1
Mathematics 1c: Solutions, Final Examination
Due: Wednesday, June 9, at 10am
1.
(a) [7 points] Let
f
:
R
3
→
R
2
be deﬁned by
f
(
x,y,z
) =
(
e

2
xy
,x
2

z
2

4
x
+ sin(
x
+
y
+
z
)
)
and let
g
:
R
2
→
R
be a function such that
g
(1
,
0) =

1, and
∇
g
(1
,
0) =
i

3
j
. Calculate the gradient of the composition
g
◦
f
at the point (0
,
0
,
0).
Solution.
Let
f
(
x,y,z
) = (
u
(
x,y,z
)
,v
(
x,y,z
)) and
h
(
x,y,z
) =
g
(
f
(
x,y,z
))
.
By the chain rule,
∂h
∂x
=
∂g
∂u
∂u
∂x
+
∂g
∂v
∂v
∂x
,
∂h
∂y
=
∂g
∂u
∂u
∂y
+
∂g
∂v
∂v
∂y
,
and
∂h
∂z
=
∂g
∂u
∂u
∂z
+
∂g
∂v
∂v
∂z
.
Thus, at (0
,
0
,
0)
,
∂h
∂x
= (1)(0) + (

3)(

4 + 1) = 9
∂h
∂y
= (1)(0) + (

3)(1) =

3
∂h
∂z
= (1)(0) + (

3)(1) =

3
.
Therefore the gradient of
g
◦
f
at (0
,
0
,
0) is 9
i

3
j

3
k
.
(b) [5 points] Find the equation of the tangent plane to the level set
g
◦
f
=

1 at the point (0
,
0
,
0), where
g
and
f
are deﬁned in part (a).
Solution.
The gradient from (a) is orthogonal to the level set
g
◦
f
=

1, so therefore the tangent plane is given by
(
x

0)(9) + (
y

0)(

3) + (
z

0)(

3) = 0
.
This may be rewritten as
3
x

y

z
= 0
.