HW 5 Solutions
Math 1920 Spring 2009
February 26, 2009
Problem 14.6.6
∇
f
= (2
x

y
)
i
+ (

2
y

x
)
j

k
∇
f

(1
,
1
,

1)
=
i

3
j

k
The tangent plane has normal
∇
f

(1
,
1
,

1)
and contains the point (1,1,1) and hence has equation
1(
x

1)

3(
y

1)

1(
z
+ 1) = 0
x

3
y

z
=

1
The normal line has direction
∇
f

(1
,
1
,

1)
and contains the point (1,1,1) and hence is given by
x
= 1 +
t
y
= 1

3
t
z
=

1

t
Problem 14.6.14
∇
f
=
yz
i
+
xz
j
+
xy
k
∇
f
(1
,
1
,
1) =
i
+
j
+
k
∇
g
= 2
x
i
+ 4
y
j
+ 6
z
k
∇
g
(1
,
1
,
1) = 2
i
+ 4
j
+ 6
k
.
The tangent line to the curve of intersection is orthogonal to both
∇
f
(1
,
1
,
1) and
∇
g
(1
,
1
,
1), so we ﬁnd its
direction by taking the cross product of these two vectors.
∇
f
(1
,
1
,
1)
x
∇
g
(1
,
1
,
1) =
±
±
±
±
±
±
i
j
k
1
1
1
2
4
6
±
±
±
±
±
±
= 2
i

4
j
+ 2
k
Therefore the equation of the tangent line is given by
x
= 1 + 2
t
y
= 1

4
t
z
= 1 + 2
t
Problem 14.6.28
a.
f
x
= 3
x
2
y
4
f
x
(1
,
1) = 3
f
y
= 4
x
3
y
3
f
x
(1
,
1) = 4
f
(1
,
1) = 1
Therefore
L
(
x, y
) = 1 + 3(
x

1) + 4(
y

1) = 3
x
+ 4
y

6
1