WEEK 4 HOMEWORK SOLUTIONS
1
a.
Our curve is parameterized by
r
(
t
) = (
x
(
t
)
, y
(
t
)) for
t
∈
R
. To calculate curvature
using the formula
(1)
κ
(
t
) =

v
(
t
)
×
a
(
t
)


v
(
t
)

3
,
we must first write our parameterization as a three dimensional vector, then compute
the associated velocity and acceleration. We can stick the plane that this curve is
stuck in at any place along the zaxis, meaning we should just set the z component
equal to the constant
c
. We then get;
r
(
t
)
=
(
x
(
t
)
, y
(
t
)
, c
)
v
(
t
)
=
(
x
(
t
)
, y
(
t
)
,
0)
a
(
t
)
=
(
x
(
t
)
, y
(
t
)
,
0)
.
To compute
v
(
t
)
×
a
(
t
) it is best to first recognize that the only nonzero component
will be
ˆ
k
. We find that
v
(
t
)
×
a
(
t
)
=
(
x
(
t
)
y
(
t
)
−
x
(
t
)
y
(
t
))
ˆ
k

v
(
t
)
×
a
(
t
)

=

x
(
t
)
y
(
t
)
−
x
(
t
)
y
(
t
)

.
Input this into 1 we get
κ
(
t
) =

x
(
t
)
y
(
t
)
−
x
(
t
)
y
(
t
)

(
x
(
t
)
2
+
y
(
t
)
2
)
(
3
/
2)
,
b.
Define the function
K
(
t
) as
K
(
t
) =
N
(
t
0
)
·
(
r
(
t
0
)
−
r
(
t
))

r
(
t
0
)
−
r
(
t
)

2
.
If we let
θ
be the angle between the two vectors in the denomiator we may writ this
expression in terms of the respective magnitudes and
θ
. Writing this out and using
1
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2
WEEK 4 HOMEWORK SOLUTIONS
the fact that
N
(
t
) is a unit vector we get that
K
(
t
)
=

N
(
t
0
)

r
(
t
0
)
−
r
(
t
)

cos(
θ
)

r
(
t
0
)
−
r
(
t
)

2
=
cos(
θ
)

r
(
t
0
)
−
r
(
t
)

c.
Now, we have oriented our coordinate system and set our initial time such that
t
0
= 0,
r
(0) = (0
,
0), and
N
(0) = (0
,
1).
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 Fall '07
 ADAMNORRIS
 Cartesian Coordinate System, Cos, Coordinate system, Polar coordinate system

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