490
C H A P T E R
14
CALCULUS OF VECTORVALUED FUNCTIONS
(ET CHAPTER 13)
x
y
1
2
1
4
y
=
1
x
1
4
⟨
1,
−
⟩
2
4
1
1
1
2
(
2,
)
To satisfy (1) we must have:
r
=
λ
1
2
+
−
1
4
2
=
λ
√
17
4
=
5
⇒
λ
=
20
√
17
Substituting in (2) we obtain the following velocity vector at
2
,
1
2
:
r
=
20
√
17
1
,
−
1
4
=
20
√
17
,
−
5
√
17
13.
A bee with velocity vector
r
(
t
)
starts out at the origin at
t
=
0 and flies around for
T
seconds. Where is the bee
located at time
T
if
T
0
r
(
u
)
du
=
0
? What does the quantity
T
0
r
(
u
)
du
represent?
SOLUTION
By the Fundamental Theorem for vectorvalued functions,
T
0
r
(
u
)
du
=
r
(
T
)
−
r
(
0
)
, hence by the
given information
r
(
T
)
=
r
(
0
)
. It follows that at time
T
the bee is located at the starting point which is at the origin. The
integral
T
0
r
(
u
)
du
is the length of the path traveled by the bee in the time interval 0
≤
t
≤
T
. Notice that there is a
difference between the displacement and the actual length traveled.
14.
Which of the following is an arc length parametrization of a circle of radius 4 centered at the origin?
(a) r
1
(
t
)
=
4 sin
t
,
4 cos
t
(b) r
2
(
t
)
=
4 sin 4
t
,
4 cos 4
t
(c) r
3
(
t
)
=
4 sin
t
4
,
4 cos
t
4
SOLUTION
The arc length parametrization is defined by the condition
r
(
t
)
=
1 for all
t
. We thus must check
whether this condition is satisfied.
(a)
The derivative vector is
r
1
(
t
)
=
4 cos
t
,
−
4 sin
t
. We compute the length of this vector:
r
1
(
t
)
=
(
4 cos
t
)
2
+
(
−
4 sin
t
)
2
=
16
(
cos
2
t
+
sin
2
t
)
=
√
16
=
4
=
1
We conclude that this parametrization is not the arc length parametrization of the circle.
(b)
We compute the derivative vector and its length:
r
2
(
t
)
=
16 cos 4
t
,
−
16 sin 4
t
r
2
(
t
)
=
(
16 cos 4
t
)
2
+
(
−
16 sin 4
t
)
2
=
16
2
(
cos
2
4
t
+
sin
2
4
t
)
=
16
2
·
1
=
16
=
1
Hence, this parametrization is not the arc length parametrization of the circle.
(c)
We find the derivative vector and its length:
r
3
(
t
)
=
4
·
1
4
cos
t
4
,
−
4
·
1
4
sin
t
4
=
cos
t
4
,
−
sin
t
4
r
3
(
t
)
=
cos
t
4
2
+
−
sin
t
4
2
=
1
Hence, this parametrization is the arc length parametrization of the circle.
15.
Let
r
(
t
)
=
3
t
+
1
,
4
t
−
5
,
2
t
.
(a)
Calculate
s
(
t
)
=
t
0
r
(
u
)
du
as a function of
t
.
(b)
Find the inverse
ϕ
(
s
)
=
t
(
s
)
and show that
r
1
(
s
)
=
r
(
ϕ
(
s
))
is an arc length parametrization.
SOLUTION
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
S E C T I O N
14.3
Arc Length and Speed
(ET Section 13.3)
491
(a)
We differentiate
r
(
t
)
componentwise and then compute the norm of the derivative vector. This gives:
r
(
t
)
=
3
,
4
,
2
r
(
t
)
=
3
2
+
4
2
+
2
2
=
√
29
We compute
s
(
t
)
:
s
(
t
)
=
t
0
r
(
u
)
du
=
t
0
√
29
du
=
√
29
u
t
0
=
√
29
t
(b)
We find the inverse
ϕ
(
s
)
=
t
(
s
)
by solving
s
=
√
29
t
for
t
. We obtain:
s
=
√
29
t
⇒
t
=
ϕ
(
s
)
=
s
√
29
We obtain the following arc length parametrization:
r
1
(
s
)
=
r
s
√
29
=
3
s
√
29
+
1
,
4
s
√
29
−
5
,
2
s
√
29
To verify that
r
1
(
s
)
is an arc length parametrization we must show that
r
1
(
s
)
=
1. We compute
r
1
(
s
)
:
r
1
(
s
)
=
d
ds
3
s
√
29
+
1
,
4
s
√
29
−
5
,
2
s
√
29
=
3
√
29
,
4
√
29
,
2
√
29
=
1
√
29
3
,
4
,
2
Thus,
r
1
(
s
)
=
1
√
29
3
,
4
,
2
=
1
√
29
3
2
+
4
2
+
2
2
=
1
√
29
·
√
29
=
1
16.
Find an arc length parametrization of the circle in the plane
z
=
9 with radius 4 and center
(
1
,
4
,
9
)
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '08
 GANGliu
 Arc Length, Derivative, Cos

Click to edit the document details