Multivariable Calculus
Self Assessment B - answers
Name:_____________________________________
1. If z=f(x, y) and xy + yz + z2= 0, find
 2Z
at (-2,1,1)
XY
Solution:
F
z
xz
y
1.


 w  f ( x, y , z )
F
y
y  2z
z
2
 z
  z  w f f y f z
2.
   


 

xy x  y  x x y x z x

1
( y  2 z )(1)  ( x  z )( 2) 
y 

0
.

 y  2 z 
y  2z
( y  2Z ) 2


1
y  2x  y 
1
y 2  2 xy



.

y  2 z ( y  2Z ) 2  y  2 z 
y  2 z ( y  2 z)3

 y 2  4 yz  4 z 2  y 2  2 xy
( y  2 z)3
444
4



27
(1  2) 3

 4 yz  2 xy  4 z 2
( y  2 z )3
another way:
 z
  xz 

  
xy x  y  2 z 
2
 y  2 z 1  z   x  z  2 z 

x 
 y  2 z 2
 x 


y 
2y 
   x  z  

y  2z 
y  2 z 
 y  2 z  y  2 z  y   x  z 2 y 





 y  2 z 2
 y  2 z 3
 y  2 z 2 z   x  z 2 y    32   12   4

27
 y  2 z 3
1  23
 y  2 z 1 
answer: _______ 
Self-Assessment page 1 Form B
4
_______________
27
2. Use the Jacobian to find
dy
at (1, -1, 1) if
dx
-2xy + 3z + x2= 6
and 2x - y2 + yz = 2
Solution:
 2 y  2x 3
( F , G)
2
y
dy
 ( x, z )



( F , G)
 2x
3
dx
 ( y, z )
 2y  z y
4
3
2 1
 10
10


2 3
7
7
3
answer:

10
7
1
3. Find the directional derivative of w= 2x2yz in the direction of v=3i-4j at he point
(-1, 1, 3, 6)
Solution :
The gradient is given by : w  4 xyzi  2 x 2 zj  2 x 2 yk  12i  6 j  2k
  3 4
The unit vector in the direction of v is u  i  j  0k
5 5
60
  36 24
Dv w  w  u 

 0    12
answer :  12
5
5
5
4. Use the Lagrangean coefficients technique to find the minimum value of
subject to the condition
w  y  z  2x 2
x  y2  z 2.
We want to find the minimum value of
w  y  z  2x 2
for point located in the paraboloid
x  y2  z 2.
Solution: using Lagrangean coefficients.
F ( x, y , z ,  )  y  z  2 x 2   ( x  y 2  z 2 )

The lagrangean equation is given by
 F
 x  4 x    0  x   4
 F
1

 1  2 y  0  y 
 y
2
 F
1

 1  2 z  0  z 
2
 z

1
1
 F
2
2
3
3
   x  y  z  0   4  2  2  0    2     2
4
4

Therefore,
2
 3 2 
1 3 4  3 4 3 4
33 4


w  y  z  2 x  3  3  2 
3 



4 
2
8
8
2 2 2 2
2 8

2
answer: ___ 
1
1
33 4
_______________
8
Self-Assessment page 2 Form B
5. Find all critical points of f(x, y) = y3 + xy2 -2xy . Determine whether each critical point yields a relative
maximum value, a relative minimum, or a saddle point.
Step 1.
 f
2
 x  y  2 y  0
 y ( y  2)  0
 y  0 and x  0 or
 2

 f
 y  2 and 2 x  12  0  x  6
  3 y 2  2 xy  2 x  0 3 y  2 xy  2 x  0
 y
Critical Points: The critical points are (0, 0, 0) and (-6, 2, 8)
Step 2.

 2 f

 2 0

 x

2 f
at x  0 and y  0  2  0    4, saddle po int

 2 f

 y
 2 0

 2 f
 x
 2


2 f
yx


 2  6 y  2x  
 2 f
 y

 2 0
 2 f

 x
 2y  2


2
 yx
 at x  6 and y  2   f  2    4, saddle po int
 2

 y

 2 f

2



y

x


Answer: both (0, 0, 0) and (-6, 2, 8) are saddle points
6) An ice cone is melting in such a way that the radius is decreasing at the rate of 2 inches per minute and
the height is decreasing at the rate of 3 inches per minute. At what rate is the volume changing when r=3
feet and h=5 feet.
1
V V dr V dh V 2
1
V   r 2h 

 


  rh 2   r 2 3
3
t
r dt h dt
t 3
3
V 2
1
.

  (36)(60) 2   (36) 2  3  2880  1296
t 3
3
V
 4176 sq  inches per min  29 sq  ft per min  91.106 ft  min
t
answer: ______  91.106
Self-Assessment page 3 Form B
ft  min _______
7. If z = x2 +y2
, x=r+2s and y = 3r +s, find
 2z
 r s
Solution:
z z x z y
     2x  2  2 y 1  4x  2 y  w
s x s y s
2z

w x w y
 ( w) 
 
  41  2(3)  10
rs r
x r y r
answer: __ 10________
8. If f(x, y) = x2 -xy + y2
find
 2z
 x y
Solution:
2
 z
  z  
     x  2 y   1
 x y x  y  x
answer:____-1 ______
9. If x=f(y, z), 4 = x3 -y3 -z3 , find
x
y
y2
answer: : ___
_____
x2
Solution:
F
x
 3y2 y2
y


 2
F
y
3x 2
x
x
3
10. Use differentials to approximate 36.3 7.6
Solution:
Let f ( x, y)  x 3 y , x  36, x  0.3,
  
f (36, 8)   36 3 8   12
y  8, y  0.4
3 y
f
f
x
x  y 
x 
y 
3 2
x
y
2 x
3 x
2
6
6
24
 18  3
 0.3   0.4  



 0.15
12
12
120 120
120 20
Therefore, 36.3 3 7.6  12  0.15  11.85
answer : 11.85
f  df 
Self-Assessment page 4 Form B
11. Find the maximum value of w= x2 +3y2 +2z2 subject to the condition 2x-3y+5z=1.
Solution: using Lagrangean coefficients.
2
2
2
The lagrangean equation is given by F ( x, y , z ,  )  x  3 y  2 z   ( 2 x  3 y  5 z  1)










F
x
F
y
F
z
F

4
 39
 4   1     39


4
 6 y  3  0  y 

x

2
39

2
5

y
 4 z  5  0  z   
39

4
5
3
25

z
 2 x  3 y  5 z  1  0  2      1  0 
39

2
4
16
12
50
78
2
Therefore, w  x 2  3 y 2  2 z 2  2  2  2  2 
answer
39
39
39
39
39
 2 x  2  0  x  
12. Find the point on the plane 2x-y +3z -5 =0 that is closest to the origin.
Solution:
We want to minimize
d  x 2  y 2  z 2 or D  d 2  x 2  y 2  z 2
subject to the constraint
that 2x - y + 3z -5 =0.
The lagrangean equation is given by
F ( x, y, z ,  )  x 2  y 2  z 2   (2 x  y  3z  5)
F
5

 2 x  2  0  x  
14  10    

x
7

F

5
 2y    0  y 

x

y
2
7

5
F
3

y
 2 z  3  0  z   
14

z
2
15
F
1
9

z
 2 x  y  3z  5  0  2      5  0 

14

2
2
5 15 
5
Therefore, the closest po int is  ,  , 
 7 14 14 










Self-Assessment page 5 Form B
13. The gradient is normal to the level surface F(x, y, z) =0. It is because of this normality of the gradient
that the maximum directional derivative ( which is ) at a point is often called the normal derivative at
Normal Derivative at that point. We use the notation
df
dn
for the normal derivative of a function f.
Find the normal derivative of z = x3 y at the point (-1, 2, 2) and the corresponding unit vector.
Solution:
z  3x 2 yi  x 3 j  6i  j 
df
 z  6 2  (1) 2  37
dn
6
1

u
i
j
37
37
14. Show that
lim
x0
y 0
lim
x0
y0
x2y
x 4  y2
0
along any line y=mx, for any m. What is
x2y
x4 
along y=x2 . Does lim
2
x0
y
y 0
x2y
x 4  y2
exist?. Explain.
Solution:
y  mx, m  0  lim
x 0
y 0
x2 y
x4  y2
m  0  y  0 and lim
y  x  lim
x2 y
x 0
y 0
x2 y
x 0
y 0
2
 lim
x4  y2
 lim
mx3
x4  m2 x2
 lim
x 0
y 0
x4
0
x4

 lim
x 0
y 0
0
1
2
x 0 x 4  y 2
x 0 x 4  x 4
y 0
y 0
Therefore the limit does not exist because it is not unique
Self-Assessment page 6 Form B
mx
x2  m2

0
m2
0