suleimenov (bs26835) – HW 09 – rusin – (55565)
1
This printout should have 22 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
I Forgot to say in class:
we are SKIP
PING all the material on curvature, torsion oF
curves, etc. You’re o± the hook! Meanwhile,
the new material we are starting (Functions
oF two or more variables) will give you lots oF
stu± to calculate and a lot less stu± to draw.
Read up on partial derivatives and ask the
TAs For help on Monday as you dive into the
second halF oF the assignment below. We’ll
discuss this on Tuesday too, oF course. Good
luck!
001
10.0 points
Determine the position vector,
r
(
t
), oF a
particle having acceleration
a
(
t
) =
−
8
k
when its initial velocity and position are given
by
v
(0) =
i
+
j
−
4
k
,
r
(0) = 2
i
+ 3
j
respectively.
1. r
(
t
) = (
t
+ 2)
i
−
(
t
−
3)
j
−
(4
t
2
−
4
t
)
k
2. r
(
t
) = (
t
+ 2)
i
+ (
t
+ 3)
j
−
(4
t
2
+ 4
t
)
k
correct
3. r
(
t
) = (
t
+ 2)
i
+ (
t
+ 3)
j
−
(4
t
2
−
4
t
)
k
4. r
(
t
) = (
t
+ 2)
i
−
(
t
−
3)
j
−
(4
t
2
+ 4
t
)
k
5. r
(
t
) = (
t
−
2)
i
+ (
t
+ 3)
j
−
(4
t
2
−
4
t
)
k
6. r
(
t
) = (
t
−
2)
i
+ (
t
+ 3)
j
−
(4
t
2
+ 4
t
)
k
Explanation:
Since
a
(
t
) =
d
v
dt
=
−
8
k
,
we see that
v
(
t
) =
−
8
t
k
+
C
where
C
is a constant vector such that
v
(0) =
C
=
i
+
j
−
4
k
.
Thus
v
(
t
) =
d
r
dt
=
i
+
j
−
(8
t
+ 4)
k
.
But then
r
(
t
) =
t
i
+
t
j
−
(4
t
2
+ 4
t
)
k
+
D
where
D
is a constant vector such that
r
(0) =
D
= 2
i
+ 3
j
.
Consequently,
r
(
t
) = (
t
+ 2)
i
+ (
t
+ 3)
j
−
(4
t
2
+ 4
t
)
k
.
002
10.0 points
Determine the minimum speed oF a particle
moving in 3space with position Function
r
(
t
) =
a
t
2
,
4
t, t
2
−
16
t
A
.
1.
min speed = 15 units/sec
2.
min speed = 12 units/sec
correct
3.
min speed = 14 units/sec
4.
min speed = 11 units/sec
5.
min speed = 13 units/sec
Explanation:
The velocity oF the particle is given by
r
′
(
t
) =
a
2
t,
4
,
2
t
−
16
A
,
so the particle has speed

r
′
(
t
)

=
r
4
t
2
+ 16 + (2
t
−
16)
2
.
This has a minimum when
d

r
′
(
t
)

dt
=
8
t
+ 4(2
t
−
16)
R
4
t
2
+ 16 + (2
t
−
16)
2
= 0
,
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2
i.e.
, when
t
= 4. Consequently, the particle
has minimum when
min speed =

r
′
(4)

= 12 units/sec
.
003
10.0 points
A ball is thrown at an angle of 39
◦
to the
ground. If the ball lands 109 meters away,
what was the initial speed of the ball?
(
Assume
g
= 9
.
8 meters/sec.)
1.
v
0
≈
33
.
3 m
/
s
2.
v
0
≈
34
.
0 m
/
s
3.
v
0
≈
33
.
0 m
/
s
correct
4.
v
0
≈
34
.
3 m
/
s
5.
v
0
≈
33
.
5 m
/
s
Explanation:
We can assume that the snowball is moving
in the
xy
plane and that it is thrown from the
origin. Then its position function is given by
r
(
t
) =
−
1
2
gt
2
j
+
t
v
0
,
where
g
is the acceleration due to gravity and
v
0
is the initial velocity. But the ball is thrown
at an elevation of 39
◦
and lands 109 meters
away. Thus we have
v
0
=

v
0

((cos 39)
i
+ (sin39)
j
)
,
so the equation becomes
r
(
t
) =
−
1
2
2
j
+
t

v
0

((cos39)
i
+ (sin 39)
j
)
=
t

v
0

(cos39)
i
+ (
t

v
0

sin 39
−
1
2
2
)
j
.
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 Spring '11
 RUSIN
 Derivative

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