1
Mathematics 1c: Solutions, Homework Set 3
Due: Monday, April 19th by 10am.
1. (10 Points)
Section 3.1, Exercise 16
Let
w
=
f
(
x, y
)
be a function of two
variables, and let
x
=
u
+
v,
y
=
u

v.
Show that
∂
2
w
∂u∂v
=
∂
2
w
∂x
2

∂
2
w
∂y
2
.
Solution.
By the chain rule,
∂w
∂v
=
∂w
∂x
·
∂x
∂v
+
∂w
∂y
·
∂y
∂v
=
w
x

w
y
.
Thus,
∂
2
w
∂u∂v
=
∂
∂u
∂w
∂v
=
∂
∂u
(
w
x

w
y
) =
∂
∂u
w
x

∂
∂u
w
y
=
∂w
x
∂x
·
∂x
∂u
+
∂w
x
∂y
·
∂y
∂u

∂w
y
∂x
·
∂x
∂u
+
∂w
y
∂y
·
∂y
∂u
=
w
xx
+
w
xy

(
w
yx
+
w
yy
) =
w
xx

w
yy
i.e.,
∂
2
w
∂u∂v
=
∂
2
w
∂x
2

∂
2
w
∂y
2
.
2. (10 Points)
Section 3.1, Exercise 22
(a)
Show that the function
g
(
x, t
) = 2 +
e

t
sin
x
satisfies the heat equation:
g
t
=
g
xx
.
[
Here
g
(
x, t
)
represents the temper
ature in a metal rod at position
x
and time
t
.
]
(b)
Sketch the graph of
g
for
t
≥
0
.
(
Hint: Look at sections by the planes
t
= 0
, t
= 1
,
and
t
= 2
.
)
(c)
What happens to
g
(
x, t
)
as
t
→ ∞
? Interpret this limit in terms of the
behavior of heat in the rod.
Solution.
(a)
Since
g
(
x, y
) = 2 +
e

t
sin
x
, then
g
t
=

e

t
sin
x, g
x
=
e

t
cos
x,
and
g
xx
=

e

t
sin
x
. Therefore,
g
t
=
g
xx
.
(b)
The graph of
g
and the times
t
= 0
,
1
,
and 2 is shown in the figure—try
this yourself on the computing site.
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2
–4
–2
0
2
4
0
1
2
3
4
0
1
2
3
4
t
x
z
t
= 0
t
= 1
t
= 2
(c)
Note that
lim
t
→∞
g
(
x, t
) = lim
t
→∞
(2 +
e

t
sin
x
) = 2
This means that the temperature in the rod at position
x
tends to be a
constant (= 2) as the time
t
is large enough.
3. (10 Points)
Section 3.2, Exercise 6
Determine the secondorder Taylor for
mula for the function
f
(
x, y
) =
e
(
x

1)
2
cos
y
expanded about the point
x
0
= 1
, y
0
= 0.
Solution.
The ingredients needed in the secondorder Taylor formula are
computed as follows:
f
x
= 2(
x

1)
e
(
x

1)
2
cos
y
f
y
=

e
(
x

1)
2
sin
y
f
xx
= 2
e
(
x

1)
2
cos
y
+ 4(
x

1)
2
e
(
x

1)
2
cos
y
f
xy
=

2(
x

1)
e
(
x

1)
2
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 Spring '08
 Ramakrishnan
 Chain Rule, Critical Point, The Chain Rule, Optimization

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