機械一丙微積分作業 14
1. Find the local extreme values and saddle points of f ( x, y )  x 4  y 4  16 xy  9 .
 y  x3 / 4
Sol: f x ( x, y)  4x 3  16 y  0
f y ( x, y)  4 y 3  16x  0  x   y 3 / 4
(1)
(2)
Substitute (1) into (2), we have
x  ( x 3 / 4) 3 / 4  x 9  256 x  0  x( x  2)( x  2)( x 2  4)( x 4  16 )  0
 x  2
x  0 x  2
, 
, 
.
 
 y  0  y  2  y  2
Therefore, (0, 0), (2,  2), (2, 2) are critical points of f .
Let ( x, y ) 
fx x
fx y
fy x
fy y

12 x 2 16
16
12 y 2
. Then
0 16
 256  0 , this implies (0, 0, 9) is a saddle points of f
16 0
48 16
(2,  2) 
 128  0 and f x x (2,  2)  48  0 .
16
48
This implies f (2,  2)  23 is a local minimum of f .
(0,0) 
48 16
 0 and f x x (2, 2)  48  0 .
16 48
This implies f (2, 2)  23 is a local minimum of f .
(2, 2) 
2. Find the local extreme values and saddles points of f ( x, y )  y 3  3x 2 y  6 x 2  6 y 2  2 .
Sol: f x ( x, y )  6 xy  12 x  0 (1)
f y ( x, y)  3 y 2  3x 2  12 y  0 (2)
From (1), we have 6 x( y  2)  0  x  0 or y  2
Substitute x  0 into (2), we have 3 y 2  12 y  0 , this implies y  0 or y  4 .
Substitute y  2 into (2), we have 3x 2  12 , this implies x  2 or x  2 .
Therefore, (0, 0), (0, 4), (2, 2), (2, 2) are critical points of f .
 ( x, y ) 
(0,0) 
fx x
fx y
fy x
fy y

6 y  12 6 x
6x
6 y  12
 12 0
 144  0 and f x x (0, 0)  12  0 . This implies
0
 12
f (0, 0)  2 is a local maximum of f .
12 0
 144  0 and f x x (0, 4)  12  0 . This implies
0 12
f (0, 4)  30 is a local minimum of f .
0 12
0  12
(2, 2) 
 144  0 ; (2, 2) 
 144  0 .
12 0
 12 0
These implies (2, 2,  14) and (2, 2,  14) are two saddle points of f .
(0,4) 
3. Find the points on the surface y 2  9  xz that are closest to the origin.
Sol: Let ( x, y, z ) be the point on the surface y 2  9  xz . Then the distance is
D  x 2  y 2  z 2  x 2  9  xz  z 2 .
Let f ( x, z )  x 2  z 2  xz  9 . Then f x ( x, z )  2 x  z , f z ( x, z )  2 z  x .
 f x ( x, z )  0
x  0
when 

z  0
 f z ( x, z )  0
Substitute
 ( x, z ) 
x  0 , z  0 into the surface y 2  9  xz , we have y  3 .
f xx
f xz
f zx
f zz

2 1
 3  0 and f xx (0,0)  2  0
1 2
These implies (0, 0,  3) is the point on the surface y 2  9  xz and the least distance is of
f (0,0,3)  3 .
4. Find the least distance from the point P(3, 0, 0) to the surface z 2  17  xy .
Sol: Let Q( x, y, z ) be the point on the surface z 2  17  xy . Then the distance PQ is
D  PQ  ( x  3) 2  y 2  z 2  ( x  3) 2  y 2  17  xy  x 2  y 2  xy  6 x  26 .
Let f ( x, y )  x 2  y 2  xy  6 x  26 . Then
 f x ( x, y )  2 x  y  6  0

 f y ( x, y )  2 y  x  0
2 x  y  6
when 
x  2 y
Substitute (2) into (1), we have
4y  y  6  y  2  x  4
(4, 2) is the only critical point of f .
(1)
(2)
Let ( x, y ) 
fx x
fx y
fy x
fy y

2 1
.
1 2
Then
(4, 2)  3  0 and f x x (4, 2)  1  0 implies f (4, 2)  14 is a local maximum value.
x  4, y  2  z  3 .
Thus, (4, 2, 3) and (4, 2,  3) are the points on the surface z 2  17  xy that are closed
to point P(3, 0, 0) and the least distance is
14
5. Calculate (1)  (6 x 2 y 3  5 y 4 ) dA where R  {(x, y) | 0  x  3, 0  y  1} ;
R
(2)
 x
D
Sol: (1)
 (6x
R
2
2y
dA where D  {(x, y) | 0  x  1, 0  y  x}
1
2
y 3  5 y 4 ) dA  
3 1
0
 (6x
0
2
y 3  5 y 4 ) dy dx
y4
y5 1
  [6 x
 5 ] | 0 dx
0
4
5
3
2
3 3
  [ x 2  1]dx
0 2
x3
21
 [  x] |30 
2
2
(2)
1
x
2y
2y
dA

D x 2  1
0 0 x 2  1 d y d x
1
1
1
y 2 |0 x d x   2
xd x
0 x 1
0 x 1


1
2
1
ln 2
ln | x 2  1 ||10 
2
2
6.
Evaluate (1)

1
3
2
e x dx dy ; (2)
0 3y
3 9

0 x2
x cos(y 2 ) dy dx .
Sol: (1) Rh  {( x, y ) | 3 y  x  3, 0  y  1}
Rv  {( x, y )} | 0  y 
1 3

e x dx dy  
2
0 3y
3 x/3

0
0
3
x
, 0  x  3}
3
2
e x dy dx
  ye x |0x / 3 dy
2
0

3
0
x x2
e dy
3
1 3 x 2 dx 2
  xe
3 0
2x
1 2
 e x |30
6
1
 (e9  1)
6
(2) Rv  {(x, y) | x 2  y  9, 0  x  3}
Rh  {(x, y)} | 0  x  y , 0  y  9}
3 9

0 x
2
x cos(y 2 ) dy dx  
9
0

9
0

y
0
x cos(y 2 ) dx dy
1 2 y
x |0 cos y 2 dy
2

1 9
y cos y 2 dy
2 0

2
1 9
2 d y
y
cos
y
2 0
2y

1
sin y 2 |90
4
1
 sin 9
4