tgo72 – Homework 9 – Gompf – (58370)
1
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001
10.0 points
Determine whether the partial derivatives
f
x
, f
y
of
f
are positive, negative or zero at the
point
P
on the graph of
f
shown in
P
x
z
y
1.
f
x
= 0
,
f
y
<
0
2.
f
x
>
0
,
f
y
>
0
3.
f
x
= 0
,
f
y
= 0
correct
4.
f
x
<
0
,
f
y
<
0
5.
f
x
<
0
,
f
y
= 0
6.
f
x
>
0
,
f
y
= 0
7.
f
x
<
0
,
f
y
>
0
8.
f
x
= 0
,
f
y
>
0
Explanation:
The value of
f
x
at
P
is the slope of the
tangent line to graph of
f
at
P
in the
x

direction, while
f
y
is the slope of the tangent
line in the
y
direction.
Thus the sign of
f
x
indicates whether
f
is increasing or decreasing
in the
x
direction, or whether the tangent line
in that direction at
P
is horizontal.
Similarly, the value of
f
y
at
P
is the slope
of the tangent line at
P
in the
y
direction,
and so the sign of
f
y
indicates whether
f
is
increasing or decreasing in the
y
direction, or
whether the tangent line in that direction at
P
is horizontal.
From the graph it thus follows that at
P
f
x
= 0
,
f
y
= 0
.
002
10.0 points
Determine
f
x
when
f
(
x, y
) = cos(4
y

x
)

x
sin(4
y

x
)
.
1.
f
x
=

x
sin(4
y

x
)
2.
f
x
=
x
cos(4
y

x
)

sin(4
y

x
)
3.
f
x
=
x
sin(4
y

x
)
4.
f
x
=

x
cos(4
y

x
)
5.
f
x
=

cos(4
y

x
)

x
sin(4
y

x
)
6.
f
x
=

2 sin(4
y

x
)

x
cos(4
y

x
)
7.
f
x
= 2 sin(4
y

x
)

x
cos(4
y

x
)
8.
f
x
=
x
cos(4
y

x
)
correct
Explanation:
From the Product Rule we see that
f
x
= sin(4
y

x
)

sin(4
y

x
)+
x
cos(4
y

x
)
.
Consequently,
f
x
=
x
cos(4
y

x
)
.
003
10.0 points
Find the slope in the
x
direction at the
point
P
(0
,
2
, f
(0
,
2)) on the graph of
f
when
f
(
x, y
) = 2(2
x
+
y
)
e

xy
.
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tgo72 – Homework 9 – Gompf – (58370)
2
1.
slope =

10
2.
slope =

12
3.
slope =

6
4.
slope =

4
correct
5.
slope =

8
Explanation:
The graph of
f
is a surface in 3space
and the slope in the
x
direction at the point
P
(0
,
2
, f
(0
,
2)) on that surface is the value of
the partial derivative
f
x
at (0
,
2). Now
f
x
= 4
e

xy

2(2
xy
+
y
2
)
e

xy
.
Consequently, at
P
(0
,
2
, f
(0
,
2))
slope =

4
.
004
10.0 points
Determine
h
=
h
(
x, y
) so that
∂f
∂x
=
h
(
x, y
)
(
x
2
+ 4
y
2
)
2
when
f
(
x, y
) =
2
x
2
y
x
2
+ 4
y
2
.
1.
h
(
x, y
) = 8
xy
3
2.
h
(
x, y
) = 16
xy
2
3.
h
(
x, y
) = 16
x
3
y
4.
h
(
x, y
) = 16
xy
3
correct
5.
h
(
x, y
) = 8
xy
2
6.
h
(
x, y
) = 8
x
3
y
Explanation:
Differentiating with respect to
x
using the
quotient rule we obtain
∂f
∂x
=
4
xy
(
x
2
+ 4
y
2
)

4
x
3
y
(
x
2
+ 4
y
2
)
2
.
Consequently,
h
(
x, y
) = 16
xy
3
.
005
10.0 points
Find
f
y
when
f
(
x, y
) =
integraldisplay
x
y
cos
(
t
4
)
dt .
1.
f
y
= 4
x
3
sin
(
y
4
)
2.
f
y
=

cos
(
y
4
)
correct
3.
f
y
= cos
(
y
4
)
4.
f
y
=

sin
(
y
4
)
5.
f
y
= sin
(
y
4
)
6.
f
y
= 4
x
3
cos
(
y
4
)
Explanation:
Now
integraldisplay
x
y
cos
(
t
4
)
dt
=

integraldisplay
y
x
cos
(
t
4
)
dt .
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 Spring '08
 RAdin
 Calculus, Derivative, Slope, marg. prod., P fx

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