Solutions
Block 2: Vector C a l c u l u s
t 2: T a n a e n t i a l and Normal V e c t o r s "
.
2.2.1(L)
-\.-
Y
L
-
8 .
The major aim of t h i s p a r t o f t h e e x e r c i s e i s t o h e l p remove a b i t
of mystery a s w e l l a s a computational s o r e s p o t t h a t many s t u d e n t s
e n c o u n t e r i n t h e formula
+
T =
dii
E.
What u s u a l l y happens i s t h a t
'n
is
not
expressed d i r e c t l y a s a
f u n c t i o n of s ( a r c l e n g t h ) . I n many c a s e s o f a k i n e m a t i c s n a t u r e ,
+
f o r example, R w i l l be e x p r e s s e d i n t e r m s of t i m e ( t ) .
For t h e s a k e of argument, l e t us suppose t h a t
'n
i s expressed a s a
f u n c t i o n of some s c a l a r q , n o t n e c e s s a r i l y e i t h e r a r c l e n g t h o r time.
Then we have a l r e a d y s e e n E x e r c i s e 2.1.4
is equal t o
c u r v e and i t s magnitude
!f
I
dii
Zi
is tangent t o t h e
6
i s e x p r e s s e d a s a f u n c t i o n of q , w e may
q u i t e c o n v e n i e n t l y simply by l e t t i n g
I n o t h e r words, t h e n , i f
find
that
Equation ( 2 ) s u p p l i e s us w i t h an e x c e l l e n t h i n t a s t o why e q u a t i o n
(1) i s c o r r e c t .
Namely, i n t h e s p e c i a l c a s e t h a t w e choose q t o
ds
= 1, whereupon e q u a t i o n ( 2 ) becomes equae q u a l s , we have t h a t
q
t i o n (1) by t -h e c h a i n r u l e (proven i n E x e- r c i s e 2.1.3).
That i s ,
-
d'n/dp
-
dS/d
I d$/dq l - 2a-
.,
- d'n
-
az-
Yet, t h e r e i s an even b e t t e r p h i l o s o p h i c a l r e a s o n f o r d e f i n i n g $
by (1) r a t h e r t h a n by ( 2 ) , even though, i n a g i v e n c a s e (2) may
be more d i r e c t t o compute t h a n i s (1). The thought i s t h a t a
u n i t t a n g e n t v e c t o r t o a c u r v e i s determined by t h e shape of t h e
c u r v e i t s e l f , n o t by e i t h e r t h e p a r t i c u l a r c o o r d i n a t e system i n
Solutions
Block 2: Vector C a l c u l u s
K n i t 2: T a n g e n t i a l and Normal Vectors
2.2.1 ( L ) c o n t i n u e d
which w e elect t o w r i t e t h e e q u a t i o n of t h e c u r v e n o r t h e v a r i a b l e used a s t h e parameter.
For example, t h e r e a r e many d i f f e r e n t p a r a m e t e r s t h a t might have been chosen t o r e p r e s e n t t h e
v e c t o r e q u a t i o n f o r t h e curve.
That i s , i n t e r m s o f o u r p r e v i o u s
n o t a t i o n , q c o u l d have been chosen i n many ways, among which a r e
However, a r c l e n g t h ( s ) i s an i n v a r i a n t of t h e
t i m e and a r c l e n g t h .
I n o t h e r words, i f w e measure t h e l e n g t h o f a given
curve.
c u r v e between two p o i n t s P and Q on t h e c u r v e , t h e l e n g t h depends
o n l y on t h e c u r v s and t h e two p o i n t s , n o t on any p a r t i c u l a r
c o o r d i n a t e system:
( a l t h o u g h t h e d e g r e e o f complexity of t h e
a c t u a l computations i n v o l v e d may w e l l depend on t h e c o o r d i n a t e
system).
dfi. g i v e s u s a d e f i n i I n summary, t h e n , t h e d e f i n i t i o n t h a t -.
T =
t i o n o f t h e u n i t t a n g e n t v e c t o r which i s independent of any part i c u l a r c o o r d i n a t e system under c o n s i d e r a t i o n .
In t h i s exercise,
however, i t i s o u r aim t o emphasize t h a t t h e a c t u a l computations
a r e b e s t done i n terms o f t h e p a r t i c u l a r v a r i a b l e w i t h r e s p e c t t o
which
6
is expressed.
With t h i s d i s c u s s i o n i n mind w e have
Therefore,
From ( 3 )
Therefore,
= t
2
+ 1.
Solutions
Block 2: Vector C a l c u l u s
Unit 2: T a n g e n t i a l and Normal V e c t o r s
2.2.1 (L) continued
-+
S i n c e $ i s a t a n g e n t v e c t o r , - $ - w i l l be a u n i t t a n g e n t v e c t o r ;
vl
hence, ( 3 ) and ( 4 ) y i e l d
I
b.
I t s h o u l d seem c l e a r t h a t no m a t t e r how c l e v e r l y w e d i s g u i s e i t ,
e q u a t i o n ( 5 ) somehow must c o n t a i n t h e a r c l e n g t h s.
The e a s i e s t way t o s e e t h a t i t does i s t o r e c a l l from t h e p r e v i o u s
unit the f a c t t h a t
v
we were
-*
Q u i t e s i m p l y , t h e n , when w e formed
I3
i n f a c t computing
I n any e v e n t , e q u a t i o n ( 6 ) combined w i t h ( 4 ) t e l l s us t h a t
whereupon
Solutions
Block 2: V e c t o r C a l c u l u s
U n i t 2: T a n g e n t i a l and Normal V e c t o r s
2.2.1 (L) c o n t i n u e d
2.2.7
W e have
f
Therefore
a.
= t
f +
(2t
+
1/2 t
1)
3
(=,
1
- ~ L 2 ( 4 )+
t2
3/2
from ( I ) , x = -5 and y = $ ( 2 t + 1)
42
A t t = 4 the p a r t i c l e is a t
From ( 2 )
+
v = 4:
.
1 1 3/2)
= (8,9),
since
,
+
=
33 t h e r e f o r e ,
= 5,
and, from (31,
b.
ds
= af
and by ( 2 )
= J t 2+ 2 t
+
1 = It + 11, b u t 0 4 t( 4
i m p l i e s t h a t It + 11 = t + 1.
ds
+
1
1 2
T h e r e f o r e , s = =t
+
Therefore
= t
4
t
10
Solutions
Block 2: Vector C a l c u l u s
U n i t 2: T a n g e n t i a l and Normal Vectors
2.2.2
continued
*+
+
c . ~t t = 4 , v = 4 1
t
~ f=l 5.
33 and
+
+
Therefore, T = d~ - d&/dt
+
1
-
-
4;
Iv I
;3 3 -- 54 1+ + 53 J 0
t
a . The r e s u l t asked f o r h e r e h a s been d e r i v e d t w i c e before--once
i n t h e l e c t u r e and once i n t h e t e x t ( i n f a c t , it i s e s s e n t i a l l y
done t w i c e i n t h e t e x t ) . C e r t a i n l y , w e c o u l d j u s t i f y t h i s p a r t
of t h e e x e r c i s e on t h e grounds t h a t t h e r e s u l t i s i m p o r t a n t
enough f o r you t o make s u r e you can d e r i v e it.
Yet,
there is
s t i l l a b e t t e r reason f o r assigning t h i s e x e r c i s e . It allows
us t o i n t r o d u c e a r a t h e r i n t e r e s t i n g s u b t l e t y concerning t h e
d i f f e r e n c e between a c u r v e and t h e p a t h t r a c e d o u t by a p a r t i cular particle.
i f w e a r e g i v e n a p a r t i c u l a r p a r t i c l e and
The i d e a i s t h i s :
w e s t u d y i t s motion, i t i s v e r y n a t u r a l t h a t a t any p o i n t on
its p a t h we w i l l p i c k a s t h e u n i t t a n g e n t v e c t o r t h e one whose
That i s , w e v i s u a l i z e
s e n s e i s t h e same a s t h e v e l o c i t y v e c t o r .
t h e c u r v e a s b e i n g g e n e r a t e d by t h e moving p a r t i c l e .
t h e curve, i t s e l f , i s inanimate.
Of c o u r s e ,
I t does n o t know how t h e
p a r t i c l e i s t r a v e r s i n g i t , and what c o u l d b e even more complic a t e d i s t h e f a c t t h a t d i f f e r e n t p a r t i c l e s may t r a v e r s e t h i s
same c u r v e i n d i f f e r e n t ways.
The key t h o u g h t i s t h a t i f we a r e mainly i n t e r e s t e d i n t h e c u r v e ,
w e a r e f r e e t o a s s i g n it a s e n s e i n any way w e p l e a s e , and, once
t h i s i s done, t h e p a r t i c l e may t r a v e r s e t h e c u r v e i n any way i t
wishes.
I t s s p e e d , however, w i l l be n e g a t i v e i f t h e p a r t i c l e
has t h e o p p o s i t e s e n s e a s t h e c u r v e a t t h a t p o i n t , and p o s i t i v e
i f i t h a s t h e same s e n s e a s t h e c u r v e a t t h a t p o i n t .
(This i s
much l i k e t h e convention we have adopted w i t h r e s p e c t t o t h e
x-axis.
I f a p a r t i c l e i s moving from l e f t - t o - r i g h t a t a given
p o i n t , we c a l l i t s speed p o s i t i v e , w h i l e i f it moves from r i g h t t o - l e f t , we c a l l i t s speed n e g a t i v e . )
d ftR i s t h a t no m a t t e r how we *
The beauty of t h e d e f i n i t i o n 3 =
e l e c t t o d e f i n e t h e s e n s e of t h e c u r v e , ? w i l l always have t h e
Solutions
Block 2: Vector C a l c u l u s
t 2:
4
Tanoential
V e c ~ r s
2 . 2 . 3 (L) c o n t i n u e d
same s e n s e a s t h e curve.
(This i s e x p l a i n e d i n t h e t e x t a s w e l l
a s i n the lecture.)
I f w e now t h i n k o f a p a r t i c l e t r a v e r s i n g
t h e c u r v e a c c o r d i n g t o some r u l e 3 = % ( t )w,e have by t h e c h a i n
rule,
+
dii
from which w e may s o l v e f o r v (= =), t o o b t a i n
where a l l w e can be s u r e o f i s t h a t
9 and
$ have t h e same
-
W e c a n n o t be s u r e t h a t t h e y have t h e same s e n s e .
direction.
-
T h a t i s , once t h e s e n s e o f t h e c u r v e i s a s s i g n e d , a p a r t i c l e
may t r a v e r s e i t s o t h a t i t s s e n s e i s o p p o s i t e t o t h a t o f t h e
ds
c u r v e , i n which c a s e
i s negative.
I f w e d i f f e r e n t i a t e (1) w i t h r e s p e c t t o t (remembering t o u s e
ds
and 9 a r e f u n c t i o n s o f t i m e [i. e.
t h e product r u l e s i n c e both
+
T h a s c o n s t a n t magnitude, b u t e x c e p t f o r s t r a i g h t l i n e s , i t s
direction varies with t i m e ] ) we obtain
To r e p l a c e
dl!
by more " f a m i l i a r n t e r m s , r e c a l l t h a t i n t h e
l e c t u r e w e saw
-b
T = cos 4
whence,
-t
1
+
sin 4
t
J,
,
Solutions
Block 2: Vector C a l c u l u s
Unit 2: T a n g e n t i a l and Normal Vectors
2.2.3(L)
d9 =
d$
continued
-sin$
i-
1
+
cos$
3=
cos ($
+
t
90°)i
+
t
s i n ( $ + 90°)]*
d*
T h i s l e d us t o t h e c o n c l u s i o n t h a t ir was the u n i t v e c t o r o b t a i n e d
4
by r o t a t i n g !b 90° i n t h e p o s i t i v e ( c o u n t e r c l o c k w i s e ) d i r e c t i o n
and t h i s u n i t v e c t o r i s d e f i n e d t o be
3,
i. e.
,N
5
d!b
a-6
That i s ,
(biote: had w e a s s i g n e d t h e s e n s e of o u r c u r v e i n t h e o p p o s i t e
d i r e c t i o n , w e would have
-+
I n o t h e r words, N may p o i n t " i n "
o r " o u t " depending on t h e
sense of t h e curve.)
To g e t t h e r i g h t s i d e of ( 2 ) i n t o t h e d e s i r e d form ( i . e . ,
dT
r e p l a c e aF by 8) , w e a g a i n u s e t h e c h a i n r u l e t o o b t a i n
-+
and s i n c e '
=
8,
to
I
I
n
Solutions
Block 2: Vector C a l c u l u s
U n i t 2: T a n g e n t i a l and Normal Vectors
2.2.3(L)
continued
With r e s p e c t t o ( 3 ) , w e o b s e r v e t h a t
3 3
=
2,
and t h a t
being
t h e measure o f how t h e d i r e c t i o n o f t h e t a n g e n t t o t h e c u r v e i s
changing a t any i n s t a n t , i s an e x c e l l e n t parameter by which t o
measure c u r v a t u r e . W e , t h e r e f o r e , l e t K =
and ( 3 ) becomes
+
d's
$
+
d s d6
ds
aTkfz) (E)
+
Equation ( 4 ) i s t h e d e s i r e d r e s u l t , b u t n o t i c e t h a t o u r proof
o f f e r s a n a l t e r n a t i v e t o t h e t e x t i n t h e s e n s e t h a t w e used t h e
d e r i v a t i o n o f (4) t o m o t i v a t e t h e " i n v e n t i o n n o f K ( w h i l e t h e
t e x t d e r i v e d K i n i t s own r i g h t f i r s t and t h e n a p p l i e d t h e res u l t t o t h e d e r i v a t i o n of ( 4 ) ) .
N o t i c e a l s o t h a t (4) i s t r u e no m a t t e r how t h e s e n s e o f t h e
Once t h e s e n s e i s chosen, o b s e r v e
c u r v e i s i n i t i a l l y chosen.
4
d
T h a t is.
t h a t =(=
K ) may be e i t h e r p o s i t i v e o r n e g a t i v e .
With t h i s i n mind, w e complete o u r problem by r e s e r v i n g o u r
c h o i c e of s e n s e of t h e c u r v e u n t i l t h i s moment ( s i n c e no harm
i s done by w a i t i n g because ( 4 ) i s t r u e f o r any s e n s e t h a t w e
choose) and w e t h e n a s s i g n t h e s e n s e t o o u r c u r v e t h a t make?
Solutions
Block 2: Vector A r i t h m e t i c
Unit 2: T a n g e n t i a l and Normal Vectors
-
-
2.2.3 ( L ) c o n t i n u e d
non-negative.
For example, given t h e c u r v e C w i t h o u t an
i n d i c a t e d s e n s e , where C i s
W e choose t h e s e n s e o f C a s shown below i n o r d e r t o make
3
> 0.
As the particle traverses C i n
the indicated direction (sense)
$I i n c r e a s e s .
I n c a s e s where no s e n s e i s a s s i g n e d t o t h e c u r v e ,
3 h a s an
ambiguous s i g n , and i n t h i s e v e n t i t i s customary t o d e f i n e
be
1%
K
to
which a g r e e s w i t h o u r d e f i n i t i o n o f s e n s e and a l s o w i t h
t h e t e x t ' s d e f i n i t i o n of
K.
There now remains o n l y one major i s s u e t h a t must be r e s o l v e d
A t t h e moment 8 has
between o u r approach and t h a t of t h e t e x t .
been d e f i n e d i n two d i f f e r e n t ways.
On t h e one hand, u s i n g o u r
-P
i n . i t i a t i v e v i s u a l model, N was a p o s i t i v e 9 0 ° r o t a t i o n of
3.
In
-t
d;+T
t h e t e x t , N i s t h e u n i t v e c t o r i n t h e d i r e c t i o n and s e n s e of E
-- .
d* - d*
d?F d
That i s ,
(&I
=
= ( l( )
K) =
NOW
=
1 l$l 131
7
lx1~*
-+
l$$l 1
Thus, i f w e r e w r i t e ( 2 ) from t h i s p o i n t of view, w e o b t a i n
Kg
Solutions
Block 2: Vector C a l c u l u s
U n i t 2 : T a n g e n t i a l and Normal Vectors
2.2.3 ( L ) c o n t i n u e d
and comparing this w i t h ( 4 ) we s e e t h a t
3
has t h e same meaning i n
either c a s e .
The b e a u t y of t h e t e x t ' s approqch i s t h a t w e need
+
dT
no knowledge o f geometry t o s e e t h a t T and
a r e perpendicular.
The drawback i s t h a t t h e approach i s s u f f i c i e n t l y a b s t r a c t ( i f you
a r e n o t used t o t h i n k i n g i n t e r m s o f v e c t o r s ) t h a t you may g e t no
feeling f o r the result.
The advantage t o o u r approach i s t h a t
i t i s s u f f i c i e n t l y v i s u a l t h a t you can see what i s happening.
On
t h e o t h e r hand, o u r approach depends on b e i n g a b l e t o d i s c o v e r a
few g e o m e t r i c r e s u l t s t h a t might n o t have seemed t h a t obvious.
I n p a r t i c u l a r , when we want t o i n v e s t i g a t e 3-dimensional s p a c e
c u r v e s , t h e geometry may be enough more complex s o t h a t t h i s
approach w i l l be even more d i f f i c u l t t o a p p l y , i n which c a s e w e
may p r e f e r t o u s e t h e more a n a l y t i c approach o f t h e t e x t .
N o t i c e t h a t t h e v a l i d i t y o f ( 4 ) does n o t r e q u i r e that w e b e a b l e
t o i n t e r p r e t t h e r e s u l t p h y s i c a l l y , b u t ( 4 ) does have a r a t h e r
easy interpretation.
Namely, t h e t a n g e n t i a l component o f t h e
a c c e l e r a t i o n i s t h e a c c e l e r a t i o n o f t h e p a r t i c l e a l o n g the c u r v e
d2s
( s i n c e t h i s i s p r e c i s e l y what
measures). The normal compo-
z
n e n t o f ' t h e a c c e l e r a t i o n i s t h e p r o d u c t of t h e c u r v a t u r e and t h e
s q u a r e o f t h e s p e e d o f the p a r t i c l e a l o n g t h e curve. While this
-
may n o t s e e m f a m i l i a r , t h e s t u d e n t o f e l e m e n t a r y p h y s i c s might
remember t h e f o r m ~ l af o r c e n t r i p e t a l a c c e l e r a t i o n i n c i r c u l a r
v
motion a s b e i n g
where v i s t h e speed o f t h e p a r t i c l e and r i s
t h e r a d i u s of t h e c i r c l e .
If w e define the radius of curvature
p t o b e t h e r e c i e o f t h e c u r v a t u r e , i.e., p =
, equation
K
v2
( 4 ) y i e l d s t h a t t h e normal component o f a c c e l e r a t i o n i s
-
b.
From (1) and ( 4 ) , w e have t h a t
-P '
Solutions Block 2: Vector Calculus Unit 2: Tangential and Normal Vectors (L) continued 2.2.3
But, since
becomes
? I d
Therefore,
K
111 = 13.1 =
and
.
b
;1
=
x
1,
.
[? x $ 1
,
= 1.
Hence, (5)
21
I ~ 1 . ~ -
'
in terms of the motion ( 2 and ); of the +
-t
path, a very practical form in many applications, since v and a may well be the only available parameters. Equation (10) yields
K
c - We had Therefore,
1
= 5
and' -+
-+
v x a =
4-,
x -t
3) + 3(5 x
5(1
4-t
= gk
-
3%
-
-t
-f
5
Therefore, Iv x a1 = 3 '
Therefore, from b. 1)
Solutions
Block 2: Vector C a l c u l u s
U n i t 2: T a n g e n t i a l and Normal Vectors
S t a r t i n g w i t h t h e " i f H p a r t , we assume t h e speed i s c o n s t a n t .
2
d s = constant.
That i s , gg
Therefore
= 0. I n t h i s c a s e ,
z implies
i s a s c a l a r m u l t i p l e of 6 s o t h a t t h e a c c e l e r a t i o n
i s normal t o t h e p a t h i n t h i s c a s e .
From (1)
Conversely, i f w e now assume t h a t t h e a c c e l e r a t i o n i s normal t o
must b e 0. But t h e
ds i s c o n s t a n t , and t h e a s s e r -
t h e path then
$-component i s
t i o n i s proved.
(From a r i g o r o u s
grammatical p o i n t of view "only i f " s h o u l d be
t r a n s l a t e d t h a t i f ds is n o t constant then t h e a c c e l e r a t i o n i s
n o t normal t o t h e p a t h .
I n t h i s c a s e t h e argument would b e t h a t
ds
d2s
s i n c e -- i s n o t constant,^ c a n n o t b e i d e n t i c a l l y 0. Hence t h e
dt
dt
+
T-component o f t h e a c c e l e r a t i o n ( i . e . d2s
is n o t always 0.
Hence the d i r e c t i o n o f
d i r e c t i o n of
the
2
a c c e l e r a t i o n i s n o t always i n the
i.)
I$(
131
a. W e have K =
=
, b u t when t h e c u r v e i s given i n t h e form
y = f ( x ) , n e i t h e r 4 n o r s may b e c o n v e n i e n t parameters.
I n s t i l l o t h e r words, when t h e e q u a t i o n i s i n the form y = f ( x )
d
d2
i t i s u s u a l l y c o n v e n i e n t t o compute
and
dx 2
Now, what w e d o know i s t h a t
4. Solutions
Block 2: Vector C a l c u l u s
U n i t 2: T a n g e n t i a l and Normal V e c t o r s
2.2.5 (L1 c o n t i n u e d Pictorially, m = tan (
D i f f e r e n t i a t i n g (1) w i t h r e s p e c t t o x, y i e l d s
sec2 (
a;;. =
Therefore,
dx
Now
, tan(
2
cos ( =
-~-
[l +
Hence,
implies t h a t
=
1
(%,'I
( 2 ) becomes
By t h e c h a i n r u l e
( i . e . , sec2 @
-
t a n 2 ( = 1).
Solutions
Block 2: Vector Calculus
Unit 2: Tangential and Normal Vectors
2.2.5 (L) continued
Therefore
IC
=
1%
=
'1.
1a/ \/-
s u b s t i t u t i n g (3) i n t o ( 4 ) y i e l d s
Had w e wished, w e could v e c t o r i z e y = f ( x ) i n t o a motion problem
by w r i t i n g
(i.e., x = t and y = f ( t )
-+
y =
f (x)).
Then
-+
t
and a = f n ( t ) j .
By our previous r e s u l t ,
1;
IKI
x
21
~f"(t,Zl
5
=
$1 ( ~ l+ [ f ' ( t )l 2
If" c t ,
)
1
*The s i g n , e v e n though i t w i l l make no d i f f e r e n c e when we compute
ds
K , depends upon w h e t h e r
i s p o s i t i v e or ne.gative, t h a t i s ,
whether s i n c r e a s e s or d e c r e a s e s i n the d i r e c t i o n o f t h e p o s i t i v e
x-axis.
Our d e f i n i t i o n - t h a t K i s p o s i t i v e a v o i d s t h i s problem.
Solution?
Block 2: Vector C a l c u l u s
U n i t 2: T a n g e n t i a l and Normal V e c t o r s
2.2.5 ( L ) c o n t i n u e d
-
Note
U n f o r t u n a t e l y t h e r e a r e s e v e r a l mathematical c o n c e p t s t h a t l e n d
themselves t o more t h a n one convention. C u r v a t u r e i s one of t h e s e .
W e have e l e c t e d t o u s e t h e approach t h a t
K
=
1
, whence
K
1 0.
Some p e o p l e d e f i n e K t o b e p o s i t i v e o r n e g a t i v e depending u p o n .
whether t h e c u r v e i s "convexn o r "concave" w i t h r e s p e c t t o t h e
position vector.
I n t h i s c a s e , one w r i t e s t h a t
= 2e2 x
Therefore,
Therefore,
IKI
4e2X
=
3/2 since
1K1
=
(1 + 4e4X)
b.
-f
2t +
? i = t l + e
-P
Therefore, v =
Therefore,
J
-t
1
+
A
2e
-
-
2t
t
J
it 3 / L
(1 + 4e )
, and
letting
K
= t , we o b t a i n
-
a.
From Exercise 2 . 2 . 5 , y = ax
+
b
+
+=
Therefore,
0.
dx
(Note:
t h i s i s what we would expect i n t u i t i v e l y f o r s t r a i g h t l i n e .
Had we wished t o see the % f o r
s t r a i g h t l i n e $ i s constant.
Therefore,
8 = +(a2 - x 2-1/2
)
curvature, n o t i c e t h a t f o r a
Hence
3= 0
(-2x) =
2
(a
.)
-X
7
- X I
2
X
a2
2
-- x2;
Solutions
Block 2: Vector Calculus
Unit 2: Tangential and Normal Vectors
2.2.7
continued
Note:
i s t h e equation of a s e m i c i r c l e of r a d i u s 2,
y =
c e n t e r e d a t (0,O) w i t h y b 0.)
For t h i s c i r c l e , of course, a i s c o n s t a n t and $I = a + 90° o r d$I = do.
da
1
In
Also s = aa hence ds = ado.
Therefore,
= -;r = -- a s before.
a a
a
t h i s c a s e , t h e c u r v a t u r e i s t h e r e c i p r o c a l of t h e r a d i u s . With t h i s
1
i n mind, given K f o r any curve, we d e f i n e i; = p t o be t h e r a d i u s of
3
c u r v a t u r e a t t h a t p o i n t . P i c t o r i a l l y , p i s t h e r a d i u s of an " i n s t a n taneous" c i r c l e c a l l e d t h e o s c u l a t o r y c i r c l e .
I.e.,
I n many kinematics problems, w e s t u d y t h e
motion of a p a r t i c l e by assuming t h e part i c l e i s on t h e o s c u l a t i n g c i r c l e a t t h e
given i n s t a n t .
2.2.8(L)
W e have
a.
-+
R = t X + ( t 2+
-+
v = 1
++ 2 t f
-f
t
a = 23.
1)s
I
I
.
Solutions
lock 2: Vector Calculus
Unit 2: Tangential and Kormal Vectors
2.2.8
continued
;i s
we a l r e a d y know t h a t
a tangent vector but not necessarily a
u n i t tangent.
This i s e a s i l y remedied by d i v i d i n g
nitude.
From (2) we- then o b t a i n
+
~ e t t i n gT =
f
by i t s mag-
r
-$-
w e have our d e s i r e d u n i t t a n g e n t vector.
1.
Namely,
We saw i n t h e previous s e c t i o n t h a t ( v l = d s and we know t h a t
-+
1
i n t h i s e x e r c i s e . See n o t e a t end of t h i s e x e r c i s e .
=
w
Hence,
ze
But, w e know t h a t aT = dLs
Therefore, aT =
Therefore, a
;d;;L2s =
d ds
8t
-
T =
From (3) w e know t h a t
1
a
and
2
(af
1;
(
d(d1
= .
+
4tL)
dt
4t
= 123 ( = 2 ,
and we a l s o know t h a t
-+
-b
(Here i t i s most important t o observe t h a t R, v,
.
-
a r e p r o p e r t i e s of t h e motion of t h e p a r t i c l e , n o t of t h e
coordinate
system.
d i f f e r e n t from t h e
i n e i t h e r case.
Thus, while t h e
%
and
8
7
'
1
and
+
t
J
-b
-b
components of a a r e
components of a , a i s t h e same v e c t o r
Pictorially,
Solutions
Block 2: Vector C a l c u l u s
U n i t 3: T a n g e n t i a l and Normal Vectors
continued
2.2.8
= aT
= a
% +
-+
a N $, byt a i s t h e same
+
1 + a
i
j
-t
J,
,.,
-.
i n e i t h e r case.)
I n any e v e n t ,
aN2 =
So
2
-
( R e c a l l a T and a N are s c a l a r s s i n c e t h e y a r e
t h e com o n e n t s o f a c c e l e r a t i o n i n t h e 9 and 8
dire*
aT from ( 3 ) and (6)
2
T h e r e f o r e , aN =
6F-Z
b. W e know t h a t aN =
K
.
d s 2 Hence
(=)
a
K
=
N
7
.ds.
Solutions
Block 2: Vector Calculus
Unit 2: Tangential and Normal Vectors
2.2.8
Note:
continued
t h e procedure used h e r e o u t l i n e s a r a t h e r common approazh
f o r f i n d i n g t h e t a n g e n t i a l and normal components of a c c e l e r a t i o n
when t h e equation of motion i s given i n C a r t e s i a n coordinates.
+
From t h e given equation we compute 2 and a. The magnitude of
2
is t h e speed of t h e p a r t i c l e along t h e curve, which i s a l s o
denoted by ds/dt.
From d s / d t we can f i n d t h e t a n g e n t i a l compoSince we
n e n t of a c c e l e r a t i o n , s i n c e t h i s i s simply d 2 s / d t 2
.
then know both t h e t o t a l a c c e l e r a t i o n
(which i s t h e same whether
+
-f
we a r e i n C a r t e s j a n c o o r d i n a t e s o r i n T and N c o o r d i n a t e s ) , we
use t h e Pythagorian Theorem t o determine t h e normal component of
acceleration.
That i s ,
' E a r l i e r w e showed how t o f i n d t h e c u r v a t u r e a s a "bonus" i f we
a l r e a d y know t h e speed of t h e p a r t i c l e and t h e normal component
of a c c e l e r a t i o n .
In p a r t
b. we want t o emphasize t h a t w e do
n o t have t o r e s o r t t o any s p e c i a l s e t of c o o r d i n a t e s t o compute
curvature
i s a p r o p e r t y of t h e p a t h and n o t t h e c o o r d i n a t e system. This
t h e curvature.
A s we mentioned i n Exercise 2.2.3
b.,
was why w e developed a formula f o r c u r v a t u r e completely i n t e r m s
of t h e v e l o c i t y and t h e a c c e l e r a t i o n of t h e p a r t i c l e a t a p a r t i cular point.
R e c a l l t h a t t h i s formula was given by
From (2) and (3) w e see t h a t
Solutions
Block 2: Vector Calculus
Unit 2: Tangential and Normal Vectors
2.2.8
continued
Moreover,
I I;
= ds
--
J
:
1 + 4 t2
lmpiies
that
P u t t i n g (10) and (11) i n t o ( 9 ) y i e l d s
which checks w i t h our answer i n s t e p 5 i n Equation ( 8 ) .
I n c l o s i n g t h i s e x e r c i s e we would l i k e t o t a k e a few moments t o
show t h e advantage, computationally,of t h e technique used i n t h i s
+
problem. We o b t a i n e d , f o r example, K and N a s n a t u r a l by-products
of r e l a t i v e l y simple computations. W e could have obtained t h e s e
q u a n t i t i e s d i r e c t l y from t h e i r d e f i n i t i o n s , and it i s probably
worthwhile t o do t h i s a t l e a s t once.
We have t h a t
=,
d$
From ( 4 ) , we can f i n d
and by t h e chain r u l e (proven i n t h e
e x e r c i s e s of t h e previous u n i t ) w e have
ds
J2
From ( 5 ) , ;IF = 1 + 4 t
and by one of our product r u l e s of v e c t o r c a l c u l u s we have from
(4
Solutions
Block. 2: Vector Calculus
U n i t 2: T a n g e n t i a l and Normal Vectors
2 . 2 . 8 continued
d$
d$ d s
1
d$
. a~
from which ( 1 4 ) y i e l d s
Z=aF/a€=,
-b
Therefore, v = 3 t
2-t
1
+
cos t
t
J.
Solutions
Block 2: Vector C a l c u l u s
U n i t 2: T a n g e n t i a l and Normal V e c t o r s
P
2.2.9
continued
+
a = 6t
-t
1
-
t
s i n t 1.
1;
1 = 4 9 t4 +
Therefore,
d s <.m.
cos2t =
From ( 4 ) ,
d2s
1
4
= 2(9t
dt -
+
-
36t3
2 J9t4
1 2
cos t)-T( 3 6 t 3
-
2 ~ 0 st s i n t ) sin 2t
+
cos2t
S i n c e t h e t a n g e n t i a l component of a c c e l e r a t i o n aT, i s e q u a l t o
,.
d~
-
36t3
I
A
sin 2t 9 A s f o r t h e normal component o f a c c e l e r a t i o n ,
a N , w e have
and from ( 3 )
P u t t i n g t h e r e s u l t s of ( 5 ) and ( 7 ) i n t o ( 6 ) , w e o b t a i n
36t
2
+
2
(36t3
sin t = 4(9t4
+
2
sin 2t)
cos2t) +
2
aN
.
Solutions Block 2: Vector Calculus Unit 2: Tangential and Normal Vectors 2.2.9
continued Theref ore,
2
=N
-
a-=
aN
--
2 6
2.
2
4
2
2
2
(36) t + 36t 4 cos t + 36t sin t + 4 cos t sin t
2
4 (9t4 + cos t)
/36tL(4 cosLt
+ tZ sin2t + 2t sin 2t
6t(2 cos t + t sin t)
2/9t 4 + cos2t
- (36t3- sin
2 2 t)
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Calculus Revisited: Multivariable Calculus
Prof. Herbert Gross
The following may not correspond to a particular course on MIT OpenCourseWare, but has been
provided by the author as an individual learning resource.
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.