MVC
Sample Quiz Solutions
SHOW ALL WORK!!
Calculator allowed – BUT you must show the set up.
NAME:
#1. Let f be a scalar field and F a vector field. Indicate which of the following are scalar fields (S),
vector fields (V), or meaningless (M).
a. div f
b. f
c.  F
d. div( f )
e. curl( f )
f.  (div F)
g. curl(curl F)
h. div(div F)
i.  ( f )
j. div(curl( f ))
k. curl(div( f ))
#2. Let r  ( x, y, z ) and let r denote r . Verify that (ln r ) 
MVC
r
.
r2
#3 Determine whether the path x (t )  (et cos t , et (sin t  cos t ), et sin t ) is a flow line for the
vector field F  ( x  z , 2 x, y ) .
#4
Explain why F  x, y, z    2 xy  z  i   x 2  y  j   x  yz  1 k cannot be the gradient of
some scalar field f without mentioning curl. (That is, show that there is no scalar function f
such that F  f ?).
MVC

1 
#5. Calculate the flow line x (t ) of the vector field F ( x, y )   x 2 , 2  that passes through the point
y 

(1,3) at t = 1. (That is, x (1)  (1,3) )
MVC
#6
Suppose F  x, y, z    2 xy  z  i   x 2  y  j   x  1 k . Determine whether or not F is a
conservative vector field (i.e. is there a scalar function f such that F  f ?). If it is, find
the function f for which F  f . If not, explain clearly.


MVC

#7 Find all local extrema and saddle points of f ( x, y)  2 x3  6 xy  y 3  2 . Characterize each point
as a local max, local min, or a saddle point.
MVC
#8 Let f ( x, y )  sin x  sin y . Find the second order Taylor polynomial for f at
a. (0,0)
  
b.  , 
2 2 
MVC
#9 Find the absolute maximum and minimum values of the function f ( x, y)  x  e y  x 2  e y on
the set S  {( x, y ) : 0  x  2, 0  y  1} .
MVC
#10 Let f ( x, y, z )  x 2 y  y 2 z  x  z . Estimate the value of f (1.01, 2.01,3.01) using
differentials. Show work!
1
1


 2

2
df   2 xy 
 dx  ( x  2 yz )dy   y 
 dz
2 xz 
2 xz 


So, letting (dx, dy, dz )  (.01,.01,.01) and ( x, y, z )  (1, 2,3) we get:
1
1


f  df   4   .01  (1  12).01   4   .01
4
4


 .0425  .13  .0425
 .215
Thus, f (1.01, 2.01,3.01)  f (1, 2,3)  .215  16.216
MVC