534
CHAPTER 13
Vector-Valued Functions
EXERCISE SET 13.1
1.
(
−∞
,
+
∞
);
r
(
π
) =
−
i
−
3
π
j
2.
[
−
1
/
3
,
+
∞
);
r
(1) =
2
,
1
3.
[2
,
+
∞
);
r
(3) =
−
i
−
ln 3
j
+
k
4.
[
−
1
,
1);
r
(0) =
2
,
0
,
0
5.
r
= 3 cos
t
i
+ (
t
+ sin
t
)
j
6.
r
= (
t
2
+ 1)
i
+
e
−
2
t
j
7.
r
= 2
t
i
+ 2 sin 3
t
j
+ 5 cos 3
t
k
8.
r
=
t
sin
t
i
+ ln
t
j
+ cos
2
t
k
9.
x
= 3
t
2
,
y
=
−
2
10.
x
= sin
2
t
,
y
= 1
−
cos 2
t
11.
x
= 2
t
−
1,
y
=
−
3
√
t
,
z
= sin 3
t
12.
x
=
te
−
t
,
y
= 0,
z
=
−
5
t
2
13.
the line in 2-space through the point (2
,
0) and parallel to the vector
−
3
i
−
4
j
14.
the circle of radius 3 in the
xy
-plane, with center at the origin
15.
the line in 3-space through the point (0
,
−
3
,
1) and parallel to the vector 2
i
+ 3
k
16.
the circle of radius 2 in the plane
x
= 3, with center at (3
,
0
,
0)
17.
an ellipse in the plane
z
=
−
1, center at (0
,
0
,
−
1), major axis of length 6 parallel to
x
-axis, minor
axis of length 4 parallel to
y
-axis
18.
a parabola in the plane
x
=
−
2, vertex at (
−
2
,
0
,
−
1), opening upward
19.
(a)
The line is parallel to the vector
−
2
i
+ 3
j
; the slope is
−
3
/
2.
(b)
y
= 0 in the
xz
-plane so 1
−
2
t
= 0,
t
= 1
/
2 thus
x
= 2 + 1
/
2 = 5
/
2 and
z
= 3(1
/
2) = 3
/
2;
the coordinates are (5
/
2
,
0
,
3
/
2).
20.
(a)
x
= 3 + 2
t
= 0,
t
=
−
3
/
2 so
y
= 5(
−
3
/
2) =
−
15
/
2
(b)
x
=
t
,
y
= 1 + 2
t
,
z
=
−
3
t
so 3(
t
)
−
(1 + 2
t
)
−
(
−
3
t
) = 2,
t
= 3
/
4; the point of intersection is
(3
/
4
,
5
/
2
,
−
9
/
4).
21.
(a)
x
y
(1, 0)
(0, 1)
(b)
x
y
(1, -1)
(1, 1)