Solutions
Block 2 : Vector Calculus
t 3 : Svace Curve and t h e Bi-Normal Vector (Optional)
2.3.1(L)
The main aim of t h i s e x e r c i s e i s t o g i v e us a b e t t e r f e e l i n g about
v e c t o r c a l c u l u s a p p l i e d t o curves and, a t t h e same time, by extendi n g our d i s c u s s i o n t o space curves, t o g e t a b e t t e r understanding
of how t h e v e c t o r concepts w e have been d i s c u s s i n g transcend t h e
2-dimensional world.
To begin w i t h , t h e equation
r e p r e s e n t s t h e locus of p o i n t s i n space described by t h e t h r e e
s c a l a r equations
I f we t h i n k (simply f o r t h e convenience of having a n i c e p h y s i c a l
i n t e r p r e t a t i o n ) of t a s denoting time, it i s n o t hard t o see t h a t ,
i n most c a s e s , t h e domain of 3 i s an i n t e r v a l . That i s , we a r e
u s u a l l y i n t e r e s t e d i n s t u d y i n g t h e path of a p a r t i c l e over some
continuous time i n t e r v a l , f o r example, from t = tl t o t = t2.
A very n a t u r a l q u e s t i o n t o ask i s whether t h e locus of p o i n t s
( x , y , z ) forms a continuous curve. Perhaps we would f e e l i n t u i t i v e l y
t h a t i f x ( t ), y ( t ), and z ( t ) were a l l continuous f u n c t i o n s of
t , then s o i s R ( t ) , and t h i s i s p r e c i s e l y what p a r t a. i s asking
us t o do. Namely,
a. We must f i r s t d e c i d e what it means f o r % t o be a continuous f u n c t i o n
of t. I n terms of our previous s t r a t e g y , i t seems f i t t i n g t h a t we
t a k e a s a f i r s t approximation t h e analogous d e f i n i t i o n i n t h e
s c a l a r c a s e . That i s ,
3 R i s s a i d t o be continuous a t t l s [ a , b ]
This, i n t u r n , s a y s t h a t given
E
+
-f
i f and only i f l i m R ( t ) = R ( t l ) .
t+tl
> 0 we can f i n d 6 > 0 such t h a t
Solutions
Block 2: Vector Calculus
Unit 3: Space Curves and t h e Bi-Normal Vector (Optional)
-
2.3.1 continued
Our n e x t s t e p i s t o v e r i f y t h a t t h i s formal d e f i n i t i o n agrees with
our i n t u i t i o n . To t h i s end we observe t h a t our d e f i n i t i o n says
t h a t 5 i s d e f i n e d a t each t i n t h e i n t e r v a l and t h a t t h e r e a r e no
3
"gapsw i n 3 s i n c e , f o r any tl i n t h e i n t e r v a l , R ( t ) can be made
s(tl)
a r b i t r a r i l y n e a r l y equal t o
simply by choosing t s u f f i c i e n t l y
3
c l o s e t o tl. Since1~lrneasurest h e d i s t a n c e from t h e o r i g i n t o a
p o i n t on t h e curve, t h e f a c t t h a t t h e r e can be no gaps i n
insures
t h a t t h e r e can be no gaps i n t h e curve.
Thus, w e may t a k e (3) a s a p r a c t i c a l , y e t p r e c i s e , working d e f i n i t i o n of c o n t i n u i t y . With t h i s i n mind,
Therefore,
- %(tl)
I r Ix(t) - x(tl) I
- z (tl)1 .
~d(t)
+ ly(t)
- y(tl) 1
+
Iz(t)
(4)
Since x, y , and z a r e a l l continuous s c a l a r f u n c t i o n s o f t , we may
6 3 such t h a t f o r a given E > 0 ,
f i n d 61,
0 < It
-
tll < 6 l
+
lx(t)
-
x(tl)
1
E
< 7
1
Solutions
Block 3 : Vector Calculus
Unit 3: Space Curves and t h e Bi-Normal Vector (Optional)
2.3.1
continued
Hence, i f w e l e t 6 = mini61,62,.631,
(5) t e l l s us t h a t
Combining (6) with ( 4 ) y i e l d s
What we have t h u s shown i n p a r t a. i s t h a t w e can d e f i n e c o n t i n u i t y
f o r a space curve i n such a way t h a t t h e d e f i n i t i o n i s independent
of our c o o r d i n a t e system.
But, i f w e a r e d e a l i n g w i t h C a r t e s i a n coor-
d i n a t e s , we have t h e "luxuryn of knowing t h a t w e can t e s t f o r c o n t i n u i t y simply by t e s t i n g each of t h e s c a l a r components f o r c o n t i n u i t y
b. (and t h i s i s something we have a l r e a d y l e a r n e d t o d o ) .
+ +
(1) Given a continuous space curve, s a y , R = R ( t ) , we know t h a t by
our r u l e s f o r d i f f e r e n t i a t i o n , i n C a r t e s i a n form, w e have
then
O u r c laim i s t h a t t h e v e c t o r d e f i n e d i n (8) i s t a n g e n t t o t h e space
curve.
Perhaps t h e e a s i e s t and t h e s a f e s t way t o s e e t h i s i s with-
5,
and $ components.
Quite i n g e n e r a l , i f %
o u t r e f e r e n c e t o I,,
+
+
and R + AR a r e two v e c t o r s drawn from t h e o r i g i n t o p o i n t s on t h e
curve, it i s easy t o s e e t h a t A f t i s t h e v e c t o r t h a t o r i g i n a t e s a t
one of t h e p o i n t s and t e r m i n a t e s a t t h e o t h e r . Thus,
Solutions
Block 2: Vector Calculus
Unit 3: Space Curves and t h e Bi-Normal Vector (Optional)
2.3.1 continued
I f we now hold one of t h e p o i n t s f i x e d , i t i S easy t o v i s u a l i z e
-+
t h a t a s A$ approaches 0 , o u r chord approaches t h e d i r e c t i o n of " t h e n
t a n g e n t l i n e t o o u r curve.
We have p u t " t h e " i n q u o t a t i o n marks t o
h i g h l i g h t a new degree o f d i f f i c u l t y t h a t i s introduced when we d e a l
w i t h space curves r a t h e r than plane curves.
I n a genuine space
curve, t h e curve does n o t always s t a y i n t h e same plane ( f o r i f i t
d i d i t would be a plane c u r v e ) . Thus, from an i n t u i t i v e p o i n t of
view, t h e curve l i e s i n some s o r t of i n s t a n t a n e o u s plane a t any
-
given t i m e , and w e t h i n k of t h e tangent l i n e a s l y i n g i n t h i s plane.
-
That is, i f we f i x t h e p o i n t Po on our space curve C, and we.look
a t t h e v e c t o r POP where P i s any o t h e r p o i n t on C, then i f our
curve i s smooth, w e s e e t h a t a s P approaches Po, t h e v e c t o r POP ap-
-
-
proaches a f i x e d d i r e c t i o n , and it i s t h i s v e c t o r t h a t we c a l l t h e
t a n g e n t v e c t o r t o t h e curve.
What r e a l l y h i g h l i g h t s t h i s d i s c u s s i o n may be seen from t h e follow+
i n g 2-dimensional case.
Suppose T i s t a n g e n t t o t h e curve C below.
I f we now l e t S denote any p l a n e t h a t i n t e r s e c t C, then by construct i o n any l i n e i n S which passes through P0 i s t a n g e n t t o t h e curve
C i n t h e s e n s e t h a t i t "touches" C a t Po.
I n t h i s r e s p e c t , then,
t h e r e a r e i n f i n i t e l y many l i n e s t h a t a r e "tangent" t o C a t Po.
( S t i l l another way of looking a t t h i s i s t h a t t o t a l k about a tang e n t p l a n e a t a p o i n t , we should be r e f e r r i n g t o a s u r f a c e n o t a
curve. )
dR'
The p o i n t is t h a t aZ p i c k s o u t t h e most " n a t u r a l " d i r e c t i o n f o r us
That i s , i f given no o t h e r
t o c a l l a t a n g e n t t o t h e curve a t Po.
i n s t r u c t i o n s , we could look a t a small p i e c e which was " s u f f i c i e n t l y
s m a l l H and w e could assume t h a t it was i n a plane. The p o i n t i s t h a t
dS
dR'
I n o t h e r words, t h e d i r e c t i o n of -6.f
-is
a v e c t o r i n t h i s plane.
dt
i s t h e one we would p i c k i n t u i t i v e l y a s t h e d i r e c t i o n of t h e curve
a t the point.
The major q u e s t i o n t h a t remains i s t h a t of c o q r e l a t i n g the s e n s e
dR
of t h e curve C with t h e s e n s e of t h e v e c t o r E. The p o i n t i s t h a t
j u s t a s i n t h e p l a n a r c a s e , i f we l e t our parameter be a r c l e n g t h
dit
( s ) , then s w i l l be a u n i t v e c t o r and i t s sense w i l l be t h e same
a s t h a t of t h e curve, no m a t t e r how we e l e c t t o choose t h e s e n s e of
t h e curve.
Solutions
Block 2: Vector Calculus
Unit 3: Space Curves and t h e Bi-Normal Vector (Optional)
2.3.1 continued
I f t h e r e i s no reason f o r otherwise choosing a s e n s e f o r our curve,
+
aft by i t s magnitude. I n any event,
we simply d e f i n e T by d i v i d i n g
+
i f we l e t R ( t ) be expressed h a C a r t e s i a n c o o r d i n a t e s , we have:
whereupon our r u l e s f o r d i f f e r e n t i a t i n g v e c t o r s y i e l d s
+
dRf be c a l l e d t h e v e l o c i t y
and we can now s e e why it i s n a t u r a l t h a t a
+
v e c t o r ( v ) . Namely, it i s t a n g e n t t o t h e curve and t h e components
dx d
d z which a r e t h e components of t h e speed of
of i t a r e
and
the particle.
+
dR
With+v now d e f i n e d t o be
w e simply d e f i n e t h e a c c e l e r a t i o n t o
dv
be
=, df,
=,
4
n.
(2)
With r e s p e c t t o o u r example,
+
R = cos 3 t
f +
+
v = -3sin 3 t
sin 3t
Z +
-j +
3 cos 3 t
+
-+
a = - 9 ~ 03 t~ 1
-
therefore,
= \/(-3sin
gsin 3 t
t j:
f +Z
j
3t)
+
(3cos 3 t )
+
2
1
or
121
=
Jnr
.
(3) From ( 1 2 ) , we s e e t h a t t h e p a r t i c l e moves w i t h c o n s t a n t speed.
From our e a r l i e r d i s c u s s i o n s , when t h e speed i s c o n s t a n t , t h e
a c c e l e r a t i o n v e c t o r must be a t r i g h t angles t o t h e v e l o c i t y v e c t o r .
A s a check, ( 1 0 ) and (11) show t h a t
Solutions
i
Block 2: Vector Calculus
Unit 3: Space Curves and t h e ~ i - N o r m a l Vector (Optional)
2.3.1
c.
continued
dR/l
Since dii i s a tangent v e c t o r ,
d H ~w i l l be a u n i t t a n g e n t v e c t o r .
?IF
aT
I n our p r e s e n t example, equation ( 1 0 ) shows t h a t
d"R
=(=
V ) = -3sin
-+
3t
-t
1
+ 3cost
t
3
+,
while (12) shows t h a t
Thus,
and s i n c e
d.
Since
df/dHl = =
-+
T =
3
-+
aZ
1$12
k T,
-+
= 1, T
==
d$
0 [c.f.,
E x e r c i s e 2.2.3(L)].
Thus,
-+
8.
i s a u n i t v e c t o r perpendicular t o T.
Then, i f we m u l t i p l y d? by t h e s c a l a r
*We
s a y "a"
r a t h e r than "the"
Call t h i s vector
1 gv/1$
,
s i n c e i f t h e curve has a s e n s e of i t s
-+
own a n d T d e n o t e s t h e u n i t t a n g e n t v e c t o r w i t h t h i s s e n s e ,
&/$
dt
(i.e.,
+
-+
=
2 %
d e p e n d i n g on w h e t h e r
whether
- is
ds
d t
2
and
dR
have
dS
then
t h e same s e n s e
positive or negative).
* * U n l e s s t h e s e n s e i s p r e v i o u s l y i m p o s e d o n t h e c u r v e C , we c h o o s e
i t s s e n s e s o t h a t t h e p o s i t i v e s i g n i n (14) i s u s e d t o d e t e r m i n e T.
*
Solutions
Block 2: Vector Calculus
Unit 3: Space Curves and t h e Bi-Normal Vector ( O p t i o n a l )
2.3.1 continued
I f w e def in.
1g1
t o be c u r v a t u r e , K; w e have from (15)
I n o u r p r e s e n t example, from ( 1 4 ) , w e have
=
Therefore,
Therefore, from (17)
1%
ad*;; =
=
9
= K.
9
+
= m(-cos 3 t i
-
sin 3t
Therefore,
9
m(-cos
3t
;-
s i n 3 t j)
.
From (161,
w e have
,
d3
z)
While w e have now s o l v e d t h i s p a r t of t h e e x e r c i s e , it might s t i l l
be b e n e f i c i a l t o see what t h e s i g n i f i c a n c e of 8 r e a l l y is. Notice
*
t h a t when w e r e s t r i c t e d o u r s t u d y t o p l a n e c u r v e s , once
was d e t e r +
mined, N was determined up t o sense. That i s , s i n c e
and 8 had t o
Solutions
Block 2: Vector Calculus
Unit 3: Space C.urves and t h e Bi-Normal Vector (Optional)
2.3.1 continued
+
be i n t h e same plane, N was determined up t o s e n s e once we knew
t h a t it had t o be perpendicular t o 5. I n 3-space, however, the
problem i s a b i t more complex. The l o c u s of a l l v e c t o r s $ such
-t
that 3
N = 0 is now a plane.
Thus, we can f i n d an i n £i n i t y of
+
+
v e c t o r s , N, each w i t h d i f f e r e n t d i r e c t i o n s such t h a t $! * N = 0.
I n t u i t i v e l y , i f w e t h i n k of t h e p l a n e which c o n t a i n s C i n a small
neighborhood of Po, we may t h i n k of f a s l y i n g i n t h i s plane, and
i n terms of t h i s plane ( i . e . , from an i n s t a n t a n e o y s p o i n t of view,
d T should a l s o l i e
we a r e back t o t h e p l a n a r c a s e ) , we expect t h a t -ds
i n t h i s same plane.
( I n o t h e r words, f o r small increments, we
f e e l t h a t 9 and ? + ~3 ake i n t h e same plane.)
The p o i n t is t h a t
3
and
8
( a s defined i n t h i s e x e r c i s e ) determine
a plane t o C a t Po ( a s would % and any o t h e r v e c t o r ) , and t h i s
p a r t i c u l a r plane i s c a l l e d t h e o s c u l a t i n g plane t o C a t Po. Phys i c a l l y , it means t h a t a t any given i n s t a n t w e can assume t h a t t h e
p a r t i c l e i s t r a v e l l i n g i n i t s o s c u l a t i n g plane r a t h e r than along
t h e space curve.
I n t h i s s e n s e , t h e o s c u l a t i n g plane i s i n
3-dimensions what t h e o s c u l a t i n g c i r c l e was i n 2-dimensions.
e.
With 3 and 8 a s d e f i n e d i n t h e previous p a r t s of t h i s e x e r c i s e , w e
may now view a plane moving along t h e curve from p o i n t t o p o i n t .
This p l a n e , a s we saw i n t h e previous p a r t of t h i s e x e r c i s e , is t h e
+
I f we now d e f i n e 8 by g = 3 X 8 , w e may view B
o s c u l a t i n g plane.
a s a u n i t t a n g e n t v e c t o r ( s i n c e it i s t h e c r o s s product of two u n i t
tangent v e c t o r s ) which i s perpendicular t o the o s c u l a t i n g plane.
a
u n i t v e c t r , we know t h a t it can only change i n d i r e c t i o l
Since 3 is
d%
must measure t h e change i n d i r e c t i o n of 3
n o t magnitude, hence
a s we move along t h e curve. Since 8 i s always normal t o t h e oscul a t i n g p l a n e , t h e change i n d i r e c t i o n of % a l s o measures t h e change
i q d i r e c t i o n of t h e o s c u l a t i n g plane. That i s , t h e magnitude of
dB measures how f a s t t h e curve i s being "twisted" o u t of t h e oscul a t i n g plane a t a given p o i n t . *
dB'
We a l s o know t h a t i n c e 5 has c o n s t a n t magnitude, a;; i s perpendicudZ
l a r t o 8. Hence
i s a v e c t o r which l i e s i n t h e o s c u l a t i n g plane
( s i n c e t h i s plane i s t h e locus of a l l v e c t o r s perpendicular t o 3 ) .
a;;
-f
+
* I n t h e s p e c i a l c a s e o f a p l a n e c u r v e , T and N a l w a y s l i e ig t h e
+
-P
aB
same p l a n e , s a y , t h e x y - p l a n e .
I n t h i s c a s e B m k whence d s = 0 .
I n o t h e r words t h e r e i s no t o r s i o n f o r a p l a n e c u r v e .
-
Solutions Block 2: Vector Calculus Unit 3: Space Curves and the Bi-Normal Vector (Optional) 2.3.1 continued More analytically, +
+
+
Since B = T x N,
d?
+
From d., a
;5.. = KN.
Therefore, di5 =
( 8x 8) + (f
d3 , and since N
+ x N
+ = 6,
x =)
From (19), dii is perpendicular to
?
(it is also perpendicular to
di$ but we do not need this for our purpose) and we already know that
di5 is also perpendicular to -b
B. .
Since dB is perpendicular to both -bB and
+ where
Hence, dii = TN
I .r 1
=
1g1
since
+ 1g1 and is called torsion
is, r =
of the curve.
Then, ih
5
-b
is parallel to N.
is a unit vector.
That
(a measure of the "twist")
Solutions Block 2: Vector Calculus Unit 3: Space Curves and the Bi-Normal Vector (Optional) 2.3.2
continued +
Since dT =
KS, (2) yields
[Note that (3) has precisely the same form as in the 2-dimensional -b
-+
case even though T and N are now space vectors. Notice that (3) agrees perfectly with the 2-dimensional case if we think in terms of the particle being in its osculating plane.] From (1) and (3), we obtain and since
3
x
-+
5
-+
= 0 while
x
8
3
-+
Therefore, Iv x a ( = Irl
positive)
.
'5
$1
=
st we have
161
=
~
1
~
(since
IB(1-+
~
= 1 and r is
Not only does (4) reestablish the validity of the recipe Solutions
Block 2: Vector Calculus
Unit 3: Space Curves and t h e Bi-Normal Vector (Optional)
2.3.2
continued
?
even f o r 3-dimensions b u t it a l s o t e l l s us t h a t
direction as
3
and
with
8
2
8
Namely, t h i s r e s u l t i s c o n s i s t e n t
both being i n t h e o s c u l a t i n g plane.
b. We now use t h e r e s u l t of a. a s follows.
+
From t h e equation f o r R, w e can f i n d 3 and
product of
has t h e same
which r e a f f i r m s why w e c a l l t h e plane determined by
the osculating
and
x
2
+
and a we g e t a v e c t o r
5 and 3
t h e plane determined by
+
-f
1
+
4t3
+
By forming t h e c r o s s
parallel t o
8,
hence normal t o
(the osculating plane).
w r i t e down t h e equation of t h e plane.
Therefore, v =
2.
We then
More s p e c i f i c a l l y ,
3t2%
Therefore, a t t h e p o i n t corresponding t o t = 1, we have
-b
-b
3
Therefore, v x a =
Therefore, 1 2 1
-
= 121
1
4
3
0
4
6
-
65
+
4%. +
6 3 + 4 2 o r , equivalently, 6 1
-
33
+
2$ i s normal
t o t h e r e q u i r e d p l a n e , while (1,2,1) i s a p o i n t i n t h e p l a n e - ( t h i s
-t
i s t h e meaning of
= 1 + 23 +
z). Therefore, t h e equation of t h e plane i s Solutions
Block 2: Vector C a l c u l u s
Unit 3: Space Curves and t h e Bi-Normal Vector (Optional)
2.3.2
continued
I;(
c. From (61,
+
=
16
+ 9 = m.
Therefore
1;13
= 263'2
= 26m,
w h i l e from ( 7 )
Therefore., from (5)
[Note t h a t from a p h y s i c a l p o i n t of view t h e meaning of p a r t s b.
and c. i s t h a t , a t t h e given i n s t a n t t = 1, w e may assume t h a t
t h e p a r t i c l e i s a t t h e p o i n t ( 1 , 2 , 1 ) i n t h e p l a n e 6x
3y + 22 = 2,
-
moving a l o n g t h e circle i n t h a t p l a n e ( t h e o s c u l a t i n g c i r c l e ) whose
radius is p = 1 = 1 3 m
-K
--I
d. Most of t h e i n f o r m a t i o n w e need f o r this problem i s a l r e a d y known.
Namely, t o a l l i n t e n t s and purposes, w e know $, s i n c e i t i s simply
.+
v
,
W e e s s e n t i a l l y know
1G1
and w e have found
K
8,
s i n c e we have a l r e a d y shown t h a t
2.
and
+.
Once 8 and
a r e known, w e f i n d 3 from t h e r e l a t i o n 8 = 3 x
+
+
( I n t h e same way t h a t $ and 8 behave s t r u c t u r a l l y l i k e i and j a t
+ +
+
t
any p o i n t on a p l a n e curve; T , N , and B behave l i k e i , 3 , and
at
+ +
+
any p o i n t on a space curve. Consequently, w e may view B = T x N
i n t h e e q u i v a l e n t forms, 3 = 8 x 8 and 8 = 8 x 3.) P i c t o r i a l l y ,
-~
- -
-
+
* N o t i c e h e r e t h a t we must u s e 121
+
m u l i i p l e such a s 61
+ +
r e q u i r e s Iv x a ] .
-
+
3j
+
+
-
+
6j
+
+
4k not another s c a l a r
2k s i n c e the r e c i p e f o r
K
specifically
-
Solutions
Block 2: Vector C a l c u l u s
U n i t 3: Space Curves and t h e Bi-Normal Vector ( O p t i o n a l )
2.3.2
continued
More s p e c i f i c a l l y , a t t = 1, w e saw e a r l i e r i n t h i s e x e r c i s e t h a t
-+
v =
f+
+
a =
43
+
3% t h e r e f o r e ,
43
+
6%
+
+
v x a = 121
Therefore,
-
K
=
65
-I.
+
If1
=
42 t h e r e f o r e
x aI
- - . -
1;13
15
x
= 14.
14
=-
2
6
~
~
(8) y i e l d s
Hence
Therefore,
6
=
7
+
-
61
-
7
Moreover ,
+
+
-+
Therefore N = B x T =
8
=
x
%
33
1
implies t h a t
- ~ 8 ,(1) becomes
-+
d 3 = KN
Since and d s =
ds
ds
+
2%
.
Solutions
B l o c k 2: Vector Calculus
Unit 3: Space Curves and the Bi-Normal Vector (Optional)
2.3.3 continued
From t h e cyclic orientation of
Hence, (2) becomes
2.3.4
a.
W
e have already seen that
and
d33
d
-+ From (2) --T= =(KN)
ds Therefore,
d3"R
d3
= u z
+ ds 3.
From our g
~ e v i o u sdiscussion,
&
-*
= -KT
- TB.
-+
Putting ( 4 ) i n t o (3) y i e l d s
9, 8,
and
8,
Solutions
Block 2: Vector Calculus
Unit 3: Space Curves and t h e Bi-Normal Vector (Optional)
2.3.4
continued
Theref o r e
d3"R
2 + a
dK; ; ~
-KT~.
, 7=-K
T
-+
ds
b.
The i n t e r e s t i n g p o i n t h e r e i s t h a t
i - t
z.
3, 8 ,
and
5
behave s t r u c t u r a l l y
j u s t t h e same a s 1, 1, and
Thus, i f two v e c t o r s a r e w r i t t e n i n
t e r m s of 9, 8, and 3 components we may compute t h e i r crossproduct
j u s t a s w e d i d i n t h e 1.
motivate one t o compute
3, and
case.
This i s perhaps what might
Namely, s i n c e t h e f i r s t v e c t o r has only a T component and t h e second
+
only an N component, we s e e t h a t t h e determinant t h a t y i e l d s t h e
t r i p l e s c a l a r product i s given by':
and expanding along t h e t o p row y i e l d s
In o t h e r words,
Finally, since
ds
8
8
= 1, we may use (2) t o commute
ds'
Putting ( 7 ) i n t o (6) y i e l d s the desired r e s u l t :
Solutions
Block 2: Vector Calculus
' U n i t 3: Space Curves and t h e ~ i - N o r m a l Vector (Optional)
continued
2.3.4
While w e make no a t t e m p t t o prove it h e r e , it i s an i n t e r e s t i n g f a c t
t h a t knowing K and r a t each p o i n t on t h e curve i s enough t o d e t e r mine t h e shape of curve uniquely up t o p o s i t i o n i n space. This i s
analogous i n t h e study of plane curves t o saying t h a t knowing t h e
s l o p e of t h e curve determines t h e shape of t h e curve up t o p o s i t i o n .
( I n f a c t , t h i s i s p r e c i s e l y what adding an a r b i t r a r y c o n s t a r t t o
t h e i n d e f i n i t e i n t e g r a l i s a l l about.)
c.
In C a r t e s i a n c o o r d i n a t e s
-t
i
'
R = xi
+
t
Y-J
+
+
zk.
Therefore,
Therefore
K~
=
While from ( 6 ) and ( 9 )
Solutions
Block 2: Vector C a l c u l u s
Unit 3: Space Curves and t h e Bi-Normal Vector ( O p t i o n a l )
2.3.4
continued
X'
-K
2
=
XI'
I'
'
y'
Y"
z'
Y'"
z"'
Z"
where t h e d i f f e r e n t i a t i o n i s w i t h r e s p e c t
t o s.
Theref o r e ,
a.
d2"R
The mimicking p r o c e d u r e w i l l have us r e p l a c e d% 7
, and ;d38
;;;5 by
ds
,,d% d2'i and a3R o r V-+ , a+, and d g ( I n o t h e r words d a w i l l p l a y
z,2;
=.
a prominent r o l e i n 3-dimensions when we a r e i n t e r e s t e d i n t o r s i o n . )
From e a r l i e r work, w e have
From ( 2 ) w e can compute d 2 (which w e do f o r p r a c t i c e , b u t w e w i l l
o n l y need the c o e f f i c i e n t of 3 s i n c e
a;:
-
-b
where d t = a 1T
+
+
a2N
+
-b
a3B )
and expanding t h i s d e t e r m i n a n t a l o n g t h e f i r s t row, w e o b t a i n
Solutions
Block 2: Vector Calculus
Unit 3: Space Curves and t h e Bi-Normal Vector (Optional)
2.3.5
continued
I n o t h e r words, n e i t h e r al nor a2 appear i n t h e value of t h e
determinant.
2
A t any r a t e , w e o b t a i n from (2) (and r e c a l l i n g t h a t d s
and
8
ds
a r e a l l functions of t )
+
d6
and a
;; =
Then s i n c e d3 = KN
-
Therefore, dt -Kt
-
r t , w e have:
ds d s ( A s a check of our e a r l i e r r e s u l t s i f w e t h i n k of t a s being an
a r b i t r a r y parameter i n ( I ) , ( 2 ) , and (3) and then l e t t = s, so t h a t
ds d2s - d3s = 0 , we obtain:
m - l t z - 2
Solutions
Block 2: Vector C a l c u l u s
U n i t 3: Space Curves and t h e Bi-Normal Vector ( O p t i o n a l )
2.3.5
continued
and t h e s e check w i t h e q u a t i o n s (1), (2)
, and
( 5 ) o f E x e r c i s e 2.3.4
Notice a l s o , i n p a s s i n g , t h a t t h e d e r i v a t i o n of e q u a t i o n (3) was
p u r e l y mathematical and r e q u i r e d n o i n s i g h t t o t h e p h y s i c s of s p a c e
dK
motion.
To be s u r e , i t might b e n i c e t o have a f e e l i n g f o r
a s t h e r a t e of change o f c u r v a t u r e e t c . , b u t t h i s i s completely
unnecessary f o r t h e problem w e a r e t r y i n g t o s o l v e .
Returning t o t h e problem a t hand w e have:
where from ( 3 )
3
d s , al -- 2
Therefore, v
(;;
X $)=-K
K
2 ds
-K
2
T,
and
2 T ( =d) s 4
(Again, a s a p a r t i a l check of
=
2
( 4 ) , i f we l e t t = s we o b t a i n
which checks w i t h o u r e a r l i e r r e s u l t . )
F i n a l l y , i f we invoke t h e p r e v i o u s l y proven r e s u l t t h a t
.
)
Solutions
Block 2: Vector Calculus
Unit 3: Space Curves and t h e Bi-Normal Vector (Optional)
2.3.5
continued
(g)4 i s t h e same a s
..
and i f we r e c a l l t h a t
I v 1 *,
then equation
( 4 ) becomes
Therefore,
-
T =
b, W e had e s t a b l i s h e d t h a t (Problem 2.3.5
Zi
= t
+ v =
-t
1
+
1
+
+
4t
+
a =
2t2j
f +
+
b.)
t3%
3t2z 4 3 + 6 t z
and w e now add t o t h e s e equations: *
In p
a r t i c u l a r a t t = 1; v =
+
a = 4 3
+
6z and
$ = 6;
-?
1
+
43
+
3;,
t h e r e f o r e , a t t = 1 w e have
*In g e n e r a l i t i s p o s s i b l e t h a t
ds
+
and Ivl d i f f e r i n s i g n , b u t i n
our c a s e , o n l y even powers o f &-occur
dt
non-negative.
s o everything i s always
Solutions
Block 2: Vector C a l c u l u s
U n i t 3: Space Curves and t h e Bi-Normal Vector ( O p t i o n a l )
2.3.5
continued
Therefore,
;1
x
21 =
a44
+
36
+
16 = 14.
P u t t i n g t h e s e r e s u l t s i n t o (5) y i e l d s
Therefore
IT I
156 .
= 7
Again, i t i s n o t o u r purpose t o have you l e a r n a s h o r t c o u r s e i n
D i f f e r e n t i a l Geometry o r t h e l i k e , b u t r a t h e r t h a t you g e t t h e
f e e l i n g o f what c u r v a t u r e and t o r s i o n mean p h y s i c a l l y .
The connect i o n between 2-dimensional and 3-dimensional s p a c e curve is t h a t
-*
-+
t h e v e c t o r s T and N a s t h e y e x i s t i n t h e 2-dimensional c a s e form
t h e o s c u l a t i n g p l a n e a t any i n s t a n t i n t h e 3-dimensional c a s e .
13 c l o s i n g , i t s h o u l d b e p o i n t e d o u t t h a t t h e q u a n t i t i e s d 3 diS
dN
o b v i o u s l y p l a y i m p o r t a n t r o l e s i n t h e s t u d y of s p a c e curves.
have a l r e a d y shown t h a t :
s,s,and
+
dN =
az
-ril
-
Kd.
These t h r e e r e s u l t s a r e known a s F r e n e t ' s formulas.
We
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Calculus Revisited: Multivariable Calculus
Prof. Herbert Gross
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