MATH 23
Sample Second Exam Solutions
was:
April, 2003
NAME
:
Dodson, B.
Section 110

313
(Last,
First)
1. (10
points
)
Sketch the level curves
. . .
.
(directly from homework.)
2. (10
points
)
Let
f
(
x, y
) =
ye
xy
.
Find each of the following.
(directly from homework.)
3. (10
points
)
Find an equation of the tangent plane to the surface
z
=
x
2

3
y
2
at the point (

3
,
2
,

3)
.
For this problem, and the “other” tangent plane problem,
it’s useful to have the correct formula. Here the surface is
a graph,
z
=
f
(
x, y
)
,
for which the normal vector of the tangent
plane is given by
< f
x
, f
y
,

1
> .
We compute
f
x
= 2
x, f
y
=

6
y.
In this case, the point on the surface has the form (
a, b, f
(
a, b
))
,
and the partial derivatives are to be evaluated at (
a, b
) = (

3
,
2)
.
So the normal is
<

6
,

12
,

1
>
and a point on the plane is the
point of tangency, so an equation for the plane is

6(
x
+ 3)

12(
y

2)

(
z
+ 3) = 0
.
In the other case (from the other exam problems, #2), for
the level surface
xyz
2
+
y
3
= 3 at the point (2
,
1
,

1)
,
we have
F
(
x, y, z
) =
c,
and the normal is
< F
x
, F
y
, F
z
>
=
< yz
2
, xz
2
+ 3
y
2
,
2
xyz >
=
<
1
,
2 + 3
,

2
>
=
<
1
,
5
,

2
>,
with partials
evaluated at (2
,
1
,

1)
.
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 Spring '06
 YUKICH
 Math, Calculus, Derivative

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