Mathematics 250
Fall 2007
Selected Solutions, Assignment 5
MVM 6.2.8
: Show that if
f
(
p
V
T
) is a smooth function of 3 variables, and
p
o
V
o
T
o
is a point where none
of the three partial derivatives vanish, so that it the level surface of
f
through
p
o
V
o
T
o
can be parametrized
by solving for any one of the variables in terms of the other two, we have the relation
(
∂p
∂V
)
T
(
∂V
∂T
)
p
(
∂T
∂p
)
V
=

1
.
Here each of the partial derivatives means the partial of the top variable considered as an implicit function
of the other two variables.
The subscripts outside the parentheses remind us what variable is being held
constant in taking the indicated partial derivative.
Response: The Implicit function Theorem tells us how to compute the partial derivatives of implicit
functions. In this case, if
p
is considered as an implicit function of
V
and
P
, then the ImpFT says that
∂p
∂V
=

∂f
∂V
∂f
∂p
.
Similarly, if we solve for V in terms of
p
and
T
, then
∂V
∂T
=

∂f
∂T
∂f
∂V
.
Now regarding
T
as a function of
p
and
V
, we get the formula
∂t
∂p
=

∂f
∂p
∂f
∂T
.
If we multiply these three expressions together, everything cancels, except the product of three (

1)s gives
us a 1 to remind us that something slightly mysterious is going on.
MVM 6.2.13
: (
The Envelope of a Family of Curves
) Suppose that
f
:
R
2
×
(
a, b
) is
C
2
, and that for
each
t
in (
a, b
), the gradient
∇
f
x
t
6
= 0, (where
∇
f
indicates the gradient with respect to
x
=
x
y
)
on the level curve
C
t
=
x
:
f
x
t
= 0
.
A curve
C
is called the
envelope
of the family of curves
{
C
t
:
t
in (
a, b
)
}
, if it is tangent to each
C
t
at the point
C
∩
C
t
of intersection.
a) Suppose that the matrix
∂f
∂x
x
o
t
o
∂f
∂y
x
o
t
o
∂
2
f
∂x∂t
x
o
t
o
∂
2
f
∂y∂t
x
o
t
o
is nonsingular. Show that for some
δ >
0, there is a
C
1
curve
g
: (
t
o

δ, t
o
+
δ
)
→
R
2
such that
f
g
(
t
)
t
= 0 =
∂f
∂t
g
(
t
)
t
.
1
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Conclude that
g
is a parametrization of the envelope near
x
o
.
b) Find the envelopes for the following families of curves:
i)
f
x
t
= (cos
t
)
x
+ (sin
t
)
y

1 = 0
.
ii)
f
x
t
=
y
+
t
2
x

t
= 0
.
iii)
f
x
t
=
(
x
t
)
2
+
y
1

t
2

1 = 0
.
Response: a) If we consider the mapping
F
:
R
2
×
(
a, b
)
→
R
2
defined by
F
(
x
t
=
f
x
t
∂f
∂t
x
t
,
then the matrix displayed in part a) is the partial Jacobian of
F
with respect to
x
and
y
, so its nonsingularity
guarantees that the zero level curve of
F
near
x
o
t
o
can be parametrized by
t
near
t
o
, that is, in an interval
(
t
o

δ, t
o
+
δ
) for some
δ >
0. If the parametrizing map is
g
, then we want to show that
g
is the envelope;
that is, that it is tangent to
C
t
o
at [
x
o
].
Since
C
t
is the zero level set of
f
t
, the gradient vector
∇
f
x
o
t
o
=
∂f
∂x
x
o
t
o
∂f
∂y
x
o
t
o
is the vector
orthogonal to the curve
C
t
o
at
x
o
. In other words, the tangent vector to
C
t
o
is orthogonal to
∇
f
x
o
t
o
.
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 Fall '06
 RogerHowe
 Derivative

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