2.4 Tangent, Normal and Binormal Vectors

[Pages:5]90

CHAPTER 2. VECTOR FUNCTIONS

2.4 Tangent, Normal and Binormal Vectors

Three vectors play an important role when studying the motion of an object along a space curve. These vectors are the unit tangent vector, the principal normal vector and the binormal vector. We have already de...ned the unit tangent vector. In this section, we de...ne the other two vectors.

Let us start by reviewing the de...nition of the unit tangent vector.

De...nition 147 (Unit Tangent Vector) Let C be a smooth curve with posi-

tion

vector

!r (t).

The

unit

tangent

vector,

denoted

! T (t)

is

de...ned

to

be

!

!r 0 (t)

T (t) = k!r 0 (t)k

2.4.1 Normal and Binormal Vectors

De...nition 148 (Normal Vector) Let C be a smooth curve with position vec-

tor !r (t). The principal normal vector or simply the normal vector, denoted

! N (t) is de...ned to be:

!

!T 0 (t)

N (t) = !T 0 (t)

(2.13)

The name of this vector suggests that it is normal to something, the question

!

!

is to what? By de...nition, T is a unit vector, that is T (t) = 1. From

proposition

120,

it

follows

that

!T 0

(t)

?

! T (t).

Thus,

!

!

N (t) ? T (t).

In fact,

!

!

N (t) is a unit vector, perpendicular to T pointing in the direction where the

curve is bending.

De...nition 149 (Binormal Vector) Let C be a smooth curve with position vector !r (t). The binormal vector, denoted !B (t), is de...ned to be

!B (t) = !T (t) !N (t)

!

!

Since both T (t) and N (t) are unit vectors and perpendicular, it follows that

!

!

!

B (t) is also a unit vector. It is perpendicular to both T (t) and N (t).

Example 150 Consider the circular helix !r (t) = hcos t; sin t; ti. Find the unit tangent, normal and binormal vectors.

Unit Tangent:

Since !T (t) =

!r 0 (t) k!r 0 (t)k ,

we need to compute !r 0 (t) and

k!r 0 (t)k.

!r 0 (t) = h sin t; cos t; 1i

2.4. TANGENT, NORMAL AND BINORMAL VECTORS

91

and

!r 0 (t)

=

p sin2 t + cos2 t + 1

p

=2

Thus

!T (t) = psin t ; cpos t ; p1 2 22

! Normal: Since N (t) =

!T 0 (t) !T 0 (t)

, we need to compute !T 0 (t) and

!T 0 (t) .

and Thus

!T 0 (t) =

pcos t ; psin t ; 0

2

2

!T 0 (t)

s

cos2 t sin2 t

=

+

2

2

= p1 2

!N (t) = h cos t; sin t; 0i

Binormal:

!B (t) = !T (t) !N (t)

=

psin t ; cpos t ; p1

2 22

!B (t) =

spin t ; pcos t ; p1 2 22

h cos t; sin t; 0i

The pictures below (...gures 2.5, 2.6 and 2.7) show the helix for t 2 [0; 2 ] as well as the three vectors !T (t), !N (t) and !B (t) plotted for various values of t. If the three vectors do not appear to be exactly orthogonal, it is because the scale is not the same in the x; y and z directions.

2.4.2 Osculating and Normal Planes

De...nition 151 (Osculating and Normal Planes) Let C be a smooth curve with position vector !r (t). Let P be a point on the curve corresponding to !r (t0) for some value of t.

!

!

1. The plane through P determined by N (t0) and B!(t0) is called the normal

plane of C at P . Note that its normal will be T (t).

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CHAPTER 2. VECTOR FUNCTIONS

Figure 2.5: Helix and the vectors !T (0), !N (0) and !B (0)

Figure 2.6: Helix and the vectors !T (1), !N (1) and !B (1)

2.4. TANGENT, NORMAL AND BINORMAL VECTORS

93

Figure 2.7: Helix and the vectors !T (4), !N (4) and !B (4)

!

!

2.

The

plane

through

P

determined

by

T

(t0)

and

N

(t0

)

is !

cal

led

the

oscu-

lating plane of C at P . Note that its normal will be B (t).

Example 152 Find the normal and osculating planes to the helix given by

!r (t) = hcos t; sin t; ti at the point

0; 1; 2

.

Earlier, we found that

!T (t) = psin t ; cpos t ; p1 2 22

!N (t) = h cos t; sin t; 0i

and

! B (t) =

spin t ;

pcos t ; p1

2 22

At the point 0; 1; , that is when t = , we have

2

2

!T

= p1 ; 0; p1

2

22

!N 2 = h0; 1; 0i

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CHAPTER 2. VECTOR FUNCTIONS

and

!B

= p1 ; 0; p1

2

22

Normal Plane: It is the plane through 0; 1; with normal !T

=

2

2

p1 ; 0; p1 . Thus its equation is

22

p1 (x 0) + p1 z

2

2

p Multiplying each side by 2 gives

=0 2

(x 0) + z

=0

2

or z x= 2

! Osculating Plane: It is the plane through 0; 1; 2 with normal B 2 =

p1 ; 0; p1 Thus its equation is 22

p1 (x 0) + p1 z

2

2

p Multiplying each side by 2 gives

2 =0

(x 0) + z

=0

2

or x+z = 2

Make sure you have read, studied and understood what was done above before attempting the problems.

2.4.3 Problems

Do # 11, 13, 37, 39, 43 at the end of 10.3 in your book

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