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Page 1: Section 11.7: Arc Length and Curvatureglahodny/Math251/Section 11.7.pdfSection 11.7: Arc Length and Curvature ... Figure 1: Unit tangent vectors for a space curve C. ... (x) = jy00j

Section 11.7: Arc Length and Curvature

Theorem: (Arc Length)

Consider the curve C defined by ~R(t) = 〈x(t), y(t), z(t)〉, a ≤ t ≤ b. If C is traversed exactlyonce as t increases from a to b, then its length is

L =

∫ b

a

||~R′(t)||dt =

∫ b

a

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt.

Example: Find the length of the curve defined by ~R(t) = 〈6t, 3√

2t2, 2t3〉, 0 ≤ t ≤ 1.

Since ~R′(t) = 〈6, 6√

2t, 6t2〉, we have

||~R′(t)|| =√

36 + 72t2 + 36t4

=√

36(1 + 2t2 + t4)

=√

36(1 + t2)2

= 6(1 + t2).

Then

L =

∫ 1

0

6(1 + t2)dt = (6t+ 2t3)∣∣10

= 4.

Definition: Suppose C is a space curve defined by ~R(t) = 〈x(t), y(t), z(t)〉, a ≤ t ≤ b. Thenits arc length function is defined by

s(t) =

∫ t

a

||~R′(u)||du =

∫ t

a

√(dx

du

)2

+

(dy

du

)2

+

(dz

du

)2

du.

Thus, s(t) is the length of the part of C between ~R(a) and ~R(t).

Note: A single curve C can be represented by more than one vector function. Differentrepresentations of curves are called parameterizations of the curve C. It is often useful toparameterize a curve with respect to arc length.

Example: Reparameterize the helix ~R(t) = 〈3 sin t, 4t, 3 cos t〉 with respect to arc lengthmeasured from the point where t = 0 in the direction of increasing t.

Since ~R′(t) = 〈3 cos t, 4,−3 sin t〉, we have

||~R′(t)|| =√

9 cos2 t+ 16 + 9 sin2 t =√

9(sin2 t+ cos2 t) + 16 =√

25 = 5.

Page 2: Section 11.7: Arc Length and Curvatureglahodny/Math251/Section 11.7.pdfSection 11.7: Arc Length and Curvature ... Figure 1: Unit tangent vectors for a space curve C. ... (x) = jy00j

The arc length function is

s = s(t) =

∫ t

0

5du = 5t.

Thus, t = s/5 and the reparameterization is given by

~R(s) =

⟨3 sin

(s5

),4s

5, 3 cos

(s5

)⟩.

Definition: The curvature of a curve defined by ~R(t) is

κ(t) =||~T ′(t)||||~R′(t)||

=||~R′(t)× ~R′′(t)||||~R′(t)||3

,

where ~T is the unit tangent vector.

Note: The curvature of a cuve C at a given point is a measure of how quickly the curvechanges direction at that point. Specifically, it measures the rate of change of the unittangent vector (which indicates the direction of C) with respect to arc length.

Figure 1: Unit tangent vectors for a space curve C.

Page 3: Section 11.7: Arc Length and Curvatureglahodny/Math251/Section 11.7.pdfSection 11.7: Arc Length and Curvature ... Figure 1: Unit tangent vectors for a space curve C. ... (x) = jy00j

Example: Find the curvature of the twisted cubic ~R(t) = 〈t, t2, t3〉 at a general point and at(0, 0, 0).

First,

~R′(t) = 〈1, 2t, 3t2〉~R′′(t) = 〈0, 2, 6t〉

||~R′(t)|| =√

1 + 4t2 + 9t4

~R′(t)× ~R′′(t) =

∣∣∣∣∣∣~i ~j ~k1 2t 3t2

0 2 6t

∣∣∣∣∣∣ = 〈6t2,−6t, 2〉.

Then||~R′(t)× ~R′′(t)|| =

√36t4 + 36t2 + 4 = 2

√9t4 + 9t2 + 1.

Thus, the curvature is

κ(t) =||~R′(t)× ~R′′(t)||||~R′(t)||3

=2√

1 + 9t2 + 9t4

(1 + 4t2 + 9t4)3/2.

At the origin the curvature is κ(0) = 2.

Theorem: (Curvature of a Plane Curve)The curvature of the plane curve y = f(x) is given by

κ(x) =|f ′′(x)|

[1 + (f ′(x))2]3/2.

Example: Find the curvature of y = sinx at a general point and at(π

2, 0)

.

Since y′ = cosx and y′′ = − sinx,

κ(x) =|y′′|

[1 + (y′)2]3/2=

| sinx|(1 + cos2 x)3/2

.

At(π

2, 0)

the curvature is κ(1) = 1.

Definition: Consider the curve defined by ~R(t). The principal unit normal vector is givenby

~N(t) =~T ′(t)

||~T ′(t)||,

Page 4: Section 11.7: Arc Length and Curvatureglahodny/Math251/Section 11.7.pdfSection 11.7: Arc Length and Curvature ... Figure 1: Unit tangent vectors for a space curve C. ... (x) = jy00j

and the binormal vector is given by

~B(t) = ~T (t)× ~N(t).

Note: The vectors ~T (t), ~N(t), and ~B(t) form a set of orthogonal vectors, called a frame, thatmoves along the curve as t varies. This frame has applications in differential geometry andmotion of spacecraft.

Figure 2: Illustration of the frame formed by the unit tangent, unit normal, and binormalvectors.

Example: Find the unit normal and binormal vectors for the helix ~R(t) = 〈sin t, cos t, t〉.

Since ~R′(t) = 〈cos t,− sin t, 1〉,

||~R′(t)|| =√

cos2 t+ sin2 t+ 1 =√

2.

Thus, the unit tangent vector is

~T (t) =~R′(t)

||~R′(t)||=

1√2〈cos t,− sin t, 1〉.

Therefore,

~T ′(t) =1√2〈− sin t,− cos t, 0〉, and ||~T ′(t)|| = 1√

2.

The unit normal vector is

~N(t) =~T ′(t)

||~T ′(t)||= 〈− sin t,− cos t, 0〉.

Page 5: Section 11.7: Arc Length and Curvatureglahodny/Math251/Section 11.7.pdfSection 11.7: Arc Length and Curvature ... Figure 1: Unit tangent vectors for a space curve C. ... (x) = jy00j

Now the binormal vector is

~B(t) = ~T (t)× ~N(t) =1√2

∣∣∣∣∣∣~i ~j ~k

cos t − sin t 1− sin t − cos t 0

∣∣∣∣∣∣ =1√2〈cos t,− sin t,−1〉.

Definition: The plane determined by the unit normal and binormal vectors ~N and ~B at apoint P on the curve C is called the normal plane of C at P . The unit tangent vector isorthogonal to the normal plane. Similarly, the plane determined by the unit tangent andunit normal vectors ~T and ~N is called the osculating plane of C at P . The binormal vectoris orthogonal to the osculating plane.

Example: Find equations for the normal and osculating planes of the helix ~R(t) = 〈sin t, cos t, t〉at the point P = (0,−1, π).

The tangent vector ~R′(t) = 〈cos t,− sin t, 1〉 is normal to the normal plane. At the point P ,~R′(π) = 〈−1, 0, 1〉. Thus, an equation of the normal plane is

−1(x− 0) + 0(y + 1) + 1(z − π) = 0

−x+ z = π.

The binormal vector is normal to the osculating plane. From the previous example,

~B(t) =1√2〈cos t,− sin t,−1〉 and so ~B(π) =

1√2〈−1, 0,−1〉.

An easier normal vector is 〈−1, 0, 1〉. So an equation of the osculating plane is

−1(x− 0) + 0(y + 1)− 1(z − π) = 0

z − x = −πx− z = π.