byrne (cmb3744) – HW 09 – rusin – (55565)
1
This
printout
should
have
22
questions.
Multiplechoice questions may continue on
the next column or page – find 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 off the hook! Meanwhile,
the new material we are starting (functions
of two or more variables) will give you lots of
stuff to calculate and a lot less stuff 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
10.0points
Determine the position vector,
r
(
t
), of a
particle having acceleration
a
(
t
) =
−
6
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
−
(3
t
2
−
4
t
)
k
2. r
(
t
) = (
t
−
2)
i
+ (
t
+ 3)
j
−
(3
t
2
+ 4
t
)
k
3. r
(
t
) = (
t
+ 2)
i
−
(
t
−
3)
j
−
(3
t
2
+ 4
t
)
k
4. r
(
t
) = (
t
+ 2)
i
+ (
t
+ 3)
j
−
(3
t
2
−
4
t
)
k
5. r
(
t
) = (
t
−
2)
i
+ (
t
+ 3)
j
−
(3
t
2
−
4
t
)
k
6. r
(
t
) = (
t
+ 2)
i
+ (
t
+ 3)
j
−
(3
t
2
+ 4
t
)
k
correct
Explanation:
Since
a
(
t
) =
d
v
dt
=
−
6
k
,
we see that
v
(
t
) =
−
6
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
−
(6
t
+ 4)
k
.
But then
r
(
t
) =
t
i
+
t
j
−
(3
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
−
(3
t
2
+ 4
t
)
k
.
002
10.0points
Determine the minimum speed of a particle
moving in 3space with position function
r
(
t
) =
(
t
2
,
6
t, t
2
−
24
t
)
.
1.
min speed = 19 units/sec
2.
min speed = 15 units/sec
3.
min speed = 16 units/sec
4.
min speed = 17 units/sec
5.
min speed = 18 units/sec
correct
Explanation:
The velocity of the particle is given by
r
′
(
t
) =
(
2
t,
6
,
2
t
−
24
)
,
so the particle has speed

r
′
(
t
)

=
radicalBig
4
t
2
+ 36 + (2
t
−
24)
2
.
This has a minimum when
d

r
′
(
t
)

dt
=
8
t
+ 4(2
t
−
24)
radicalbig
4
t
2
+ 36 + (2
t
−
24)
2
= 0
,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
byrne (cmb3744) – HW 09 – rusin – (55565)
2
i.e.
, when
t
= 6.
Consequently, the particle
has minimum when
min speed =

r
′
(6)

= 18 units/sec
.
003
10.0points
A ball is thrown at an angle of 42
◦
to the
ground.
If the ball lands 111 meters away,
what was the initial speed of the ball?
(
Assume
g
= 9
.
8 meters/sec.)
1.
v
0
≈
33
.
1 m
/
s
correct
2.
v
0
≈
34
.
7 m
/
s
3.
v
0
≈
33
.
4 m
/
s
4.
v
0
≈
33
.
7 m
/
s
5.
v
0
≈
34
.
0 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 42
◦
and lands 111 meters
away. Thus we have
v
0
=

v
0

((cos 42)
i
+ (sin 42)
j
)
,
so the equation becomes
r
(
t
) =
−
1
2
gt
2
j
+
t

v
0

((cos 42)
i
+ (sin 42)
j
)
=
t

v
0

(cos 42)
i
+ (
t

v
0

sin 42
−
1
2
gt
2
)
j
.
Now, the ball hits the ground when
t

v
0

sin 42
−
1
2
gt
2
= 0
.
i.e.
, after
t
= 2(

v
0

sin 42)
/g
seconds, and it
will then be a distance of

v
0

2
sin 84
g
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Chu
 Derivative, Byrne

Click to edit the document details