1

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

2

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

4

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

10

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

11

Math 114 – Rimmer

13.3/13.4 Arc Length and Curvature

( )

( )

( )( )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