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680
CHAPTER 15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
(a)
Find
d
dt
f
(
c
(
t
))
without making any calculations. Explain.
(b)
Verify your answer to (a) using the Chain Rule.
SOLUTION
(a)
The level curves of
f
(
x
,
y
)
are the circles
x
2
+
y
2
=
c
2
.S
ince
c
(
t
)
is a parametrization of the unit circle,
f
has
constant value 1 on
c
.Thatis,
f
(
c
(
t
))
=
1, which implies that
d
dt
f
(
c
(
t
))
=
0.
(b)
We now ±nd
d
dt
f
(
c
(
t
))
using the Chain Rule:
d
dt
f
(
c
(
t
))
=
∂
f
∂
x
dx
dt
+
∂
f
∂
y
dy
dt
(1)
We compute the derivatives involved in (1):
∂
f
∂
x
=
∂
∂
x
³
x
2
+
y
2
´
=
2
x
,
∂
f
∂
y
=
∂
∂
y
³
x
2
+
y
2
´
=
2
y
dx
dt
=
d
dt
(
cos
t
)
=−
sin
t
,
dy
dt
=
d
dt
(
sin
t
)
=
cos
t
Substituting the derivatives in (1) gives
d
dt
f
(
c
(
t
))
=
2
x
(
−
sin
t
)
+
2
y
cos
t
Finally, we substitute
x
=
cos
t
and
y
=
sin
t
to obtain
d
dt
f
(
c
(
t
))
=−
2cos
t
sin
t
+
2sin
t
cos
t
=
0
.
4.

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