1
. Suppose that the temperature
z
at the point (
x, y
) is
z
=
f
(
x, y
) = 40

3
x
2
+ 2
y
2
.
a) In what direction should a bug at (1
,
2) to move to get warmer the
most quickly? What is the value of the maximal temperature increase rate at
(1
,
2)? What is the temperature increase rate in the direction of
~v
=
h
3
,
4
i
?
Answer
. In the direction of
∇
f
(1
,
2). Since
∇
f
(
x, y
) =
h
6
x,
4
y
i
,
∇
f
(1
,
2) =
h
6
,
8
i
. The maximal rate is
∇
f
(1
,
2)

= 10
.
The unit vector in the direction of
~v
is
~u
=
D
3
5
,
4
5
E
, and the increase rate
is
D
~u
f
(1
,
2) =
∇
f
(1
,
2)
·
D
3
5
,
4
5
E
=

18
5
+
32
5
=
14
5
.
b) Assume the position of the bug at time
t
is
~
r
(
t
) =
D
sin
π
2
t
,
2
t
E
.
Find
at what rate the temperature increases on the bug’s path at
t
= 1
.
Answer
. The temperature on the bug’s path at
t
is
z
=
f
(
x
(
t
)
, y
(
t
))
.
We have
~
r
(1) =
h
x
(1)
, y
(1)
i
=
h
1
,
2
i
,
d~
r
dt
=
D
π
2
cos
π
2
t
,
2
E
and by chain rule,
dz
dt

t
=1
=
f
x
(
x
(1)
, y
(1))
dx
dt

t
=1
+
f
y
(
x
(1)
, y
(1))
dy
dt

t
=1
=
f
x
(1
,
2)
dx
dt

t
=1
+
f
y
(1
,
2)
dy
dt

t
=1
= (

6)(0) + 8(2) = 16
.
2
. Consider the surface
S
:
z
=
y
+
e
xy
.
a) Write an equation of the tangent plane to
S
at
A
(0
,
1
,
2)
.
Find the
distance between that plane and the point
B
(1
,
2
,
3)
.
Answer
. The vector
~n
orthogonal to
S
at (0
,
1
,
2) is the gradient
∇
(
z

y

e
xy
) =
h
ye
xy
,

1

xe
xy
,
1
i
at (0
,
1
,
2) :
~n
=
h
1
,

1
,
1
i
and the equation of the tangent plane is

x

(
y

1) +
z

2 = 0 or

x

y
+
z

1 = 0
.
The distance is
d
=

A
~
B
·
~n


~n

=
1
q
(

1)
2
+ (

1)
2
+ 1
2
=
1
√
3
.