3.
When the trace cursor is moving upward the particle is
moving to the right, and when the cursor is moving
downward the particle is moving to the left. Again we find
the same values of
t
for when the particle reverses
direction.
4.
We can represent the velocity by graphing the parametric
equations
x
4
(
t
)
5
x
1
9
(
t
)
5
12
t
2
2
32
t
1
15,
y
4
(
t
)
5
2 (part 1),
x
5
(
t
)
5
x
1
9
(
t
)
5
12
t
2
2
32
t
1
15,
y
5
(
t
)
5
t
(part 2),
x
6
(
t
)
5
t
,
y
6
(
t
)
5
x
1
9
(
t
)
5
12
t
2
2
32
t
1
15 (part 3)
[
2
8, 20] by [
2
3, 5]
(
x
4
,
y
4
)
[
2
8, 20] by [
2
3, 5]
(
x
5
,
y
5
)
[
2
2, 5] by [
2
10, 20]
(
x
6
,
y
6
)
For (
x
4
,
y
4
) and (
x
5
,
y
5
), the particle is moving to the right
when the
x
-coordinate of the graph (velocity) is positive,
moving to the left when the
x
-coordinate of the graph
(velocity) is negative, and is stopped when the
x
-coordinate
of the graph (velocity) is 0. For (
x
6
,
y
6
), the particle is
moving to the right when the
y
-coordinate of the graph
(velocity) is positive, moving to the left when the
y
-coordinate of the graph (velocity) is negative, and is
stopped when the
y
-coordinate of the graph (velocity) is 0.
Exploration 3
Seeing Motion on a Graphing
Calculator
1.
Let
t
Min
5
0 and
t
Max
5
10.
2.
Since the rock achieves a maximum height of 400 feet, set
y
Max to be slightly greater than 400, for example
y
Max
5
420.
4.
The grapher proceeds with constant increments of
t
(time),
so pixels appear on the screen at regular time intervals.
When the rock is moving more slowly, the pixels appear
closer together. When the rock is moving faster, the pixels
appear farther apart. We observe faster motion when the
pixels are farther apart.
Quick Review 3.4
1.
The coefficient of
x
2
is negative, so the parabola opens
downward.
Graphical support:
[
2
1, 9] by [
2
300, 200]
2.
The
y
-intercept is
f
(0)
52
256.
See the solution to Exercise 1 for graphical support.
3.
The
x
-intercepts occur when
f
(
x
)
5
0.
2
16
x
2
1
160
x
2
256
5
0
2
16(
x
2
2
10
x
1
16)
5
0
2
16(
x
2
2)(
x
2
8)
5
0
x
5
2 or
x
5
8
The
x
-intercepts are 2 and 8. See the solution to Exercise 1
for graphical support.
4.
Since
f
(
x
)
16(
x
2
2
10
x
1
16)
16(
x
2
2
10
x
1
25
2
9)
16(
x
2
5)
2
1
144,
the range is (
2‘
, 144].
See the solution to Exercise 1 for graphical support.
5.
Since
f
(
x
)
16(
x
2
2
10
x
1
16)
16(
x
2
2
10
x
1
25
2
9)
16(
x
2
5)
2
1
144,
the vertex is at (5, 144). See the solution to Exercise 1 for
graphical support.
6.
f
(
x
)
5
80
2
16
x
2
1
160
x
2
256
5
80
2
16
x
2
1
160
x
2
336
5
0
2
16(
x
2
2
10
x
1
21)
5
0
2
16(
x
2
3)(
x
2
7)
5
0
x
5
3 or
x
5
7
f
(
x
)
5
80 at
x
5
3 and at
x
5
7.
See the solution to Exercise 1 for graphical support.
7.
}
d
d
y
x
}
5
100
2
32
x
1
160
5
100
60
5
32
x
x
5 }
1
8
5
}
}
d
d
y
x
} 5
100 at
x
5 }
1
8
5
}
Graphical support: the graph of NDER
f
(
x
) is shown.
[
2
1, 9] by [
2
200, 200]
8.
}
d
d
y
x
}
.
0
2
32
x
1
160
.
0
2
32
x
. 2
160
x
,
5
}
d
d
y
x
} .
0 when
x
,
5.
See the solution to Exercise 7 for graphical support.
Section 3.4
81