3,???? =0.01,?? =4,???? =0.01
By definition
?? =v?? 2
+?? 2
?? =tan
-1
?? ??
By total derivative
???? =
??? ??? ???? +
??? ??? ????
=
?? v?? 2
+?? 2
???? +
?? v?? 2
+?? 2
????
=
3×0.01
5
+
4
5
×0.01=
7
5
×0.01
Percentage error in ??
=
??? ?? ×100=
7
5
×0.01
v3
2
+4
2
×100
=
7
25
=0.28%
Similarly, ?? =tan
-1
?? ?? =tan
-1
4
3
Page 2
Edurev123
2. Functions of Two and Three Variables
2.1 If ?? =?? ±?? .???? and ?? =?? ±?? .???? with approximately what accuracy can you
calculate the polar coordinates ?? and ?? of the point ?? (?? ,?? ) ? Express your
estimates as percentage changes of the value that ?? and ?? have at point (?? ,?? ) .
(2009 : 20 Marks)
Solution:
Approach : ?? and ?? are functions of two variables ?? and ?? . We write the total derivative of
each to calculate error from the error in the variables.
We have ?? =3,???? =0.01,?? =4,???? =0.01
By definition
?? =v?? 2
+?? 2
?? =tan
-1
?? ??
By total derivative
???? =
??? ??? ???? +
??? ??? ????
=
?? v?? 2
+?? 2
???? +
?? v?? 2
+?? 2
????
=
3×0.01
5
+
4
5
×0.01=
7
5
×0.01
Percentage error in ??
=
??? ?? ×100=
7
5
×0.01
v3
2
+4
2
×100
=
7
25
=0.28%
Similarly, ?? =tan
-1
?? ?? =tan
-1
4
3
???? =
??? ??? ???? +
??? ??? ????
=
1
1+
?? 2
?? 2
-
?? ?? 2
???? +
1
1+
?? 2
?? 2
-
1
?? ????
=
-?? ?? 2
+?? 2
???? +
?? ?? 2
+?? 2
????
? ???? =
-4×0.01
25
+
3
25
×0.01=
-1
25
×0.01
? Percentage error =
1
2.5
tan
-1
4
3
×0.01×100=0.043%
2.2 Let ?? :?? ?? ??? be defined as
?? (?? ,?? )={
????
v?? ?? +?? ?? if (?? ,?? )?(?? ,?? )
?? if (?? ,?? )=(?? ,?? )
Is ?? continuous at (?? ,?? ) ? Compute partial derivatives of ?? at any point (?? ,?? ) if
exist.
(2009 : 20 Marks)
Solution:
Approach : Using definition of continuity and partial derivative.
Function being rational is continuous at all (?? ,?? ) except possibly at (0,0) .
Continuity at (0,0) :
To check whether lim
(?? ,?? )?(0,0)
?
????
v?? 2
+?? 2
=0.
Converting to polar coordinates with ?? =?? cos ?? ,?? =?? sin ?? ,lim
(?? ,?? )?(0,0)
?~lim
?? ?0
?
lim
?? ?0
?
?? 2
sin ?? cos ?? v?? 2
cos
2
?? +?? 2
sin
2
?? =lim
?? ?0
?
?? 2
sin ?? cos ?? ?? (?? is always +???? )
=lim
?? ?0
??? sin ?? cos ?? =0 for all values of ??
??? (?? ,?? ) is cont. at (0,0) as well.
Partial derivative at (?? ,?? )?(0,0)
Page 3
Edurev123
2. Functions of Two and Three Variables
2.1 If ?? =?? ±?? .???? and ?? =?? ±?? .???? with approximately what accuracy can you
calculate the polar coordinates ?? and ?? of the point ?? (?? ,?? ) ? Express your
estimates as percentage changes of the value that ?? and ?? have at point (?? ,?? ) .
(2009 : 20 Marks)
Solution:
Approach : ?? and ?? are functions of two variables ?? and ?? . We write the total derivative of
each to calculate error from the error in the variables.
We have ?? =3,???? =0.01,?? =4,???? =0.01
By definition
?? =v?? 2
+?? 2
?? =tan
-1
?? ??
By total derivative
???? =
??? ??? ???? +
??? ??? ????
=
?? v?? 2
+?? 2
???? +
?? v?? 2
+?? 2
????
=
3×0.01
5
+
4
5
×0.01=
7
5
×0.01
Percentage error in ??
=
??? ?? ×100=
7
5
×0.01
v3
2
+4
2
×100
=
7
25
=0.28%
Similarly, ?? =tan
-1
?? ?? =tan
-1
4
3
???? =
??? ??? ???? +
??? ??? ????
=
1
1+
?? 2
?? 2
-
?? ?? 2
???? +
1
1+
?? 2
?? 2
-
1
?? ????
=
-?? ?? 2
+?? 2
???? +
?? ?? 2
+?? 2
????
? ???? =
-4×0.01
25
+
3
25
×0.01=
-1
25
×0.01
? Percentage error =
1
2.5
tan
-1
4
3
×0.01×100=0.043%
2.2 Let ?? :?? ?? ??? be defined as
?? (?? ,?? )={
????
v?? ?? +?? ?? if (?? ,?? )?(?? ,?? )
?? if (?? ,?? )=(?? ,?? )
Is ?? continuous at (?? ,?? ) ? Compute partial derivatives of ?? at any point (?? ,?? ) if
exist.
(2009 : 20 Marks)
Solution:
Approach : Using definition of continuity and partial derivative.
Function being rational is continuous at all (?? ,?? ) except possibly at (0,0) .
Continuity at (0,0) :
To check whether lim
(?? ,?? )?(0,0)
?
????
v?? 2
+?? 2
=0.
Converting to polar coordinates with ?? =?? cos ?? ,?? =?? sin ?? ,lim
(?? ,?? )?(0,0)
?~lim
?? ?0
?
lim
?? ?0
?
?? 2
sin ?? cos ?? v?? 2
cos
2
?? +?? 2
sin
2
?? =lim
?? ?0
?
?? 2
sin ?? cos ?? ?? (?? is always +???? )
=lim
?? ?0
??? sin ?? cos ?? =0 for all values of ??
??? (?? ,?? ) is cont. at (0,0) as well.
Partial derivative at (?? ,?? )?(0,0)
??? ??? =
?? v?? 2
+?? 2
-???? ·
2?? 2
v?? 2
+?? 2
(?? 2
+?? 2
)
=
?? (?? 2
+?? 2
)-?? ?? 2
(?? 2
+?? 2
)
3/2
=
?? 3
(?? 2
+?? 2
)
3/2
??? ??? =
?? v?? 2
+?? 2
-???? ·
2?? 2v?? 2
+?? 2
(?? 2
+?? 2
)
=
?? 3
(?? 2
+?? 2
)
3/2
(Simply ?? ??? as it is symmetrical function in ??
and ?? )
Partial derivative at (0,0)
??? ??? =lim
h?0
?
?? (h,0)-?? (0,0)
h
=lim
h?0
?
0-0
h
=0
??? ??? =lim
?? ?0
?
?? (0,?? )-?? (0,0)
?? =lim
?? ?0
?
0-0
?? =0
2.3 A space probe in the shape of the ellipsoid ?? ?? ?? +?? ?? +?? ?? ?? =???? enters the
earth's atmosphere and its surface begins to heat. After one hour the temperature
at the point (?? ,?? ,?? ) on the probe surface is given by
?? (?? ,?? ,?? )=?? ?? ?? +?? ???? -???? ?? +??????
Find hottest point on the probe surface.
(2009 : 20 Marks)
Solution:
Approach: Use Lagrange's multipliers.
Max.?? (?? ,?? ,?? )=8?? 2
+4???? -16?? +600 …(i)
?????????????? ???? 4?? 2
+?? 2
+4?? 2
=1 …(???? )
?????? ?? =(8?? 2
+4???? -16?? +600)+?? (4?? 2
+?? 2
+4?? 2
-16)
Taking total derivative of ??
???? =(16+8?? )?????? +(4?? +2???? )???? +(8???? -16)???? +4??
For maximum/minimum
???? =0?(16+8?? )?? =0;4?? +2???? =0;8???? +4?? -16=0 …(iii)
Let ?? ?0?
?? =-
1
2
Page 4
Edurev123
2. Functions of Two and Three Variables
2.1 If ?? =?? ±?? .???? and ?? =?? ±?? .???? with approximately what accuracy can you
calculate the polar coordinates ?? and ?? of the point ?? (?? ,?? ) ? Express your
estimates as percentage changes of the value that ?? and ?? have at point (?? ,?? ) .
(2009 : 20 Marks)
Solution:
Approach : ?? and ?? are functions of two variables ?? and ?? . We write the total derivative of
each to calculate error from the error in the variables.
We have ?? =3,???? =0.01,?? =4,???? =0.01
By definition
?? =v?? 2
+?? 2
?? =tan
-1
?? ??
By total derivative
???? =
??? ??? ???? +
??? ??? ????
=
?? v?? 2
+?? 2
???? +
?? v?? 2
+?? 2
????
=
3×0.01
5
+
4
5
×0.01=
7
5
×0.01
Percentage error in ??
=
??? ?? ×100=
7
5
×0.01
v3
2
+4
2
×100
=
7
25
=0.28%
Similarly, ?? =tan
-1
?? ?? =tan
-1
4
3
???? =
??? ??? ???? +
??? ??? ????
=
1
1+
?? 2
?? 2
-
?? ?? 2
???? +
1
1+
?? 2
?? 2
-
1
?? ????
=
-?? ?? 2
+?? 2
???? +
?? ?? 2
+?? 2
????
? ???? =
-4×0.01
25
+
3
25
×0.01=
-1
25
×0.01
? Percentage error =
1
2.5
tan
-1
4
3
×0.01×100=0.043%
2.2 Let ?? :?? ?? ??? be defined as
?? (?? ,?? )={
????
v?? ?? +?? ?? if (?? ,?? )?(?? ,?? )
?? if (?? ,?? )=(?? ,?? )
Is ?? continuous at (?? ,?? ) ? Compute partial derivatives of ?? at any point (?? ,?? ) if
exist.
(2009 : 20 Marks)
Solution:
Approach : Using definition of continuity and partial derivative.
Function being rational is continuous at all (?? ,?? ) except possibly at (0,0) .
Continuity at (0,0) :
To check whether lim
(?? ,?? )?(0,0)
?
????
v?? 2
+?? 2
=0.
Converting to polar coordinates with ?? =?? cos ?? ,?? =?? sin ?? ,lim
(?? ,?? )?(0,0)
?~lim
?? ?0
?
lim
?? ?0
?
?? 2
sin ?? cos ?? v?? 2
cos
2
?? +?? 2
sin
2
?? =lim
?? ?0
?
?? 2
sin ?? cos ?? ?? (?? is always +???? )
=lim
?? ?0
??? sin ?? cos ?? =0 for all values of ??
??? (?? ,?? ) is cont. at (0,0) as well.
Partial derivative at (?? ,?? )?(0,0)
??? ??? =
?? v?? 2
+?? 2
-???? ·
2?? 2
v?? 2
+?? 2
(?? 2
+?? 2
)
=
?? (?? 2
+?? 2
)-?? ?? 2
(?? 2
+?? 2
)
3/2
=
?? 3
(?? 2
+?? 2
)
3/2
??? ??? =
?? v?? 2
+?? 2
-???? ·
2?? 2v?? 2
+?? 2
(?? 2
+?? 2
)
=
?? 3
(?? 2
+?? 2
)
3/2
(Simply ?? ??? as it is symmetrical function in ??
and ?? )
Partial derivative at (0,0)
??? ??? =lim
h?0
?
?? (h,0)-?? (0,0)
h
=lim
h?0
?
0-0
h
=0
??? ??? =lim
?? ?0
?
?? (0,?? )-?? (0,0)
?? =lim
?? ?0
?
0-0
?? =0
2.3 A space probe in the shape of the ellipsoid ?? ?? ?? +?? ?? +?? ?? ?? =???? enters the
earth's atmosphere and its surface begins to heat. After one hour the temperature
at the point (?? ,?? ,?? ) on the probe surface is given by
?? (?? ,?? ,?? )=?? ?? ?? +?? ???? -???? ?? +??????
Find hottest point on the probe surface.
(2009 : 20 Marks)
Solution:
Approach: Use Lagrange's multipliers.
Max.?? (?? ,?? ,?? )=8?? 2
+4???? -16?? +600 …(i)
?????????????? ???? 4?? 2
+?? 2
+4?? 2
=1 …(???? )
?????? ?? =(8?? 2
+4???? -16?? +600)+?? (4?? 2
+?? 2
+4?? 2
-16)
Taking total derivative of ??
???? =(16+8?? )?????? +(4?? +2???? )???? +(8???? -16)???? +4??
For maximum/minimum
???? =0?(16+8?? )?? =0;4?? +2???? =0;8???? +4?? -16=0 …(iii)
Let ?? ?0?
?? =-
1
2
4?? -?? =0
-4z+4?? =16
}?? =
16
3
;?? =
4
3
Putting this in (iii)
?? 2
=4-
1
4
(
256
9
+
64
9
)=
-44
9
<0
So this value is not possible.
? x=0
4?? +2???? =0
8???? +4?? =16
(4?? 2
-4)?? =-16
? y=
4
1-?? 2
And z=
-??
?? 2
=
2?? ?? 2
-1
Substituting in (ii)
16
(1-?? 2
)
2
+
16?? 2
(1-?? 2
)
2
=16
? 1+?? 2
=(1-?? 2
)
2
? 1+?? 2
=?? 4
-2?? 2
+1
? ?? 2
(?? 2
-3)=0
? ?? =0;?? 2
=3
?? =0??? =0,?? =4,?? =0
Let us consider ?? and ?? as independent and ?? as dependent variable.
(Note: We can do this because we have been given a relation between ?? ,?? and ?? )
From (ii) partially differentiating w.r.t. ?? .
8?? +8?? ??? ??? =0?
??? ??? =-
?? ?? ?
??? ??? =16?? +4?? ??? ??? -16
??? ??? =16?? -
4????
?? +
16?? ?? ?
2
?? ??? 2
=16+(16-4?? )(
1
?? -
?? ?? 2
·
??? ??? )
=16+(16-4?? )(
1
?? +
?? 2
?? 3
)
Page 5
Edurev123
2. Functions of Two and Three Variables
2.1 If ?? =?? ±?? .???? and ?? =?? ±?? .???? with approximately what accuracy can you
calculate the polar coordinates ?? and ?? of the point ?? (?? ,?? ) ? Express your
estimates as percentage changes of the value that ?? and ?? have at point (?? ,?? ) .
(2009 : 20 Marks)
Solution:
Approach : ?? and ?? are functions of two variables ?? and ?? . We write the total derivative of
each to calculate error from the error in the variables.
We have ?? =3,???? =0.01,?? =4,???? =0.01
By definition
?? =v?? 2
+?? 2
?? =tan
-1
?? ??
By total derivative
???? =
??? ??? ???? +
??? ??? ????
=
?? v?? 2
+?? 2
???? +
?? v?? 2
+?? 2
????
=
3×0.01
5
+
4
5
×0.01=
7
5
×0.01
Percentage error in ??
=
??? ?? ×100=
7
5
×0.01
v3
2
+4
2
×100
=
7
25
=0.28%
Similarly, ?? =tan
-1
?? ?? =tan
-1
4
3
???? =
??? ??? ???? +
??? ??? ????
=
1
1+
?? 2
?? 2
-
?? ?? 2
???? +
1
1+
?? 2
?? 2
-
1
?? ????
=
-?? ?? 2
+?? 2
???? +
?? ?? 2
+?? 2
????
? ???? =
-4×0.01
25
+
3
25
×0.01=
-1
25
×0.01
? Percentage error =
1
2.5
tan
-1
4
3
×0.01×100=0.043%
2.2 Let ?? :?? ?? ??? be defined as
?? (?? ,?? )={
????
v?? ?? +?? ?? if (?? ,?? )?(?? ,?? )
?? if (?? ,?? )=(?? ,?? )
Is ?? continuous at (?? ,?? ) ? Compute partial derivatives of ?? at any point (?? ,?? ) if
exist.
(2009 : 20 Marks)
Solution:
Approach : Using definition of continuity and partial derivative.
Function being rational is continuous at all (?? ,?? ) except possibly at (0,0) .
Continuity at (0,0) :
To check whether lim
(?? ,?? )?(0,0)
?
????
v?? 2
+?? 2
=0.
Converting to polar coordinates with ?? =?? cos ?? ,?? =?? sin ?? ,lim
(?? ,?? )?(0,0)
?~lim
?? ?0
?
lim
?? ?0
?
?? 2
sin ?? cos ?? v?? 2
cos
2
?? +?? 2
sin
2
?? =lim
?? ?0
?
?? 2
sin ?? cos ?? ?? (?? is always +???? )
=lim
?? ?0
??? sin ?? cos ?? =0 for all values of ??
??? (?? ,?? ) is cont. at (0,0) as well.
Partial derivative at (?? ,?? )?(0,0)
??? ??? =
?? v?? 2
+?? 2
-???? ·
2?? 2
v?? 2
+?? 2
(?? 2
+?? 2
)
=
?? (?? 2
+?? 2
)-?? ?? 2
(?? 2
+?? 2
)
3/2
=
?? 3
(?? 2
+?? 2
)
3/2
??? ??? =
?? v?? 2
+?? 2
-???? ·
2?? 2v?? 2
+?? 2
(?? 2
+?? 2
)
=
?? 3
(?? 2
+?? 2
)
3/2
(Simply ?? ??? as it is symmetrical function in ??
and ?? )
Partial derivative at (0,0)
??? ??? =lim
h?0
?
?? (h,0)-?? (0,0)
h
=lim
h?0
?
0-0
h
=0
??? ??? =lim
?? ?0
?
?? (0,?? )-?? (0,0)
?? =lim
?? ?0
?
0-0
?? =0
2.3 A space probe in the shape of the ellipsoid ?? ?? ?? +?? ?? +?? ?? ?? =???? enters the
earth's atmosphere and its surface begins to heat. After one hour the temperature
at the point (?? ,?? ,?? ) on the probe surface is given by
?? (?? ,?? ,?? )=?? ?? ?? +?? ???? -???? ?? +??????
Find hottest point on the probe surface.
(2009 : 20 Marks)
Solution:
Approach: Use Lagrange's multipliers.
Max.?? (?? ,?? ,?? )=8?? 2
+4???? -16?? +600 …(i)
?????????????? ???? 4?? 2
+?? 2
+4?? 2
=1 …(???? )
?????? ?? =(8?? 2
+4???? -16?? +600)+?? (4?? 2
+?? 2
+4?? 2
-16)
Taking total derivative of ??
???? =(16+8?? )?????? +(4?? +2???? )???? +(8???? -16)???? +4??
For maximum/minimum
???? =0?(16+8?? )?? =0;4?? +2???? =0;8???? +4?? -16=0 …(iii)
Let ?? ?0?
?? =-
1
2
4?? -?? =0
-4z+4?? =16
}?? =
16
3
;?? =
4
3
Putting this in (iii)
?? 2
=4-
1
4
(
256
9
+
64
9
)=
-44
9
<0
So this value is not possible.
? x=0
4?? +2???? =0
8???? +4?? =16
(4?? 2
-4)?? =-16
? y=
4
1-?? 2
And z=
-??
?? 2
=
2?? ?? 2
-1
Substituting in (ii)
16
(1-?? 2
)
2
+
16?? 2
(1-?? 2
)
2
=16
? 1+?? 2
=(1-?? 2
)
2
? 1+?? 2
=?? 4
-2?? 2
+1
? ?? 2
(?? 2
-3)=0
? ?? =0;?? 2
=3
?? =0??? =0,?? =4,?? =0
Let us consider ?? and ?? as independent and ?? as dependent variable.
(Note: We can do this because we have been given a relation between ?? ,?? and ?? )
From (ii) partially differentiating w.r.t. ?? .
8?? +8?? ??? ??? =0?
??? ??? =-
?? ?? ?
??? ??? =16?? +4?? ??? ??? -16
??? ??? =16?? -
4????
?? +
16?? ?? ?
2
?? ??? 2
=16+(16-4?? )(
1
?? -
?? ?? 2
·
??? ??? )
=16+(16-4?? )(
1
?? +
?? 2
?? 3
)
At ?? =4,
?
2
?? ??? 2
>0, so this cannot be maxima.
?? 2
= 3,?? =±v3
?? =v3,?? =v3,?? =-2,?? =0
Again
?
2
?? ??? 2
>0 so this can not be a maxima.
?? =-v3,?? =-v3,?? =-2,?? =0
?
2
?? ??? 2
=16-
24
v3
<0
3o, this is a maxima.
? Hottest point is (0,-2,-v3) .
2.4 Evaluate :?? =?
?? ??????????? +???????? +?? ?? ?? ???????? where ?? is the outer surface of the
part of the sphere ?? ?? + ?? ?? +?? ?? =?? in the first quadrant.
(2009 : 20 Marks)
Solution:
Approach: Surface integral with two changing variables can be calculated by taking the
projection of the surface in the required plane. Use of polar coordinates can simplify
integration.
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Importance of Functions of Two and Three Variables
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Functions of Two and Three Variables Notes
Functions of Two and Three Variables Notes offer in-depth insights into the specific topic to help you master it with ease.
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Functions of Two and Three Variables UPSC Questions
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The content of Functions of Two and Three Variables is prepared as per the latest UPSC syllabus.
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