20 Moving Trihedral.pdf

Moving Trihedral

Mathematics 54 - Elementary Analysis 2

The Moving Trihedral

Consider a particle moving along a curve, defined by~R(t).

The particle’s trajectory (path) can be characterized by three unit vectors, namely

~T(t): unit tangent vector,

~

N(t): unit normal vector, and

Unit Tangent Vector

LetCbe a smooth curve defined by~R(t).The vector~R0_{(t) is non-zero and is} tangent to the curveC. We define theunit tangent vectorto the curveCby

~T(t)₌ ~R 0_(t) k~R0_(t)k.

Unit Tangent Vector

Example

Consider a moving particle whose path is given by~R(t)= 〈t2−1,t〉.

The path of the particle is shown below. Now,~R0(t)= 〈2t, 1〉. Hence,~T(t)=p 1

4t2₊₁〈2t, 1〉=

¿ _2t p

4t2₊₁,

1 p

4t2₊₁

When are

~

R

(

t

) and

~

R

(

t

) Orthogonal

In general,~R(t) and~R0_{(t) are not perpendicular.}

k

~

R

(

t

)

k =

constant

Theorem.

If~R(t) has constant magnitude for allt, then

~R(t)·~R0(t)=0.

That is,~R(t) and~R0_{(t) are perpendicular for allt.} Proof.Letk~R(t)k =k, wherekis some constant.

Consider

0 = d dt(k

2₎

= d dt(k~Rk

2₎

= d

dt £~

R(t)_·~R(t)¤

k

~

R

(

t

)

k =

constant

Consider the circle~R(t)_{= 〈}cost, sint〉. Note that_k~R(t)_{k =}1. Now,~R0(t)_{= 〈−}sint, cost〉.

Unit Normal Vector

Now, consider the vector~T0(t).

By the previous theorem,~T(t) and~T0(t) are perpendicular sincek~T(t)k =1 for allt.

Note, however, that~T0(t) is not always a unit vector. We define the vector

~

N(t)₌ ~T 0_(t) k~T0_(t)k as theunit normal vectorto the space curve~R(t).

Remark

Unit Normal Vector

Example

Consider~R(t)_{= 〈}lnt,t_〉. Find~T(2) and~N(2).

Solution.Differentiating~R, we have~R0_(t)=­1

t, 1

® .

Moreover,°°~R0(t) ° °=

q

1

t2+1=1t

p 1+t2_.

Hence,~T(t)₌Dp1

1+t2,

t

p

1+t2

E .

Thus,~T(2)₌Dp1

5, 2 p 5 E .

Now,~T0(t)=D−_(1+tt2₎3/2,_(1+t12₎3/2

E .

And~T0(2)=D−p2

125, 1 p 125 E Finally, ~

N(2)=

D

−p2

125, 1 p 125 E ° ° ° D

−p2

125, 1 p 125 E° ° °

=D−p2

5, 1

p

5

E

Remark.In 2Dspace,

~

Unit Normal Vector

Example

Let~R(t)= 〈2 cost, 2 sint,t〉. Find~T(t) and~N(t) att=3₄π.

Solution.We have~R0_{(t)= 〈−2 sint, 2 cost, 1〉, and}° °~R0(t)

° °=

p 5.

Thus,~T(t)₌D₋p2

5sint, 2

p

5cost, 1

p

5

E .

And,~T¡₃π

4 ¢ = D − p 2 p

5,−

p 2 p 5, 1 p 5 E .

Now,~T0_(t)=D_−p2 5cost,−

2

p

5sint, 0

E .

And,°°~T0(t) ° °=p2

5.

Hence,~N(t)_{= 〈−}cost,₋sint, 0〉.

And~N¡₃π

4

¢ =

Dp

2 2 ,−

p

2 2 , 0

E .

Unit Binormal Vector

Consider a space curve~R(t) in 3Dspace.

Suppose~T(t) and~N(t) are its unit tangent and unit normal vectors at any t.

Define~B(t) by

~B(t)₌~T(t)_×~N(t).

Since~T(t) and~N(t) are perpendicular unit vectors, then

k~B(t)_{k = k}~T(t)_×~N(t)_{k = k}~T(t)_kk~N(t)_{k =}1.

We call~B(t) theunit binormal vectorto the space curve~R(t)

Unit Binormal

Example

Consider the space curve~R(t)_{= 〈}2 cost, 2 sint,t〉. Compute~B¡₃π

4

¢ .

Solution.From the previous example, we have

~T µ₃π

4 ¶ = * − p 2 p

5,− p 2 p 5, 1 p 5 + and ~ N µ₃π

4 ¶

= *p

2 2 ,−

p 2 2 , 0

+ .

Thus,~B¡3π

4

¢ =

Dp

2 2p5,

p

2 2p5,

2

p

5

TNB Frame

TNB Frame

Other Formulas for TNB

The vectors~T(t),~N(t), and~B(t) follow the right-hand rule of the thumb. And it can be shown that

~B(t)₌~T(t)_×~N(t)

~

N(t)=~B(t)×~T(t)

~T(t)₌~N(t)_×~B(t) Also,~B(t)₌ ~R

0_(t)×~_R00_(t) °

Another Formula for

~

B

(

t

)

Unit Binormal Vector

If~R(t) is a space curve in 3D space, then the unit binormal vector is also given by

~B(t)= ~R

0_(t)×~_R00_(t) °

°~R0(t)×~R00(t) ° ° .

Proof.Recall that

~B(t) ₌ ~T(t)_×N~(t)

= ~R 0_(t) ° °~R0(t)

° °

× ~T 0_(t) ° °~T0(t)

° °

= ~R

0_(t)×~_T0_(t) °

°~R0(t)×~T0(t) ° °

Unit Binormal Vector

Meanwhile,

~T(t)= ~R 0_(t) ° °~R0(t)

° °

=⇒ ~T0(t)= ~R 00_(t) ° °~R0(t)

° °

−

~R0(t)h_dtd °°~R0(t) ° ° i

° °~R0(t)

° °

2

Thus,

~R0(t)_×~T0(t) ₌ ~R

0_(t)×~_R00_(t) °

°~R0(t) ° °

°

°~R0(t)×~T0(t) °

° =

°

°~R0(t)×~R00(t) ° ° °

°~R0(t) ° °

Hence,

~B(t)₌ ~R

0_(t)×~_R00_(t) °

Moving Trihedral

Example

Let~R(t)₌­ t,1₂t2_,1

3t3

®

. Compute the vectors~T,~N, and~Batt₌1.

Solution.We have

~R0(t)= 〈1,t,t2〉

~R0_{(1)= 〈1, 1, 1〉} °

°~R0(1) ° °=

p 3

~R00(t)= 〈0, 1, 2t〉

~R00_{(1)= 〈0, 1, 2〉} °

°~R00(1) ° °=

p 5

Therefore,~T(1)₌p1

3〈1, 1, 1〉 =

D 1 p 3, 1 p 3, 1 p 3 E .

And,~B(1)₌°~R0(1)×~R00(1) °~R0(1)×~R00(1) ° °=

〈1,1,1〉×〈0,1,2〉 k〈1,1,1〉×〈0,1,2〉k= 〈

1,−2,1〉 k〈1,−2,1〉k=

D 1 p 6, −2 p 6, 1 p 6 E .

Finally,~N(1)=~B(1)×~T(1)=Dp1

6, −2 p 6, 1 p 6 E ×Dp1

3, 1 p 3, 1 p 3 E =D−p1

2, 0, 1

p

2

Exercises

1 _{Find the moving trihedral att=1 to}~_R(t)=

¿_t3

3,t

2_{, 2t}

À .

2 Find the equation of the rectifying plane att₌0 to the curve

 

 x=et y=2e−t z=2t

.

3 _{Find the equation of the normal plane to the curve given by}