1
. (a) Find the equation of the tangent plane to the surface
S
:
yz
=
ln(
x
+
z
) at
A
(0
,
0
,
1)
.
What is the distance from the tangent plane to the
point
B
(3
,
1
,
2)?
Answer.
S
:
f
(
x,y,z
) = ln(
x
+
z
)

yz
= 0. We ﬁnd
∇
f
(
x,y,z
) =
±
1
x
+
z
,

z,
1
x
+
z

y
²
.
The normal vector at (0
,
0
,
1) is
~n
=
h
1
,

1
,
1
i
.
The equation of the tangent plane is
x

y
+
z

1 = 0
.
The distance
D
=
³
³
³
→
AB
·
~n

~n

³
³
³
=
h
3
,
1
,
1
i · h
1
,

1
,
1
i
1
√
3
=
√
3
.
(b) Write a parametric equation for the line of intersection of the tangent
plane in (a) and the plane
z
=
x
+ 2
y

1
.
Write the equation of the line in
symmetric form. What is the distance from the line to the point
B
(3
,
1
,
2)?
Answer. Normal vector to the second plane is
~n
2
=
∇
(
z

x

2
y
) =
h
1
,

2
,
1
i
. We choose a point in the intersection of two planes
z
=
x
+2
y

1 =
y

x
+ 1
,
2
x
+
y
= 2: for example,
C
(1
,
0
,
0)
.
The line passes through
it in the direction of
~
τ
=
~n
×
~n
2
=
h
1
,

2
,

3
i
. The equation in parametric
form:
x
= 1 +
t,y
=

2
t,z
=

3
t
. Symmetric form (eliminating
t
):
x

1 =

y/
2 =

z/
3
.
Since
~a
=
C
~
B
=
h
2
,
1
,
2
i
,

~
τ

=
√
14
,~a
·
~
τ
=

6, the distance to
B
(1
,
2
,
3) is
D
=
v
u
u
t

~a

2

~a
·
~
τ

~
τ

!
2
=
s
9

36
14
=
s
45
7
.
2.
(a) Determine whether
f
(
x,y,z
) =
√
x
2
+ 3
yz
is increasing or de
creasing at (

1
,
1
,
1) in the direction of
~a
=
h
1
,

1
,
1
i
.
At what rate?
Answer. We ﬁnd the gradient
∇
f
(
x,y,z
) =
*
x
√
x
2
+ 3
yz
,
3
z
2
√
x
2
+ 3
yz
,
3
y
2
√
x
2
+ 3
yz
+
and
∇
f
(

1
,
1
,
1) =
D

1
2
,
3
2
,
3
2
E
. The rate of change in the direction of
~a
is
D
~u
f
(

1
,
1
,
1) =
∇
f
(

1
,
1
,
1)
·
~a

~a

=
D

1
2
,
3
2
,
3
2
E
· h
1
,

1
,
1
i
1
√
3
=

1
2
√
3
<
0.
So,
f
is decreasing at the rate

1
2
√
3
.