S E C T I O N
14.4
Curvature
(ET Section 13.4)
531
Hint:
Observe that
T
(
t
)
=
cos
θ
(
t
),
sin
θ
(
t
)
.
y
x
r
(
t
)
(
t
)
T
(
t
)
= ⟨
cos
(
t
), sin
(
t
)
⟩
FIGURE 19
The curvature is the rate of change of
θ
(
t
)
.
SOLUTION
Since
T
(
t
)
is a unit vector that makes an angle
θ
(
t
)
with the positive
x
axis, we have
T
(
t
)
=
cos
θ
(
t
),
sin
θ
(
t
) .
Differentiating this vector using the Chain Rule gives:
T
(
t
)
= −
θ
(
t
)
sin
θ
(
t
),
θ
(
t
)
cos
θ
(
t
)
=
θ
(
t
)
−
sin
θ
(
t
),
cos
θ
(
t
)
We compute the norm of the vector
T
(
t
)
:
T
(
t
)
=
θ
(
t
)
−
sin
θ
(
t
),
cos
θ
(
t
)
= 
θ
(
t
)

(
−
sin
θ
(
t
))
2
+
(
cos
θ
(
t
))
2
= 
θ
(
t
)
 ·
1
= 
θ
(
t
)

When
r
(
s
)
is a parametrization by arc length we have:
κ
(
s
)
=
d
T
ds
=
d
T
dt
dt
d
θ
d
θ
ds
=
θ
(
t
)
1

θ
(
t
)

d
θ
ds
=
d
θ
ds
as desired.
60.
A particle moves along the path
y
=
x
3
with unit speed. How fast is the tangent turning (i.e., how fast is the angle
of inclination changing) when the particle passes through the point
(
2
,
8
)
?
SOLUTION
The particle has unit speed hence the parametrization is in arc length parametrization. Therefore, the change
in the angle of inclination is, by Exercise 59:
d
θ
ds
=
κ
(
s
)
In particular at the point
(
2
,
8
)
we have:
d
θ
ds
(
2
,
8
)
=
κ
(
s
)
(
2
,
8
)
(1)
We, thus, must find the curvature at the given point. We use the formula for the curvature of a graph in the plane:
κ
(
x
)
=

y
(
x
)

(
1
+
y
(
x
)
2
)
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 Winter '08
 GANGliu
 Rate Of Change, Dot Product, Euclidean space, Angle of inclination, Arc length parametrization, unit speed

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