2
Ans:
∇
f
(
x, y, z
)
=
b
yz xz yx
B
so that
∇
f
(8
,
64
,
2)
=
b
128 16 512
B
df
=
∇
f
(
x, y, z
)
dx
y
p
(
x
)
dx
z
p
(
x
)
dx
p
=
b
128 16 512
B
0
.
1
16
×
0
.
1
0
.
0833
×
0
.
1
=
12
.
8 + 16
×
0
.
16 + 512
×
0
.
025
=
42
.
7
(d) Identify the direction
h
*
that (
x, y, z
) moves in, starting from (8
,
64
,
2), when
x
in-
creases. Write down the directional derivative of
f
in the direction
h
*
, i.e.,
f
h
*
(
·
,
·
,
·
),
and evaluate this derivative at (8
,
64
,
2). Using
f
h
*
(8
,
64
,
2), approximate the change
in
f
when
x
increases by 0.1 units, starting from (8
,
64
,
2).
Ans:
When
x
increases by one, the vector
(
x, y, z
)
increases in the direction
(
dx, y
p
(
x
)
dx, z
p
(
x
)
dx
)
.
When
x
= 8
and
dx
= 1
, therefore,
(
x, y, z
)
increases in the direction
h
*
= (1
,
16
,
0
.
0833)
.
The
unit length
vector pointing in this direction is
h
*
/
||
h
*
||
= (0
.
0624
,
0
.
9980
,
0
.
0052)
.
Therefore, using the diFerential to compute the directional derivative,
f
h
*
(8
,
64
,
2) =
b
128 16 512
B
0
.
0624
0
.
9980
0
.
0052
= 26
.
613
. Now a
dx
of 0.1 induces a shift in
R
3
of
0
.
1
h
*
,
which has length
0
.
1
||
h
*
||
.
df
=
f
h
*
(8
,
64
,
2)
×
0
.
1
||
h
*
||
= 42
.
67
(e) Check to see that all four of these distinct methods give you the same answer!
Ans:
Amazingly, they do!
(2) Recall that a function
f
:
R
n
→
R
m
is nothing more than
m
functions,
f
1
...f
m
, each
mapping
R
n
→
R
, and stacked on top of each other.
(a) Using this fact, write down a formal deFnition of the directional derivative of
f
at
x
0
in the direction
h
∈
R
n
, for a function
f
:
R
n
→
R
m
. Your deFnition should be of the