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466
CHAPTER 14
CALCULUS OF VECTORVALUED FUNCTIONS
(ET CHAPTER 13)
Let us look, instead, at
(
PF
1
+
2
)
2
, and show that this is equal to
K
2
. Since everything is positive, this will imply
that
1
+
2
=
K
, as desired.
(
1
+
2
)
2
=
2
r
2
+
4
a
2
r
2
+
2
b
2
r
2
+
2
p
r
4
+
2
b
2
r
4
+
b
4
r
4
=
2
r
2
+
4
a
2
r
2
+
2
b
2
r
2
+
2
(
1
+
b
2
)
r
2
=
4
r
2
(
1
+
a
2
+
b
2
)
=
K
2
14.2 Calculus of VectorValued Functions
(ET Section 13.2)
Preliminary Questions
1.
State the three forms of the Product Rule for vectorvalued functions.
SOLUTION
The Product Rule for scalar multiple
f
(
t
)
of a vectorvalued function
r
(
t
)
states that:
d
dt
f
(
t
)
r
(
t
)
=
f
(
t
)
r
0
(
t
)
+
f
0
(
t
)
r
(
t
)
The Product Rule for dot products states that:
d
r
1
(
t
)
·
r
2
(
t
)
=
r
1
(
t
)
·
r
0
2
(
t
)
+
r
0
1
(
t
)
·
r
2
(
t
)
Finally, the Product Rule for cross product is
d
r
1
(
t
)
×
r
2
(
t
)
=
r
1
(
t
)
×
r
0
2
(
t
)
+
r
0
1
(
t
)
×
r
2
(
t
).
In Questions 2–6, indicate whether true or false and if false, provide a correct statement.
2.
The derivative of a vectorvalued function is de±ned as the limit of the difference quotient, just as in the scalarvalued
case.
The statement is true. The derivative of a vectorvalued function
r
(
t
)
is de±ned a limit of the difference
quotient:
r
0
(
t
)
=
lim
t
→
0
r
(
t
+
h
)
−
r
(
t
)
h
in the same way as in the scalarvalued case.
3.
There are two Chain Rules for vectorvalued functions, one for the composite of two vectorvalued functions and
one for the composite of a vectorvalued and scalarvalued function.
This statement is false. A vectorvalued function
r
(
t
)
is a function whose domain is a set of real numbers
and whose range consists of position vectors. Therefore, if
r
1
(
t
)
and
r
2
(
t
)
are vectorvalued functions, the composition
“
(
r
1
·
r
2
)(
t
)
=
r
1
(
r
2
(
t
))
” has no meaning since
r
2
(
t
)
is a vector and not a real number. However, for a scalarvalued
function
f
(
t
)
, the composition
r
(
f
(
t
))
has a meaning, and there is a Chain Rule for differentiability of this vectorvalued
function.
4.
The terms “velocity vector” and “tangent vector” for a path
r
(
t
)
mean one and the same thing.
This statement is true.
5.
The derivative of a vectorvalued function is the slope of the tangent line, just as in the scalar case.
The statement is false. The derivative of a vectorvalued function is again a vectorvalued function, hence
it cannot be the slope of the tangent line (which is a scalar). However, the derivative,
r
0
(
t
0
)
is the direction vector of the
tangent line to the curve traced by
r
(
t
)
,at
r
(
t
0
)
.
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 Winter '08
 GANGliu

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