Graphics Reference
In-Depth Information
where F(x,y) = f(u(x,y),v(x,y)). Sloppiness may be acceptable, but only as long as one
knows how to express things correctly when needed.
4.3.16. Example.
Given z = x
2
+ 3xy, x = sin u, and y = u
2
, to find dz/du. Note that
this is really just a restatement of the problem in Example 4.3.8.
Solution.
We have
∂
∂
z
u
∂
∂
z
x
∂
∂
x
u
∂
∂
z
y
dy
du
2
(
)
+
()
=
=
+
=
23
xy u xu
+
cos
32 2
sin
u uu
cos
+
3
cos
u
+
6
sin
u
.
This answer clearly agrees with the earlier one.
4.3.17. Example.
Let u = f(x - ct) + g(x + ct). To show that
2
2
∂
∂
u
∂
∂
u
2
c
=
.
2
2
x
t
Solution.
Define r(x,t) = x - ct and s(x,t) = x + ct. Then u(x,t) = f(r(x,t)) + g(s(x,t)).
The result follows from the following computations:
∂
∂
u
x
∂
∂
r
x
∂
∂
s
x
()
=¢
()
(
)
(
)
+¢
()
(
)
(
)
=¢
()
(
)
+¢
()
(
)
xt
,
f rxt
,
xt
,
g rxt
,
xt
,
f rxt
,
g rxt
,
∂
∂
u
t
∂
∂
r
t
∂
∂
s
t
()
=¢
()
(
)
(
)
+¢
()
(
)
()
=-
¢
()
(
)
+¢
(
()
)
xt
,
f rxt
,
xt
,
g rxt
,
x
,
t
cf
r x t
,
cg r x t
,
and
2
∂
∂
u
∂
∂
r
x
∂
∂
s
x
()
= ¢
()
(
)
(
)
+ ¢
()
(
)
(
)
= ¢
()
(
)
+ ¢
()
(
)
xt
,
f rxt
,
xt
,
g rxt
,
xt
,
f rxt
,
g rxt
,
2
x
2
∂
∂
u
t
∂
∂
r
t
∂
∂
s
t
()
=-
¢¢
()
(
)
(
)
+¢¢
(
()
)
(
)
=¢¢
2
(
()
)
+¢¢
2
(
()
)
x t
,
cf
r x t
,
x t
,
cg
r
xxt
,
xt
,
cf rxt
,
cg rxt
,
2
Definition.
Let f :
R
n
Æ
R
be a function that is differentiable at a point
p
. Then the
gradient of f at
p
, denoted by —f (
p
), is defined to be the Jacobian matrix f¢(
p
), that is,
—
()
=
(
()
()
)
f
p
f
p
,...,
Df
p
.
1
n
A generalization of partial derivatives is the directional derivative.
Definition.
Let
U
be an open subset of
R
n
and let f :
U
Æ
R
. Let
v
be an arbitrary
nonzero vector. If
p
Œ
U
and if the limit
(
)
-
()
f
pv p
+
hf
h
lim
h
Æ
0
