362 Higher Engineering Mathematics
Following the procedure:
(i)
∂z
∂x
= 2 (x^2 +y^2 ) 2 x− 16 x and
∂z
∂y
= 2 (x^2 +y^2 ) 2 y+ 16 y
(ii) for stationary points,
2 (x^2 +y^2 ) 2 x− 16 x= 0
i.e. 4x^3 + 4 xy^2 − 16 x=0(1)
and 2(x^2 +y^2 ) 2 y+ 16 y= 0
i.e. 4 y(x^2 +y^2 + 4 )=0(2)
(iii) From equation (1),y^2 =
16 x− 4 x^3
4 x
= 4 −x^2
Substitutingy^2 = 4 −x^2 in equation (2) gives
4 y(x^2 + 4 −x^2 + 4 )= 0
i.e. 32y=0andy= 0
Wheny=0 in equation (1), 4x^3 − 16 x= 0
i.e. 4 x(x^2 − 4 )= 0
from which,x=0orx=± 2
The co-ordinates of the stationary points are
(0, 0), (2, 0) and (−2, 0).
(iv)
∂^2 z
∂x^2
= 12 x^2 + 4 y^2 − 16 ,
∂^2 z
∂y^2
= 4 x^2 + 12 y^2 +16 and
∂^2 z
∂x∂y
= 8 xy
(v) For the point (0, 0),
∂^2 z
∂x^2
=− 16 ,
∂^2 z
∂y^2
=16 and
∂^2 z
∂x∂y
= 0
For the point (2, 0),
∂^2 z
∂x^2
= 32 ,
∂^2 z
∂y^2
=32 and
∂^2 z
∂x∂y
= 0
For the point (−2, 0),
∂^2 z
∂x^2
= 32 ,
∂^2 z
∂y^2
=32 and
∂^2 z
∂x∂y
= 0
(vi)
(
∂^2 z
∂x∂y
) 2
=0 for each stationary point
(vii) ( 0 , 0 ) =( 0 )^2 −(− 16 )( 16 )= 256
( 2 , 0 ) =( 0 )^2 −( 32 )( 32 )=− 1024
(− 2 , 0 )=( 0 )^2 −( 32 )( 32 )=− 1024
(viii) Since( 0 , 0 )>0,the point (0, 0) is a saddle
point.
Since( 0 , 0 )<0and
(
∂^2 z
∂x^2
)
( 2 , 0 )
>0,the point
(2, 0) is a minimum point.
Since (− 2 , 0 )<0and
(
∂^2 z
∂x^2
)
(− 2 , 0 )
>0, the
point (−2, 0) is a minimum point.
Looking down towards thex-yplane from above, an
approximate contour map can be constructed to repre-
sent the value ofz. Such a map is shown in Fig. 36.9.
To produce a contour map requires a large number of
x-yco-ordinates to be chosen and the values ofzat
each co-ordinate calculated. Here are a few examples of
points used to construct the contour map.
Whenz= 0 , 0 =(x^2 +y^2 )^2 − 8 (x^2 −y)^2
In addition, when, say,y=0 (i.e. on thex-axis)
0 =x^4 − 8 x^2 ,i.e.x^2 (x^2 − 8 )= 0
from which, x=0orx=±
√
8
Hencethecontourz=0 crossesthex-axisat 0and±
√
8,
i.e. at co-ordinates (0, 0), (2.83, 0) and (−2.83, 0) shown
as points,S,aandbrespectively.
Whenz=0andx= 2 then
0 =( 4 +y^2 )^2 − 8 ( 4 −y^2 )
i.e. 0= 16 + 8 y^2 +y^4 − 32 + 8 y^2
i.e. 0=y^4 + 16 y^2 − 16
Let y^2 =p,thenp^2 + 16 p− 16 =0and
p=
− 16 ±
√
162 − 4 ( 1 )(− 16 )
2
=
− 16 ± 17. 89
2
= 0 .945 or− 16. 945
Hence y=
√
p=
√
( 0. 945 )or
√
(− 16. 945 )
=± 0 .97 or complex roots.