Higher Engineering Mathematics

360 DIFFERENTIAL CALCULUS

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)<0 and

(

∂^2 z

∂x^2

)

(2, 0)

>0,the point

(2, 0) is a minimum point.

Since (−2, 0)<0 and

(

∂^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 represent the value ofz. Such a map is shown

in Fig. 36.9. To produce a contour map requires a

large number ofx-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

Hence the contourz=0 crosses thex-axis at 0

and±

√

8, i.e. at co-ordinates (0, 0), (2.83, 0) and

(−2.83, 0) shown as points,S,aandbrespectively.

Whenz=0 andx= 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 =0 and

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.

Hence the z=0 contour passes through the

co-ordinates (2, 0.97) and (2,−0.97) shown as a

canddin Fig. 36.9.

Similarly, for thez= 9 contour, wheny=0,

9 =(x^2 + 02 )^2 −8(x^2 − 02 )

i.e. 9 =x^4 − 8 x^2

i.e. x^4 − 8 x^2 − 9 = 0

Hence (x^2 −9)(x^2 +1)=0.

from which,x=±3 or complex roots.

Thus thez=9 contour passes through (3, 0) and

(−3, 0), shown aseandfin Fig. 36.9.

Ifz=9 andx=0, 9 =y^4 + 8 y^2

i.e. y^4 + 8 y^2 − 9 = 0

i.e. (y^2 +9)(y^2 −1)= 0

from which,y=±1 or complex roots.