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Math 1920
Solutions for 13.1
4.
We know
x
= cos 2
t
y
= 3 sin 2
t
By trigonometry, we obtain:
x
2
+
1
9
y
2
= 1
which is the equation of an ellipse centered at zero.
By diﬀerentiating each component, we obtain the velocity and acceleration
vectors:
~v
=
d~
r
dt
= (

2 sin 2
t
)
~
i
+ (6 cos 2
t
)
~
j
~a
=
d~v
dt
= (

4 cos 2
t
)
~
i
+ (

12 sin 2
t
)
~
j
Thus at
t
= 0, we get
~v
t
=0
= 6
~
j
~a
t
=0
=

4
~
i
8.
By diﬀerentiating each component, we obtain the following velocity and
acceleration vectors:
~v
=
d~
r
dt
=
~
i
+ 2
t
~
j
~a
=
d~v
dt
= 2
~
j
Hence at
t
=

1
,
0 and 1, we have:
1
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View Full Document ~v
t
=

1
=
~
i

2
~
j
~v
t
=0
=
~
i
~v
t
=1
=
~
i
+ 2
~
j
~a
t
=

1
= 2
~
j
~a
t
=0
= 2
~
j
~a
t
=1
= 2
~
j
17.
By diﬀerentiating each component, we obtain the following velocity and
acceleration vectors:
~v
=
d~
r
dt
=
2
t
t
2
+ 1
~
i
+
1
t
2
+ 1
~
j
+
t
√
t
2
+ 1
~
k
~a
=
d~v
dt
=

2
t
2
+ 2
(
t
2
+ 1)
2
~
i

2
t
(
t
2
+ 1)
2
~
j
+
1
(
t
2
+ 1)
3
/
2
~
k
At
t
= 0, the vectors and norms are:
~v
t
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