4.1
Chapter Four
Derivatives
4.1 Derivatives
Suppose
f
is a vector function and
t
0
is a point in the interior of the domain of
f
(
t
0
in the interior of a set
S
of real numbers means there is an interval centered at
t
0
that
is a subset of
S
.).
The derivative is defined just as it is for a plain old everyday real
valued function, except, of course, the derivative is a vector.
Specifically, we say that
f
is
differentiable
at t
0
if there is a vector
v
such that
lim
[
(
)
(
)]
t
t
h
t
h
t
→
+

=
0
1
0
0
f
f
v
.
The vector
v
is called the
derivative of f at t
0
.
Now, how would we find such a thing?
Suppose
f
i
j
k
( )
t
a t
b t
c t
=
+
+
.
Then
1
0
0
0
0
0
0
0
0
h
t
h
t
a t
h
a t
h
b t
h
b t
h
c t
h
c t
h
[
(
)
(
)]
(
)
(
)
(
)
(
)
(
)
(
)
f
f
i
j
k
+

=
+

+
+

+
+

.
It should now be clear that the vector function
f
is differentiable at
t
0
if and only if each
of the coordinate functions
a t
b t
( ), ( ), and
c t
( ) is.
Moreover, the vector derivative
v
is
v
i
j
k
=
+
+
'( )
'( )
'( )
.
Now we “know” what the derivative of a vector function is, and we know how to
compute it, but what is it, really?
Let’s see.
Let
f
i
j
( )
t
t
t
=
+
3
This is, of course, a
vector function which describes the graph of the function
y
x
=
3
. Let’s look at the
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derivative of
f
at
t
0
:
v
i
j
=
+
3
0
2
t
.
Convince yourself that the direction of the vector
v
is
the direction tangent to the graph of
y
x
=
3
at the point (
,
)
t
t
0
0
3
.
It is not so clear what
we should define to be the tangent to a curve other than a plane curve.
Again, vectors
come to our rescue.
If
f
is a vector description of a space curve, the direction of the
derivative
f
'( )
t
vector is the
tangent
direction at the point
f
( )
t
the derivative
f
'( )
t
is
said to be tangent to the curve at
f
( )
t
.
If
f
( )
t
specifies the position of a particle at time
t
, then, of course, the derivative
is the
velocity
of the particle, and its length 
'( )
f
t
is the
speed
Thus the distance the
particle travels from time
t
a
=
to time
t
b
=
is given by the integral of the speed:
d
t dt
a
b
=
∫

'( )
f
.
If the particle behaves nicely, this distance is precisely the length of the arc of the curve
from
f
( )
a
to
f
( )
b
It should be clear what we mean by “behaves nicely”. .
For the
distance traveled by the particle to be the same as the length of its path, there must be no
“backtracking”, or reversing direction.
This means we must not allow the velocity to be
zero for any
t
between
a
and
b
Example
Consider the function
r
i
j
( )
cos
sin
t
t
t
=
+
Then the derivative, or velocity, is
r
i
j
'( )
sin
cos
t
t
t
= 
+
.
This vector is indeed tangent to the curve described by
r
(which
we already know to be a circle of radius 1 centered at the origin.) at
r
( )
t
Note that the
scalar product
r
r
( )
'( )
sin cos
sin cos
t
t
t
t
t
t
⋅
= 
+
=
0, and so the tangent vector and the
vector from the center of the circle to the point on the circle are perpendiculara well
known fact you learned from Mrs. Turner in 4
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 Fall '09
 Peter
 Derivatives

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