MATH 241, 2nd Examination
Solutions and Grading Key
Prof. Jonathan Rosenberg
Friday, October 28, 2005
1. (25 points, divided as indicated)
(a) (15 points) Find the tangent plane to the surface
x
sin
y
cos
z
+
e
xy
+
z
2
= 1
at the point
(1
,
0
,
0)
.
Solution.
Let
f
(
x, y, z
) =
x
sin
y
cos
z
+
e
xy
+
z
2
; then
∇
f
= (sin
y
cos
z
+
ye
xy
, x
cos
y
cos
z
+
xe
xy
,
−
x
sin
y
sin
z
+ 2
z
)
.
Thus
∇
f
(1
,
0
,
0) = (0
,
2
,
0)
, and this must be normal to the level surface of
f
through the point
(1
,
0
,
0)
. So the
equation of the tangent plane is
(0
,
2
,
0)
·
(
x
−
1
, y, z
) = 0
or
2
y
= 0
or
y
= 0
.
(b) (10 points) Is it possible to locally solve for
z
as a smooth function of
x
and
y
(in a neighborhood of
(1
,
0)
), and
if so, what are
∂z/∂x
and
∂z/∂y
when
x
= 1
and
y
= 0
?
Solution.
No, it is not possible to locally solve for
z
as a smooth function of
x
and
y
(in a neighborhood of
(1
,
0)
), since the tangent plane
y
= 0
is perpendicular to the
x

y
plane. (Or to phrase it another way, there are
no points in the tangent plane over points with
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 Spring '08
 Wolfe
 Math, Calculus, Critical Point, 2J, Prof. Jonathan Rosenberg

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