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Page 1
Edurev123
7. Functions of Several (Two or Three)
Variables
7.1 Find the maxima, minima and saddle points of the surface ?? =(?? ?? -
?? ?? )?? (-?? ?? -?? ?? )/?? .
(2010 : 15 Marks)
Solution:
Given, the surface is ?? =(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
At extremum,
?? ?? =?? ?? =0
???????????????????????????????????????????????????????????????????????????? ?? =2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)
??????????????????? (
-?? 2
-?? 2
2
)
{2?? +(?? 2
+?? 2
)(-?? )}=0 ???????? ?
??????????????????????????????????????????????2?? -?? (?? 2
-?? 2
)=0 ???????? ?
??????????????????????????????????????????????????? (2-?? 2
+?? 2
)=0????????????????????????????????????????????????????????????????????????????(1)?
?????????????????????????????????????????????? ?? =-2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
×(
-2?? 2
)=0
????????????????-2?? +(?? 2
-?? 2
)(-?? )=0?????????
?????????????????????????-?? (2+?? 2
-?? 2
)=0?????????
????????????????????????????? (2+?? 2
-?? 2
)=0?????????????????????????????????????????????????????????????????????????????????????????????????(2)
Solving (1) and (2), we get solutions as
(?? ,?? )=(0,0),(±v2,0),(0,±v2)
?? ????
=2?? (-?? 2
-?? 2
)/2
+2?? ?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)-(3?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
-(?? 3
-?? 2
?? )?? (
-?? 2
-?? 2
2
)
×(-?? )
=?? (
-?? 2
-?? 2
2
)
(2-2?? 2
-3?? 2
+?? 2
+?? 4
-?? 2
?? 2
)
Page 2
Edurev123
7. Functions of Several (Two or Three)
Variables
7.1 Find the maxima, minima and saddle points of the surface ?? =(?? ?? -
?? ?? )?? (-?? ?? -?? ?? )/?? .
(2010 : 15 Marks)
Solution:
Given, the surface is ?? =(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
At extremum,
?? ?? =?? ?? =0
???????????????????????????????????????????????????????????????????????????? ?? =2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)
??????????????????? (
-?? 2
-?? 2
2
)
{2?? +(?? 2
+?? 2
)(-?? )}=0 ???????? ?
??????????????????????????????????????????????2?? -?? (?? 2
-?? 2
)=0 ???????? ?
??????????????????????????????????????????????????? (2-?? 2
+?? 2
)=0????????????????????????????????????????????????????????????????????????????(1)?
?????????????????????????????????????????????? ?? =-2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
×(
-2?? 2
)=0
????????????????-2?? +(?? 2
-?? 2
)(-?? )=0?????????
?????????????????????????-?? (2+?? 2
-?? 2
)=0?????????
????????????????????????????? (2+?? 2
-?? 2
)=0?????????????????????????????????????????????????????????????????????????????????????????????????(2)
Solving (1) and (2), we get solutions as
(?? ,?? )=(0,0),(±v2,0),(0,±v2)
?? ????
=2?? (-?? 2
-?? 2
)/2
+2?? ?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)-(3?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
-(?? 3
-?? 2
?? )?? (
-?? 2
-?? 2
2
)
×(-?? )
=?? (
-?? 2
-?? 2
2
)
(2-2?? 2
-3?? 2
+?? 2
+?? 4
-?? 2
?? 2
)
?? ????
=(-2?? -?? 2
?? +?? 3
)?? (
?? 2
-?? 2
2
)
×(-?? )+?? (
-?? 2
-?? 2
2
)
(-2-?? 2
+3?? 2
)
=?? (
-?? 2
-?? 2
2
)
(2?? 2
+?? 2
?? 2
-?? 4
-2-?? 2
+3?? 2
)
?? ????
=(-2?? -?? 2
?? +?? 3
)(?? -?? 2
-?? 2
2
)×(-?? )+?? (
-?? 2
-?? 2
2
)
(-2???? )
=?? (
-?? 2
-?? 2
2
)
(2???? +?? 3
?? -?? ?? 3
-2???? )
=?? (
-?? 2
-?? 2
2
)
(?? 3
?? -?? ?? 3
)
At (±v2,0):??? ????
=
-?? ?? ,?? ????
=
?? ?? ,?? ????
=0
???? ????
·?? ????
-?? ????
2
=
-8
?? 2
At (0,±v2):???? ????
=
?? ?? ,?? ????
=
-2
?? ,?? ????
=0
???? ????
·?? ????
-?? ?? ?? 2
=
-8
?? 2
At all extremum points, ?? ????
·?? ????
-?? ????
2
<0.
? All extremum points viz. (0,0),(±v2,0) and (0,±v2) are saddle points.
7.2 Find the shortest distance from the origin (?? ,?? ) to the hyperbola
?? ?? +?? ???? +?? ?? ?? =??????
(2011 : 15 Marks)
Solution:
Let (?? ,?? ) be any point on the given hyperbola.
We need to minimize v?? 2
+?? 2
or equivalently (?? 2
+?? 2
) .
Consider
?? (?? ,?? ,?? )=?? 2
+?? 2
-?? (?? 2
+8???? +7?? 2
-225)
???????????????????????????????????????? ?? (?? ,?? ,?? )=2?? -2???? -8???? =0
Or (?? -1)?? +4???? =0??????????????????????????????????????????????????????????????????????????????????(i)
Also,?????????????????????????????????? ?? (?? ,?? ,?? )=2?? -8???? -14???? =0?????????????????????????????????
Or 4???? +(7?? -1)?? =0
Page 3
Edurev123
7. Functions of Several (Two or Three)
Variables
7.1 Find the maxima, minima and saddle points of the surface ?? =(?? ?? -
?? ?? )?? (-?? ?? -?? ?? )/?? .
(2010 : 15 Marks)
Solution:
Given, the surface is ?? =(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
At extremum,
?? ?? =?? ?? =0
???????????????????????????????????????????????????????????????????????????? ?? =2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)
??????????????????? (
-?? 2
-?? 2
2
)
{2?? +(?? 2
+?? 2
)(-?? )}=0 ???????? ?
??????????????????????????????????????????????2?? -?? (?? 2
-?? 2
)=0 ???????? ?
??????????????????????????????????????????????????? (2-?? 2
+?? 2
)=0????????????????????????????????????????????????????????????????????????????(1)?
?????????????????????????????????????????????? ?? =-2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
×(
-2?? 2
)=0
????????????????-2?? +(?? 2
-?? 2
)(-?? )=0?????????
?????????????????????????-?? (2+?? 2
-?? 2
)=0?????????
????????????????????????????? (2+?? 2
-?? 2
)=0?????????????????????????????????????????????????????????????????????????????????????????????????(2)
Solving (1) and (2), we get solutions as
(?? ,?? )=(0,0),(±v2,0),(0,±v2)
?? ????
=2?? (-?? 2
-?? 2
)/2
+2?? ?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)-(3?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
-(?? 3
-?? 2
?? )?? (
-?? 2
-?? 2
2
)
×(-?? )
=?? (
-?? 2
-?? 2
2
)
(2-2?? 2
-3?? 2
+?? 2
+?? 4
-?? 2
?? 2
)
?? ????
=(-2?? -?? 2
?? +?? 3
)?? (
?? 2
-?? 2
2
)
×(-?? )+?? (
-?? 2
-?? 2
2
)
(-2-?? 2
+3?? 2
)
=?? (
-?? 2
-?? 2
2
)
(2?? 2
+?? 2
?? 2
-?? 4
-2-?? 2
+3?? 2
)
?? ????
=(-2?? -?? 2
?? +?? 3
)(?? -?? 2
-?? 2
2
)×(-?? )+?? (
-?? 2
-?? 2
2
)
(-2???? )
=?? (
-?? 2
-?? 2
2
)
(2???? +?? 3
?? -?? ?? 3
-2???? )
=?? (
-?? 2
-?? 2
2
)
(?? 3
?? -?? ?? 3
)
At (±v2,0):??? ????
=
-?? ?? ,?? ????
=
?? ?? ,?? ????
=0
???? ????
·?? ????
-?? ????
2
=
-8
?? 2
At (0,±v2):???? ????
=
?? ?? ,?? ????
=
-2
?? ,?? ????
=0
???? ????
·?? ????
-?? ?? ?? 2
=
-8
?? 2
At all extremum points, ?? ????
·?? ????
-?? ????
2
<0.
? All extremum points viz. (0,0),(±v2,0) and (0,±v2) are saddle points.
7.2 Find the shortest distance from the origin (?? ,?? ) to the hyperbola
?? ?? +?? ???? +?? ?? ?? =??????
(2011 : 15 Marks)
Solution:
Let (?? ,?? ) be any point on the given hyperbola.
We need to minimize v?? 2
+?? 2
or equivalently (?? 2
+?? 2
) .
Consider
?? (?? ,?? ,?? )=?? 2
+?? 2
-?? (?? 2
+8???? +7?? 2
-225)
???????????????????????????????????????? ?? (?? ,?? ,?? )=2?? -2???? -8???? =0
Or (?? -1)?? +4???? =0??????????????????????????????????????????????????????????????????????????????????(i)
Also,?????????????????????????????????? ?? (?? ,?? ,?? )=2?? -8???? -14???? =0?????????????????????????????????
Or 4???? +(7?? -1)?? =0
Since (?? ,?? )?(0,0) (as hyperbola does not pass through the origin), then solving for ?? ,
we have
|
?? -1 4?? 4?? 7?? -1
|=0?9?? 2
+8?? -1=0
???????????????????????????????????????????????????????????????????????????? =-1,
1
9
If ?? =-1, then -2?? -4?? =0 or ?? =-2?? .
? From ?? 2
+8???? +7?? 2
=225, we have
-5?? 2
=225 for which no real solution exists.
If ?? =
1
9
, then from (i), ???????? =2??
? From ?? 2
+8???? +7?? 2
=225, we have
?? 2
=5 and ?? 2
=20
??????????????????????????????????????????????????????????? 2
+?? 2
=25
Thus, the required shortest distance is v25=5.
7.3 Let ?? (?? ,?? )={
(?? +?? )
?? ?? ?? +?? ?? ; if (?? ,?? )?(?? ,?? )
?? , if (?? ,?? )=(?? ,?? )
. Show that
?? ?? ?? ?? and
?? ?? ?? ?? exist at (?? ,?? ) though
?? (?? ,?? ) is not continuous at (?? ,?? ) .
(2012 : 15 Marks)
Solution:
?
?? (?? ,?? )?={
(?? +?? )
2
?? 2
+?? 2
, if (?? ,?? )?(0,0)
1, if (?? ,?? )?(0,0;
?? ?? (0,0)?=lim
h?0
?
?? (h,0)-?? (0,0)
h
?=lim
h?0
?
h
2
h
-1
h
=0
?? ?? (0,0)?=lim
?? ?0
?
?? (0,?? )-?? (0,0)
?? ?=lim
?? ?0
?
?? 2
?? 2
-1
?? =0
??? ?? (0,0) and ?? ?? (0,0) exists.
Page 4
Edurev123
7. Functions of Several (Two or Three)
Variables
7.1 Find the maxima, minima and saddle points of the surface ?? =(?? ?? -
?? ?? )?? (-?? ?? -?? ?? )/?? .
(2010 : 15 Marks)
Solution:
Given, the surface is ?? =(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
At extremum,
?? ?? =?? ?? =0
???????????????????????????????????????????????????????????????????????????? ?? =2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)
??????????????????? (
-?? 2
-?? 2
2
)
{2?? +(?? 2
+?? 2
)(-?? )}=0 ???????? ?
??????????????????????????????????????????????2?? -?? (?? 2
-?? 2
)=0 ???????? ?
??????????????????????????????????????????????????? (2-?? 2
+?? 2
)=0????????????????????????????????????????????????????????????????????????????(1)?
?????????????????????????????????????????????? ?? =-2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
×(
-2?? 2
)=0
????????????????-2?? +(?? 2
-?? 2
)(-?? )=0?????????
?????????????????????????-?? (2+?? 2
-?? 2
)=0?????????
????????????????????????????? (2+?? 2
-?? 2
)=0?????????????????????????????????????????????????????????????????????????????????????????????????(2)
Solving (1) and (2), we get solutions as
(?? ,?? )=(0,0),(±v2,0),(0,±v2)
?? ????
=2?? (-?? 2
-?? 2
)/2
+2?? ?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)-(3?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
-(?? 3
-?? 2
?? )?? (
-?? 2
-?? 2
2
)
×(-?? )
=?? (
-?? 2
-?? 2
2
)
(2-2?? 2
-3?? 2
+?? 2
+?? 4
-?? 2
?? 2
)
?? ????
=(-2?? -?? 2
?? +?? 3
)?? (
?? 2
-?? 2
2
)
×(-?? )+?? (
-?? 2
-?? 2
2
)
(-2-?? 2
+3?? 2
)
=?? (
-?? 2
-?? 2
2
)
(2?? 2
+?? 2
?? 2
-?? 4
-2-?? 2
+3?? 2
)
?? ????
=(-2?? -?? 2
?? +?? 3
)(?? -?? 2
-?? 2
2
)×(-?? )+?? (
-?? 2
-?? 2
2
)
(-2???? )
=?? (
-?? 2
-?? 2
2
)
(2???? +?? 3
?? -?? ?? 3
-2???? )
=?? (
-?? 2
-?? 2
2
)
(?? 3
?? -?? ?? 3
)
At (±v2,0):??? ????
=
-?? ?? ,?? ????
=
?? ?? ,?? ????
=0
???? ????
·?? ????
-?? ????
2
=
-8
?? 2
At (0,±v2):???? ????
=
?? ?? ,?? ????
=
-2
?? ,?? ????
=0
???? ????
·?? ????
-?? ?? ?? 2
=
-8
?? 2
At all extremum points, ?? ????
·?? ????
-?? ????
2
<0.
? All extremum points viz. (0,0),(±v2,0) and (0,±v2) are saddle points.
7.2 Find the shortest distance from the origin (?? ,?? ) to the hyperbola
?? ?? +?? ???? +?? ?? ?? =??????
(2011 : 15 Marks)
Solution:
Let (?? ,?? ) be any point on the given hyperbola.
We need to minimize v?? 2
+?? 2
or equivalently (?? 2
+?? 2
) .
Consider
?? (?? ,?? ,?? )=?? 2
+?? 2
-?? (?? 2
+8???? +7?? 2
-225)
???????????????????????????????????????? ?? (?? ,?? ,?? )=2?? -2???? -8???? =0
Or (?? -1)?? +4???? =0??????????????????????????????????????????????????????????????????????????????????(i)
Also,?????????????????????????????????? ?? (?? ,?? ,?? )=2?? -8???? -14???? =0?????????????????????????????????
Or 4???? +(7?? -1)?? =0
Since (?? ,?? )?(0,0) (as hyperbola does not pass through the origin), then solving for ?? ,
we have
|
?? -1 4?? 4?? 7?? -1
|=0?9?? 2
+8?? -1=0
???????????????????????????????????????????????????????????????????????????? =-1,
1
9
If ?? =-1, then -2?? -4?? =0 or ?? =-2?? .
? From ?? 2
+8???? +7?? 2
=225, we have
-5?? 2
=225 for which no real solution exists.
If ?? =
1
9
, then from (i), ???????? =2??
? From ?? 2
+8???? +7?? 2
=225, we have
?? 2
=5 and ?? 2
=20
??????????????????????????????????????????????????????????? 2
+?? 2
=25
Thus, the required shortest distance is v25=5.
7.3 Let ?? (?? ,?? )={
(?? +?? )
?? ?? ?? +?? ?? ; if (?? ,?? )?(?? ,?? )
?? , if (?? ,?? )=(?? ,?? )
. Show that
?? ?? ?? ?? and
?? ?? ?? ?? exist at (?? ,?? ) though
?? (?? ,?? ) is not continuous at (?? ,?? ) .
(2012 : 15 Marks)
Solution:
?
?? (?? ,?? )?={
(?? +?? )
2
?? 2
+?? 2
, if (?? ,?? )?(0,0)
1, if (?? ,?? )?(0,0;
?? ?? (0,0)?=lim
h?0
?
?? (h,0)-?? (0,0)
h
?=lim
h?0
?
h
2
h
-1
h
=0
?? ?? (0,0)?=lim
?? ?0
?
?? (0,?? )-?? (0,0)
?? ?=lim
?? ?0
?
?? 2
?? 2
-1
?? =0
??? ?? (0,0) and ?? ?? (0,0) exists.
Now,
lim
(?? ,?? )?(0,0)
??? (?? ,?? )?= lim
(?? ,?? )?(0,0)
?
(?? +?? )
2
?? 2
+?? 2
?= lim
(?? ,?? )?(0,0)
?
?? 2
+?? 2
+2????
?? 2
+?? 2
Taking ?? =????
???????????????????????????????????????????????????=lim
?? ?0
?
?? 2
+?? 2
?? 2
+2?? ·????
?? 2
+?? 2
?? 2
???????????????????????????????????????????????????=lim
?? ?0
?
?? +?? 2
+2?? 1+?? 2
, which is different for the different values of ?? .
?lim
(?? ,?? )?(0,0)
??? (?? ,?? ) does not exist.
7.4 Find the minimum distance of the line given by the planes ?? ?? +?? ?? +?? ?? =??
and ?? -?? =?? from the origin, by the method of Lagrange's multipliers.
(2012 : 15 Marks)
Solution:
Let
?? =?? (?? ,?? ,?? )=?? 2
+?? 2
+?? 2
???????????????????????????????????????????(??)
?? (?? ,?? ,?? ) =3?? +4?? +5?? -7=0????????????????????????????????????????????????(ii)
h(?? ,?? ,?? ) =?? -?? -9=0???????????????????????????????????????????????????????????????(iii)
??? =?? ??? +?? ?h
<2?? ,2?? ,2?? > =?? (3,4,5)+?? (1,0,-1)
2?? =3?? +?? 2?? =4?? ??2?? =5?? -?? ????? =
3?? +?? 2
,?? =2?? ,?? =
5?? -?? 2
From (ii),
3[
3?? +?? 2
+2?? +
5?? -?? 2
]=7?????????
????????????????????????????????????????????????????????????????????????????????????????????25?? -?? =7???????????????????????????????????????????????(v)??
???????? (?????? )????????????????????????????????????????????????????????
3?? +?? 2
-
5?? -?? 2
=9
??????????????????????????????????????????????????????????????????????????????????????????????-?? +?? =9?????????????????????????????????????????????????(vi)
Solving (v) and (vi)
Page 5
Edurev123
7. Functions of Several (Two or Three)
Variables
7.1 Find the maxima, minima and saddle points of the surface ?? =(?? ?? -
?? ?? )?? (-?? ?? -?? ?? )/?? .
(2010 : 15 Marks)
Solution:
Given, the surface is ?? =(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
At extremum,
?? ?? =?? ?? =0
???????????????????????????????????????????????????????????????????????????? ?? =2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)
??????????????????? (
-?? 2
-?? 2
2
)
{2?? +(?? 2
+?? 2
)(-?? )}=0 ???????? ?
??????????????????????????????????????????????2?? -?? (?? 2
-?? 2
)=0 ???????? ?
??????????????????????????????????????????????????? (2-?? 2
+?? 2
)=0????????????????????????????????????????????????????????????????????????????(1)?
?????????????????????????????????????????????? ?? =-2?? ?? (-?? 2
-?? 2
)/2
+(?? 2
-?? 2
)?? (-?? 2
-?? 2
)/2
×(
-2?? 2
)=0
????????????????-2?? +(?? 2
-?? 2
)(-?? )=0?????????
?????????????????????????-?? (2+?? 2
-?? 2
)=0?????????
????????????????????????????? (2+?? 2
-?? 2
)=0?????????????????????????????????????????????????????????????????????????????????????????????????(2)
Solving (1) and (2), we get solutions as
(?? ,?? )=(0,0),(±v2,0),(0,±v2)
?? ????
=2?? (-?? 2
-?? 2
)/2
+2?? ?? (
-?? 2
-?? 2
2
)
×(
-2?? 2
)-(3?? 2
-?? 2
)?? (
-?? 2
-?? 2
2
)
-(?? 3
-?? 2
?? )?? (
-?? 2
-?? 2
2
)
×(-?? )
=?? (
-?? 2
-?? 2
2
)
(2-2?? 2
-3?? 2
+?? 2
+?? 4
-?? 2
?? 2
)
?? ????
=(-2?? -?? 2
?? +?? 3
)?? (
?? 2
-?? 2
2
)
×(-?? )+?? (
-?? 2
-?? 2
2
)
(-2-?? 2
+3?? 2
)
=?? (
-?? 2
-?? 2
2
)
(2?? 2
+?? 2
?? 2
-?? 4
-2-?? 2
+3?? 2
)
?? ????
=(-2?? -?? 2
?? +?? 3
)(?? -?? 2
-?? 2
2
)×(-?? )+?? (
-?? 2
-?? 2
2
)
(-2???? )
=?? (
-?? 2
-?? 2
2
)
(2???? +?? 3
?? -?? ?? 3
-2???? )
=?? (
-?? 2
-?? 2
2
)
(?? 3
?? -?? ?? 3
)
At (±v2,0):??? ????
=
-?? ?? ,?? ????
=
?? ?? ,?? ????
=0
???? ????
·?? ????
-?? ????
2
=
-8
?? 2
At (0,±v2):???? ????
=
?? ?? ,?? ????
=
-2
?? ,?? ????
=0
???? ????
·?? ????
-?? ?? ?? 2
=
-8
?? 2
At all extremum points, ?? ????
·?? ????
-?? ????
2
<0.
? All extremum points viz. (0,0),(±v2,0) and (0,±v2) are saddle points.
7.2 Find the shortest distance from the origin (?? ,?? ) to the hyperbola
?? ?? +?? ???? +?? ?? ?? =??????
(2011 : 15 Marks)
Solution:
Let (?? ,?? ) be any point on the given hyperbola.
We need to minimize v?? 2
+?? 2
or equivalently (?? 2
+?? 2
) .
Consider
?? (?? ,?? ,?? )=?? 2
+?? 2
-?? (?? 2
+8???? +7?? 2
-225)
???????????????????????????????????????? ?? (?? ,?? ,?? )=2?? -2???? -8???? =0
Or (?? -1)?? +4???? =0??????????????????????????????????????????????????????????????????????????????????(i)
Also,?????????????????????????????????? ?? (?? ,?? ,?? )=2?? -8???? -14???? =0?????????????????????????????????
Or 4???? +(7?? -1)?? =0
Since (?? ,?? )?(0,0) (as hyperbola does not pass through the origin), then solving for ?? ,
we have
|
?? -1 4?? 4?? 7?? -1
|=0?9?? 2
+8?? -1=0
???????????????????????????????????????????????????????????????????????????? =-1,
1
9
If ?? =-1, then -2?? -4?? =0 or ?? =-2?? .
? From ?? 2
+8???? +7?? 2
=225, we have
-5?? 2
=225 for which no real solution exists.
If ?? =
1
9
, then from (i), ???????? =2??
? From ?? 2
+8???? +7?? 2
=225, we have
?? 2
=5 and ?? 2
=20
??????????????????????????????????????????????????????????? 2
+?? 2
=25
Thus, the required shortest distance is v25=5.
7.3 Let ?? (?? ,?? )={
(?? +?? )
?? ?? ?? +?? ?? ; if (?? ,?? )?(?? ,?? )
?? , if (?? ,?? )=(?? ,?? )
. Show that
?? ?? ?? ?? and
?? ?? ?? ?? exist at (?? ,?? ) though
?? (?? ,?? ) is not continuous at (?? ,?? ) .
(2012 : 15 Marks)
Solution:
?
?? (?? ,?? )?={
(?? +?? )
2
?? 2
+?? 2
, if (?? ,?? )?(0,0)
1, if (?? ,?? )?(0,0;
?? ?? (0,0)?=lim
h?0
?
?? (h,0)-?? (0,0)
h
?=lim
h?0
?
h
2
h
-1
h
=0
?? ?? (0,0)?=lim
?? ?0
?
?? (0,?? )-?? (0,0)
?? ?=lim
?? ?0
?
?? 2
?? 2
-1
?? =0
??? ?? (0,0) and ?? ?? (0,0) exists.
Now,
lim
(?? ,?? )?(0,0)
??? (?? ,?? )?= lim
(?? ,?? )?(0,0)
?
(?? +?? )
2
?? 2
+?? 2
?= lim
(?? ,?? )?(0,0)
?
?? 2
+?? 2
+2????
?? 2
+?? 2
Taking ?? =????
???????????????????????????????????????????????????=lim
?? ?0
?
?? 2
+?? 2
?? 2
+2?? ·????
?? 2
+?? 2
?? 2
???????????????????????????????????????????????????=lim
?? ?0
?
?? +?? 2
+2?? 1+?? 2
, which is different for the different values of ?? .
?lim
(?? ,?? )?(0,0)
??? (?? ,?? ) does not exist.
7.4 Find the minimum distance of the line given by the planes ?? ?? +?? ?? +?? ?? =??
and ?? -?? =?? from the origin, by the method of Lagrange's multipliers.
(2012 : 15 Marks)
Solution:
Let
?? =?? (?? ,?? ,?? )=?? 2
+?? 2
+?? 2
???????????????????????????????????????????(??)
?? (?? ,?? ,?? ) =3?? +4?? +5?? -7=0????????????????????????????????????????????????(ii)
h(?? ,?? ,?? ) =?? -?? -9=0???????????????????????????????????????????????????????????????(iii)
??? =?? ??? +?? ?h
<2?? ,2?? ,2?? > =?? (3,4,5)+?? (1,0,-1)
2?? =3?? +?? 2?? =4?? ??2?? =5?? -?? ????? =
3?? +?? 2
,?? =2?? ,?? =
5?? -?? 2
From (ii),
3[
3?? +?? 2
+2?? +
5?? -?? 2
]=7?????????
????????????????????????????????????????????????????????????????????????????????????????????25?? -?? =7???????????????????????????????????????????????(v)??
???????? (?????? )????????????????????????????????????????????????????????
3?? +?? 2
-
5?? -?? 2
=9
??????????????????????????????????????????????????????????????????????????????????????????????-?? +?? =9?????????????????????????????????????????????????(vi)
Solving (v) and (vi)
?? =
2
3
,?? =
29
3
? ?? -
35
6
,?? -
4
3
,?? -
-19
6
? Minimum distance =v?? 2
+?? 2
+?? 2
=5v
?? 6
???????? (???? )
7.5 Let ?? (?? ,?? )=?? ?? +?? ???? +?? ?? ?? +?? ?? +?? . At what points will ?? (?? ,?? ) be maximum or
minimum?
(2013 : 10 Marks)
Solution:
?? (?? ,?? )=?? 2
+4???? +3?? 2
+?? 3
+1
For stationary points
??? ??? ?=0;
??? ??? =0
????????????????????????????????????????????????4?? +6?? +3?? 2
?=0????????????????????????????????????????????????????????????????????????????????????????????(??)
·2?? +4?? ?=0??? =-2?? ????????????????????????????????????????????????????????????????????(???? )
From (i) and (ii)
3?? 2
-2?? =0??? =0;?? =
2
3
So, (0,0) and (
2
3
,
-4
3
) are stationary points.
To test for maxima or minima we calculate 2 and partial derivative
??????????????????????????????????????????????????
?
2
?? ??? 2
=6+6?? ;?
?
2
?? ??? 2
=2;?
?
2
?? ??? ??? =4
??????????????
?
2
?? ??? 2
×
?
2
?? ??? 2
-(
?
2
?? ??? ??? )
2
=2(6+6?? )-4
2
???????????????????????????????????????????????????????????=12?? -4
????????????????????????????????????????????? -?? 2
=-4<0 at (0,0)
So, no extremum at this point and ???? -?? 2
=4>0 at (
2
3
,
-4
3
) and since
?
2
?? ??? 2
=2>0 so
minima at (
2
3
,
-4
3
) .
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FAQs on Functions of Several (Two or Three) Variables - Mathematics Optional Notes for UPSC
| 1. What are the key concepts to understand in Functions of Several Variables for the UPSC exam? |  |
Ans. Key concepts to understand in Functions of Several Variables for the UPSC exam include partial derivatives, gradient, critical points, optimization, and Lagrange multipliers.
| 2. How are partial derivatives calculated in Functions of Several Variables for the UPSC exam? | |
Ans. Partial derivatives in Functions of Several Variables are calculated by taking the derivative of a function with respect to one variable while holding all other variables constant.
| 3. What is the significance of the gradient in Functions of Several Variables for the UPSC exam? | |
Ans. The gradient in Functions of Several Variables indicates the direction of steepest ascent of a function at a given point and is crucial for optimization problems.
| 4. How are critical points identified and used in Functions of Several Variables for the UPSC exam? | |
Ans. Critical points in Functions of Several Variables are identified by finding where the gradient is zero and are used to determine local extrema of a function.
| 5. How do Lagrange multipliers come into play in Functions of Several Variables for the UPSC exam? | |
Ans. Lagrange multipliers are utilized in Functions of Several Variables to optimize a function subject to one or more constraints by incorporating the constraints into the optimization process.
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