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Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
x
y
z
t a=
t b=
( ) ( ) ( ) ( ), ,t f t g t h t=r
( ) ( )
( ) ( )
( ) ( )
2
2 2
In the plane (2-d):
:
parametric
Ar
equations : and :
c Length 1
Arc Length
b
a
b
a
y f x
x f t
f x d
y g
x
f t g t dt
t
′ = +
′ ′ = +
=
= =
∫
∫
( ) ( ) ( )2 2 2
Arc Length
b
a
f t g t h t dt′ ′ ′ = + + ∫
( )Arc Length
b
a
t dt′= ∫ r
( ) ( )
:
t
a
s t u du′= ∫
Arc Length Function
r
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
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Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
3
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
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Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature( ) ( )t
a
s t u du′= ∫ r
distance and times t= =
( ) gives position on the curve as a function of timetr
like driving and looking at the clock to tell where
you are after you have traveled a certain time t
we would like to have as a function of
so that we could reparametrize in terms of .
t s
s
its more natural to look at the odometer or the
mile markers to tell where you are after you
have traveled a certain distance s
( )( ) ( )
Using the arc length function we can find as a function of
giving us or just
t s
t s sr r
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature( ) 3sin ,4 ,3cost t t t=r
Reparametrize the curve with respect to arc length measured
from the point where 0 in the direction of increasing t t=
( ) 3cos ,4, 3sint t t′ = −r
( ) 2 29cos 16 9sint t t′ = + +r ( )2 29 cos sin 16t t= + + 9 16= + 25= 5=
( )0
t
s u du′= ∫ r0
5
t
du= ∫ 5t=
55
ss t t= ⇒ =
( ) ( ) ( ) ( )5 5 53sin ,4 ,3coss s ss =r
( )5 is the position vector of the point 5 units
of length along the curve from its starting point.
r
5
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
( )tT( )tT
( )t t+T �s∆
∆T
( )In order to calculate curvature ,we need
to use the unit tangent vector and
the arc length parameter . s
κ
T
- a measure how quickly a curve changes direction at a pointCurvature
another way to look at curvature is how sharply the curve bends at a point
P
Q
curvature @ curvature @ P Q>
- the magnitude of the rate of change of the unit
tangent vector with respect to the arc length parameter .s
Curvature
T
( )d
sds
κ ′= =T
T
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
( ) ( ) ( ) ( )5 5 53sin ,4 ,3coss s ss =r
( ) ( ) ( )3 345 5 5 5 5cos , , sins ss −′ =r
( ) ( ) ( )2 29 16 925 5 25 25 5
cos sins ss′ = + +r ( ) ( )2 29 1625 5 5 25
cos sins s = + + 9 1625 25
= +
( ) 1s′ =r
( )( )
( )( ) ( )3 34
5 5 5 5 5cos , , sins s
ss
s
−′
= =′
rT
r
( ) ( ) ( )3 325 5 25 5
sin ,0, coss ss − −′ =T
( ) ( ) ( )2 29625 5 5
sin coss ssκ ′ = = + T
3
25κ =
�
( )d d d
sds ds ds
κ
′′= = =
T
T rr
( )sκ ′′= r
6
Math 114 – Rimmer
13.3/13.4 Arc Length and CurvatureCurvature in terms of t
d
dsκ =
Td
dtds
dt
=
T ( ) ( )t
a
s t u du′= ∫ r( )
( )
t
t
′=
′
T
r
( ) 3sin ,4 ,3cost t t t=r
( ) 5t′ =r
( ) 3cos ,4, 3sint t t′ = −r
( )( )
( )
tt
t
′=
′
rT
r3 345 5 5cos , , sint t−=
( ) 2 2925
3sin cos
5t t t′ ⇒ = + = T( ) 3 3
5 5sin ,0, cost t t− −′ =T
( )
( )
t
tκ
′=
′
T
r
3
5
5=
3
25=
( )
( )
t
tκ
′=
′
T
r
( )Fund. Theorem
of Calculus
dst
dt⇒ ′= r
Math 114 – Rimmer
13.3/13.4 Arc Length and CurvatureCurvature in terms of t
( )( )
( )
tt
t
′=
′
rT
r( ) ( ) ( )t t t′ ′⇒ =r r T ( ) ( )
dst t
dt′⇒ =r T
scalar
functionvector
( )Take a derivative to find :t′′r
( ) ( ) ( )2
2
d s dst t t
dt dt′′ ′⇒ = +r T T
( ) ( )Consider: t t′ ′′×r r
2
2
ds d s ds
dt dt dt
′= × +
T T T
2
2
ds d s ds ds
dt dt dt dt
′= × + ×
T T T T
[ ] [ ]22
2
ds d s ds
dt dt dt
′= ⋅ × + ×
T T T T
( ) ( ) ( ) [ ]2
t t t′ ′′ ′ ′× = ×r r r T T
( ) ( ) ( )2
scalar funct.
t t t′ ′′ ′ ′× = ×r r r T T���
0
7
Math 114 – Rimmer
13.3/13.4 Arc Length and CurvatureCurvature in terms of t
sinθ′ ′× =T T T T
sin 1θ′⊥ ⇒ =T T
′ ′× =T T T T
( ) ( ) ( )2
t t t′ ′′ ′ ′× =r r r T ( )( )
3
tt
′′=
′
Tr
r
1=T
( ) ( )
( )3
t t
tκ
′ ′′×=
′
r r
r
( ) ( ) ( )Show that if (constant), then for all .t c t t t′= ⊥
Section 14.2
r r r
( ) ( ) ( )1 so for all .t t t t′= ⊥T T T
′ ′× =T T T
( ) ( ) ( )2
t t t′ ′′ ′ ′× = ×r r r T T
( )
( )
t
t
′⋅
′
r
r
κ�
( ) ( ) ( )3
t t t κ′ ′′ ′× =r r r
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
( ) 3sin ,4 ,3cost t t t=r
( ) 3cos , 4, 3sint t t′ = −r
( ) 3sin ,0, 3cost t t′′ = − −r
( ) ( ) 3cos 4 3sin
3sin 0 3cos
t t t t
t t
′ ′′× = − =
− −
i j k
r r ( )2 212cos , 9 cos sin ,12sint t t t − − − +
( ) ( ) 12cos ,9,12sint t t t′ ′′× = −r r
( ) 5t′ =r
( ) ( ) 2 2144 cos sin 81t t t t′ ′′ × = + + r r
( ) ( ) 15t t′ ′′× =r r
225=
( ) ( )
( )3
t t
tκ
′ ′′×=
′
r r
r
15
125=
3
25κ =
8
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
( ) ( ) ( ) ( )
( ) ( )3/ 2
2 2
Let be a smooth curve in the plane. ,
Show
t xy t f t g t
f g f g
f g
κ
=
′ ′′ ′′ ′−= ′ ′+
r r
( ) ( )
( )3
t t
tκ
′ ′′×=
′
r r
r
( ) ( ) ( ), , 0 t f t g t=r
( ) ( ) ( ), ,0 t f t g t′ ′ ′=r
( ) ( ) ( ), ,0 t f t g t′′ ′′ ′′=r
( ) ( ) 0,0,t t f g f g′ ′′ ′ ′′ ′′ ′× = −r r
( ) ( ) ( )2
t t f g f g′ ′′ ′ ′′ ′′ ′× = −r r f g f g′ ′′ ′′ ′= −
( ) ( )( ) ( )( )2 2
t f t g t′ ′ ′= +r
( ) ( )( ) ( )( )( )3/ 23 2 2
t f t g t′ ′ ′= +r
( ) ( )
( )3
t t
tκ
′ ′′×=
′
r r
r ( ) ( )3/ 2
2 2
f g f g
f g
′ ′′ ′′ ′−= ′ ′+
9
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
( ) ( ) ( )
( ) ( )3/ 2
2 2
, t f t g t
f g f g
f g
κ
=
′ ′′ ′′ ′−= ′ ′+
Planar parametric curve
r( )
( )
( ) ( )
( )3
t
t
t
t t
t
κ
κ
′=
′
′ ′′×=
′
In terms of
T
r
r r
r
( )
( )
s
s
s
κ
κ
′=
′′=
In terms of
T
r
( )
( )
( )( )3/ 2
2
1
y f x
f x
f x
κ
=
′′= ′+
Planar function
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
1=T ′⇒ ⊥T T but is not necessarily a unit vector.′T
( )
making a unit vector gives a new vector
called the , t
′
′=
′
T
Tprincipal unit normal vector N
T
( )remember is called the tT principal unit tangent vector
( ) ( ) ( )The vector = is called the .t t t×B T N binormal vector
( ) ( ) ( ) is orthogonal to both and .t t tB T N
( )tB
1=B
( ) frame Frenet frame moves along the curve,
has applications in differential geometry
TNB
( ) is useful for finding the degree of twisting of the curvetorsionB
( )tN
( )tT
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Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature
The plane determined by the vectors and is called the .T N osculating plane
The plane determined by the vectors and is called the .N B normal plane
The circle that :
)a lies in the osculating plane of the curve at the point C P
)b has the same tangent as at C P
) ( )c lies on the concave side of side towards which pointsC N
)d has the reciprocal of curvature as its radius
is called the .osculating circle
Math 114 – Rimmer
13.3/13.4 Arc Length and Curvature