VECTOR FUNCTIONS

12.3Arc Length

and Curvature

In this section, we will learn how to find:

The arc length of a curve and its curvature.

VECTOR FUNCTIONS

PLANE CURVE LENGTH

In Section 10.2, we defined the length of a plane curve with parametric equations

x = f(t), y = g(t), a ≤ t ≤ b

as the limit of lengths of inscribed polygons.

[ ] [ ]2 2

2 2

'( ) '( )b

a

b

a

L f t g t dt

dx dy dtdt dt

= +

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫

∫

Formula 1 PLANE CURVE LENGTH

For the case where f’ and g’ are continuous, we arrived at the following formula:

SPACE CURVE LENGTH

The length of a space curve is defined in exactly the same way.

Suppose that the curve has the vector equation

r(t) = <f(t), g(t), h(t)>, a ≤ t ≤ b

Equivalently, it could have the parametric equations

x = f(t), y = g(t), z = h(t)

where f’, g’ and h’ are continuous.

SPACE CURVE LENGTH

If the curve is traversed exactly once as tincreases from a to b, then it can be shown that its length is:

[ ] [ ] [ ]2 2 2

2 2 2

'( ) '( ) '( )b

a

b

a

L f t g t h t dt

dx dy dz dtdt dt dt

= + +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∫

∫

Formula 2 SPACE CURVE LENGTH

ARC LENGTH

Notice that both the arc length formulas 1 and 2 can be put into the more compact form

'( )b

aL t dt= ∫ r

Formula 3

That is because:

For plane curves r(t) = f(t) i + g(t) j

For space curves r(t) = f(t) i + g(t) j + h(t) k

[ ] [ ]2 2'( ) '( ) '( ) '( ) '( )t f t g t f t g t= + = +r i j

[ ] [ ] [ ]2 2 2

'( ) '( ) '( ) '( )

'( ) '( ) '( )

t f t g t h t

f t g t h t

= + +

= + +

r i j k

ARC LENGTH

Find the length of the arc of the circular helix with vector equation

r(t) = cos t i + sin t j + t k

from the point (1, 0, 0) to the point (1, 0, 2π).

Example 1 ARC LENGTH

Since r’(t) = -sin t i + cos t j + k, we have:

2 2'( ) ( sin ) cos 1

2

t t t= − + +

=

r

Example 1 ARC LENGTH

The arc from (1, 0, 0) to (1, 0, 2π) is described by the parameter interval 0 ≤ t ≤ 2π.

So, from Formula 3, we have:2

0

2

0

'( )

2

2 2

L t dt

dt

π

π

π

=

=

=

∫∫

r

Example 1 ARC LENGTH

ARC LENGTH

A single curve C can be represented by more than one vector function.

For instance, the twisted cubic r1(t) = <t, t 2, t 3> 1 ≤ t ≤ 2

could also be represented by the function r2(u) = <eu, e2u, e3u> 0 ≤ u ≤ ln 2

The connection between the parameters t and u is given by t = eu.

Equations 4 & 5 ARC LENGTH

We say that Equations 4 and 5 are parametrizations of the curve C.

PARAMETRIZATION

If we were to use Equation 3 to compute the length of C using Equations 4 and 5, we would get the same answer.

In general, it can be shown that, when Equation 3 is used to compute arc length, the answer is independent of the parametrization that is used.

PARAMETRIZATION

Now, we suppose that C is a curve given by a vector function

r(t) = f(t) i + g(t) j + h(t) k a ≤ t ≤ b

where: r’ is continuous.C is traversed exactly once as t increases from a to b.

ARC LENGTH

ARC LENGTH FUNCTION

We define its arc length function sby:

2 2 2

( ) '( )t

a

t

a

s t u du

dx dy dz dudu du du

=

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∫

∫

r

Equation 6

Thus, s(t) is the length of the part of Cbetween r(a) and r(t).

ARC LENGTH FUNCTION

If we differentiate both sides of Equation 6 using Part 1 of the Fundamental Theorem of Calculus (FTC1), we obtain:

'( )ds tdt

= r

Equation 7 ARC LENGTH FUNCTION

It is often useful to parametrize a curve with respect to arc length.

This is because arc length arises naturally from the shape of the curve and does not depend on a particular coordinate system.

PARAMETRIZATION

If a curve r(t) is already given in terms of a parameter t and s(t) is the arc length function given by Equation 6, then we may be able to solve for t as a function of s:

t = t(s)

PARAMETRIZATION

Then, the curve can be reparametrized in terms of s by substituting for t:

r = r(t(s))

REPARAMETRIZATION

Thus, if s = 3 for instance, r(t(3)) is the position vector of the point 3 units of length along the curve from its starting point.

REPARAMETRIZATION

Reparametrize the helix

r(t) = cos t i + sin t j + t k

with respect to arc length measured from (1, 0, 0) in the direction of increasing t.

Example 2 REPARAMETRIZATION

The initial point (1, 0, 0) corresponds to the parameter value t = 0.

From Example 1, we have:

So,

'( ) 2ds tdt

= =r

0 0( ) '( ) 2 2

t ts s t u du du t= = = =∫ ∫r

Example 2 REPARAMETRIZATION

Therefore, and the required reparametrization is obtained by substituting for t:

/ 2t s=

( ) ( ) ( )( ( ))

cos / 2 sin / 2 / 2

t s

s s s= + +

r

i j k

Example 2 REPARAMETRIZATION

A parametrization r(t) is called smoothon an interval I if:

r’ is continuous.

r’(t) ≠ 0 on I.

SMOOTH PARAMETRIZATION

A curve is called smooth if it has a smooth parametrization.

A smooth curve has no sharp corners or cusps.

When the tangent vector turns, it does so continuously.

SMOOTH CURVE

If C is a smooth curve defined by the vector function r, recall that the unit tangent vector T(t) is given by:

This indicates the direction of the curve.

'( )( )'( )ttt

=rTr

SMOOTH CURVES

You can see that T(t) changes direction:

Very slowly when C is fairly straight.More quickly when C bends or twists more sharply.

SMOOTH CURVES

The curvature of C at a given point is a measure of how quickly the curve changes direction at that point.

CURVATURE

Specifically, we define it to be the magnitude of the rate of change of the unit tangent vector with respect to arc length.

We use arc length so that the curvature will be independent of the parametrization.

CURVATURE

The curvature of a curve is:

where T is the unit tangent vector.

dds

κ =T

Definition 8 CURVATURE—DEFINITION

The curvature is easier to compute if it is expressed in terms of the parameter t instead of s.

CURVATURE

So, we use the Chain Rule (Theorem 3 in Section 13.2, Formula 6) to write:

/and/

d d ds d d dtdt ds dt ds ds dt

κ= = =T T T T

CURVATURE

However, ds/dt = |r’(t)| from Equation 7.

So, '( )

( )'( )

tt

tκ =

Tr

Equation/Formula 9 CURVATURE

Show that the curvature of a circle of radius a is 1/a.

We can take the circle to have center the origin.

Then, a parametrization is:

r(t) = a cos t i + a sin t j

Example 3 CURVATURE

Therefore, r’(t) = –a sin t i + a cos t j

and |r’(t)| = a

So,

and

Example 3 CURVATURE

'( )( ) sin cos'( )

'( ) cos sin

tt t tt

t t t

= = − +

= − −

rT i jr

T i j

This gives |T’(t)| = 1.

So, using Equation 9, we have:

'( ) 1( )'( )

tt

t aκ = =

Tr

Example 3 CURVATURE

The result of Example 3 shows—in accordance with our intuition—that:

Small circles have large curvature.

Large circles have small curvature.

CURVATURE

We can see directly from the definition of curvature that the curvature of a straight line is always 0—because the tangent vector is constant.

CURVATURE

Formula 9 can be used in all cases to compute the curvature.

Nevertheless, the formula given by the following theorem is often more convenient to apply.

CURVATURE

The curvature of the curve given by the vector function r is:

3

'( ) ''( )( )

'( )t t

tt

κ×

=r r

r

Theorem 10 CURVATURE

T = r’/|r’| and |r’| = ds/dt.

So, we have:

' ' dsdt

= =r r T T

Proof CURVATURE

Hence, the Product Rule (Theorem 3 in Section 13.2, Formula 3) gives:

2

2'' 'd s dsdt dt

= +r T T

Proof CURVATURE

Using the fact that T x T = 0 (Example 2 in Section 12.4), we have:

( )2

' '' 'dsdt

⎛ ⎞× = ×⎜ ⎟⎝ ⎠

r r T T

Proof CURVATURE

Now, |T(t)| = 1 for all t.

So, T and T’ are orthogonal by Example 4 in Section 13.2

Proof CURVATURE

Hence, by Theorem 6 in Section 12.4,

2

2

2

' '' '

'

'

dsdt

dsdt

dsdt

⎛ ⎞× = ×⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

r r T T

T T

T

Proof CURVATURE

Thus,

and

( )2 2

3

' '' ' '''

/ '

' ' ''' '

ds dt

κ

× ×= =

×= =

r r r rT

r

T r rr r

Proof CURVATURE

Find the curvature of the twisted cubic r(t) = <t, t2, t3>

at:A general point

(0, 0, 0)

Example 4 CURVATURE

First, we compute the required ingredients:

2

2 4

'( ) 1, 2 ,3 ''( ) 0, 2,6

'( ) 1 4 9

t t t t t

t t t

= =

= + +

r r

r

Example 4 CURVATURE

Example 4 CURVATURE

2

2

4 2

4 2

'( ) ''( ) 1 2 30 2 6

6 6 2

'( ) ''( ) 36 36 4

2 9 9 1

t t t tt

t t

t t t t

t t

× =

= − +

× = + +

= + +

i j kr r

i j k

r r

Then, Theorem 10 gives:

At the origin, where t = 0, the curvature is:

ĸ(0) = 2

( )2 4

3 3/ 22 4

'( ) ''( ) 2 1 9 9( )'( ) 1 4 9

t t t ttt t t

κ× + +

= =+ +

r rr

Example 4 CURVATURE

For the special case of a plane curve with equation y = f(x), we choose x as the parameter and write:

r(x) = x i + f(x) j

CURVATURE

Then,

r’(x) = i + f’(x) j

andr’’(x) = f’’(x) j

CURVATURE

Since i x j = k and j x j = 0, we have:

r’(x) x r’’(x) = f’’(x) k

CURVATURE

We also have:

[ ]2'( ) 1 '( )x f x= +r

CURVATURE

So, by Theorem 10,

( )3/ 22

''( )( )

1 '( )

f xx

f xκ =

⎡ ⎤+⎣ ⎦

Formula 11 CURVATURE

Find the curvature of the parabola y = x2

at the points (0, 0), (1, 1), (2, 4)

Example 5 CURVATURE

Since y’ = 2x and y’’ = 2, Formula 11 gives:

( )

3/ 22

3/ 22

''( )

1 ( ')

2

1 4

yx

y

x

κ =⎡ ⎤+⎣ ⎦

=+

Example 5 CURVATURE

At (0, 0), the curvature is κ(0) = 2.

At (1, 1), it is κ(1) = 2/53/2 ≈ 0.18

At (2, 4), it is κ(2) = 2/173/2 ≈ 0.03

Example 5 CURVATURE

Observe from the expression for κ(x) or the graph of κ here that:

κ(x) → 0 as x → ±∞

This corresponds to the fact that the parabola appears to become flatter as x → ±∞

Example 5 CURVATURE

NORMAL AND BINORMAL VECTORS

At a given point on a smooth space curve r(t), there are many vectors that are orthogonal to the unit tangent vector T(t).

We single out one by observing that, because |T(t)| = 1 for all t, we have T(t) · T’(t) by Example 4 in Section 13.2.

So, T’(t) is orthogonal to T(t).

Note that T’(t) is itself not a unit vector.

NORMAL VECTORS

However, if r’ is also smooth, we can define the principal unit normal vector N(t) (simply unit normal) as:

'( )( )'( )ttt

=TNT

NORMAL VECTOR

NORMAL VECTORS

We can think of the normal vector as indicating the direction in which the curve is turning at each point.

BINORMAL VECTOR

The vector B(t) = T(t) x N(t)

is called the binormal vector.

BINORMAL VECTORS

It is perpendicular to both T and Nand is also a unit vector.

NORMAL & BINORMAL VECTORS

Find the unit normal and binormal vectors for the circular helix

r(t) = cost i + sin t j + t k

Example 6

First, we compute the ingredients needed for the unit normal vector:

( )

'( ) sin cos '( ) 2

'( ) 1( ) sin cos'( ) 2

t t t t

tt t tt

= − + + =

= = − + +

r i j k r

rT i j kr

Example 6 NORMAL & BINORMAL VECTORS

( )1 1'( ) cos sin '( )2 2

'( )( ) cos sin'( )

cos , sin ,0

t t t t

tt t tt

t t

= − − =

= = − −

= − −

T i j T

TN i jT

Example 6 NORMAL & BINORMAL VECTORS

This shows that the normal vector at a point on the helix is horizontal and points toward the z-axis.

Example 6 NORMAL & BINORMAL VECTORS

The binormal vector is:

1( ) ( ) ( ) sin cos 12 cos sin 01 sin , cos ,12

t t t t tt t

t t

⎡ ⎤⎢ ⎥= × = −⎢ ⎥⎢ ⎥− −⎣ ⎦

= −

i j kB T N

Example 6 NORMAL & BINORMAL VECTORS

The figure illustrates Example 6 by showing the vectors T, N, and Bat two locations on the helix.

NORMAL & BINORMAL VECTORS

In general, the vectors T, N, and B, starting at the various points on a curve, form a set of orthogonal vectors—called the TNBframe—that moves along the curve as t varies.

TNB FRAME

This TNB frame plays an important role in:

The branch of mathematics known as differential geometry.

Its applications to the motion of spacecraft.

TNB FRAME

NORMAL PLANE

The plane determined by the normal and binormal vectors N and B at a point P on a curve C is called the normal plane of C at P.

It consists of all lines that are orthogonal to the tangent vector T.

OSCULATING PLANE

The plane determined by the vectors T and N is called the osculating planeof C at P.

The name comes from the Latin osculum, meaning ‘kiss.’

It is the plane that comes closest to containing the part of the curve near P.

For a plane curve, the osculating plane is simply the plane that contains the curve.

OSCULATING PLANE

OSCULATING CIRCLE

The osculating circle (the circle of curvature) of C at P is the circle that:

Lies in the osculating plane of C at P.

Has the same tangent as C at P.

Lies on the concave side of C (toward which N points).

Has radius ρ = 1/ĸ (the reciprocal of the curvature).

It is the circle that best describes how C behaves near P.

It shares the same tangent, normal, and curvature at P.

OSCULATING CIRCLE

NORMAL & OSCULATING PLANES

Find the equations of the normal plane and osculating plane of the helix in Example 6 at the point

P(0, 1, π/2)

Example 7

The normal plane at P has normal vector r’(π/2) = <–1, 0, 1>.

So, an equation is:

or

( ) ( )1 0 0 1 1 02

2

x y z

z x

π

π

⎛ ⎞− − + − + − =⎜ ⎟⎝ ⎠

= +

Example 7 NORMAL & OSCULATING PLANES

The osculating plane at P contains the vectors T and N.

So, its normal vector is:

T x N = B

Example 7 NORMAL & OSCULATING PLANES

From Example 6, we have:

1( ) sin , cos ,12

1 1,0,2 2 2

t t t

π

= −

⎛ ⎞ =⎜ ⎟⎝ ⎠

B

B

Example 7 NORMAL & OSCULATING PLANES

A simpler normal vector is <1, 0, 1>.

So, an equation of the osculating plane is:

or

( ) ( )1 0 0 1 1 02

2

x y z

z x

π

π

⎛ ⎞− + − + − =⎜ ⎟⎝ ⎠

= − +

Example 7 NORMAL & OSCULATING PLANES

The figure shows the helix and the osculating plane in Example 7.

NORMAL & OSCULATING PLANES

OSCULATING CIRCLES

Find and graph the osculating circle of the parabola y = x2 at the origin.

From Example 5, the curvature of the parabola at the origin is ĸ(0) = 2.

So, the radius of the osculating circle at the origin is 1/ĸ = ½ and its center is (0, ½).

Example 8

Therefore, its equation is:

( )22 1 12 4x y+ − =

Example 8 OSCULATING CIRCLES

For the graph, we use parametric equations of this circle:

x = ½ cos t y = ½ + ½ sin t

Example 8 OSCULATING CIRCLES

The graph is displayed.Example 8 OSCULATING CIRCLES

SUMMARY

We summarize the formulas for unit tangent, unit normal and binormal vectors, and curvature.

3

'( ) '( )( ) ( ) ( ) ( ) ( )'( ) '( )

'( ) '( ) ''( )'( ) '( )

t tt t t t tt t

t t tdds t t

κ

= = = ×

×= = =

r TT N B T Nr T

T r rTr r