Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3:
D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.1(L)
I n o u r s t u d y o f t h e c a l c u l u s o f a s i n g l e r e a l v a r i a b l e , w e saw
t h a t i f f ' ( a ) e x i s t e d t h e n f was continuous a t x = a.
In other
words, w e saw t h a t t h e p r o p e r t y o f b e i n g d i f f e r e n t i a b l e i m p l i e d
( I n g e o m e t r i c terms, w e a r e s a y i n g
t h e p r o p e r t y of c o n t i n u i t y .
t h a t i f a c u r v e i s smooth t h e n it must be unbroken.)
Analogously, one might e x p e c t t h a t , i n two v a r i a b l e s , i f f x ( a , b )
and f
( a , b ) e x i s t t h e n f must be continuous a t t h e p o i n t ( a r b ) .
Y
The aim o f t h i s e x e r c i s e i s t o show t h a t t h i s need n o t be t r u e .
Before w e a c t u a l l y do t h e e x e r c i s e , l e t u s observe t h a t , i n t h e
c a s e of one independent v a r i a b l e , t h e f a c t t h a t f ' ( a ) e x i s t e d
g u a r a n t e e d t h a t t h e d i r e c t i o n a l d e r i v a t i v e e x i s t e d i n each d i r e c t i o n ,
s i n c e i n t h i s c a s e t h e r e a r e o n l y two d i r e c t i o n s .
I n t h e case
o f two independent v a r i a b l e s , however, t h e f a c t t h a t t h e d i r e c t i o n a l
d e r i v a t i v e s e x i s t i n b o t h t h e h o r i z o n t a l and v e r t i c a l d i r e c t i o n s
t e l l s u s n o t h i n g about what i s happening i n o t h e r d i r e c t i o n s .
R a t h e r t h a n c o n t i n u e w i t h an a b s t r a c t d i s c u s s i o n , w e s h a l l show
i n t h i s e x e r c i s e t h a t t h e function g has t h e property t h a t
gx(O , O )
and gy ( 0 . 0 ) b o t h e x i s t
p o i n t (0,O).
-
yet g is n
o t continuous a t t h e
T h i s , i n t u r n , i s enough t o show t h a t t h e assumption
t h a t g i s c o n t i n u o u s a t a p o i n t , a f t e r w e know t h a t both gx
e x i s t a t t h a t p o i n t , i s n o t redundant.
Y
A s f a r a s t h e s o l u t i o n of o u r e x e r c i s e i s concerned, p a r t of it
i s t r i v i a l , s i n c e w e have a l r e a d y proven t h a t g was n o t continuous
and g
a t (0,O) i n E x e r c i s e 3.1.8 ( L )
.
Thus, w e need o n l y show t h a t g,(O,O) and g (0,O) e x i s t . The
Y
e a s i e s t approach i s t o work d i r e c t l y from t h e d e f i n i t i o n o f gx
and g
Y'
For example g x ( a , b ) means
have
lim
g(a+Ax,b)-g(a,b)
Ax+O
Ax
that
we
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.1 c o n t i n u e d
S i n c e g i s symmetric i n x and y [ i . e . ,
g ( x , y ) = g ( y , x ) l , it follows
a t once t h a t
also.
( I f t h e symmetry p r o p e r t y d o e s n ' t s t r i k e your f a n c y , mimic o u r
procedure f o r showing gx(O,O) = 0 t o prove g (0,O) = 0.)
Y
T h i s e x e r c i s e sets t h e s t a g e f o r a d e e p e r a n a l y s i s of what happens
when w e d e a l w i t h f u n c t i o n s of s e v e r a l v a r i a b l e s .
Without going
i n t o d e t a i l a b o u t why it i s s o ( t h e proof of Theorem 1 i n S e c t i o n
15.4 o f t h e t e x t shows t h i s a s does o u r o p t i o n a l e x e r c i s e 3 . 3 . 9 ) ,
t h e f a c t remains t h a t i f f x ( a , b ) and f
(arb) exist and i f b o t h
Y
f X and f Y a r e c o n t i n u o u s a t ( a , b ) , t h e n f w i l l be c o n t i n u o u s a t
( a , b ) ( a l t h o u g h i n t h e t e x t t h i s c o n c l u s i o n i s s t a t e d a s p a r t of
the hypotheses).
From a somewhat i n t u i t i v e p o i n t o f view, t h e f a c t
*Do n o t b e t e m p t e d t o s a y t h a t s i n c e g ( O , O ) = O ,
T h i s would b e t r u e i f g ( x , y ) = O .
a
ao
g(x,y)l
=--0.
ax
(0,O)
I n t h e 1-dimensional c a s e , n o t i c e
3-.
L
t h a t i f g ( x ) = x -2x
then ft(x)=2x-2.
Hence, g t ( 0 ) = - 2 even though
( P i c t o r i a l l y , g (x)=x2-2x s a y s t h a t t h e c u r v e y=g ( x ) p a s s e s
through (0,O).
I t d o e s n o t s a y t h a t t h e s l o p e of t h e c u r v e a t
t h i s point is 0).
P e r h a p s a b e t t e r e x a m p l e m i g h t b e t h e o n e we
u s e d i n P a r t 1 o f t h e c o u r s e w h e r e f was d e f i n e d by
g (O)=O.
T h i s means t h a t f ( x ) 5 x + 3 .
t h a t while f ( 3 ) = 6 , f t ( 3 ) = 1 n o t 0.
Notice
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.1 c o n t i n u e d
t h a t f x and f
a r e c o n t i n u o u s a t (a.b) means t h a t t h e f u n c t i o n i s
Y
" w e l l behaved" i n a neighborhood of ( a r b ) . To go one s t e p f u r t h e r , and t h i s w i l l be d i s c u s s e d i n somewhat
more d e t a i l i n t h i s and t h e n e x t u n i t , i t t u r n s o u t t h a t i f we
want t o t h i n k o f f a s b e i n g d i f f e r e n t i a b l e , it i s n o t enough t o
t h i n k o n l y of f x and f e x i s t i n g ; t h e y must a l s o be continuous.
Y
Quite i n general, i f f(xl,
xn) h a s t h e p r o p e r t y t h a t f
X1
and f
e x i s t and a r e continuous a t ( a l , . .
,a n ) t h e n w e s a y t h a t
xn
a n ) . I n t h i s e v e n t , f i s continuous
f i s d i f f e r e n t i a b l e a t ( a-l ,
...
,...,
.
...,
a t (al,...,an).
f need n o t be continuous i f f
a r e n o t a l l continuous a t (al,
...,a n ) .
,..., and
f
X1
xn
To t r y t h i s i d e a o u t i n terms of o u r p r e s e n t example, s i n c e
gx(O,O) and g (0.0) e x i s t b u t g i s n o t continuous a t ( 0 . 0 ) . it
Y i s n o t continuous a t ( 0 , O ) .
L e t u s look a t g x ( x , y ) and see whether l i m
g (x,y) e x i s t s .
(x,y)-+(O,O) x Since must be t h a t e i t h e r gx o r g,
we obtain
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U nit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
3.3.1 continued
NOW, one way of viewing l i m
g (x,y) i s by f i r s t l e t t i n g
(x,y)+(O,O) x
x+O (holding y#O f i x e d ) and then l e t t i n g y+O.
I f we do t h i s , then
and from (1) t h i s y i e l d s
=
+
(depending on whether y+0+ o r YO-)
Solutions
B l o c k 3: P a r t i a l Derivatives
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
-
--
-
-
3 . 3 . 1 continued
Since g
X
(0,O) = 0 ,
( 2 ) shows u s t h a t
*
lim
( x , y ) + ( O , O ) sx ( x l ~ ) # s ~ ( o l o )
s o t h a t b y t h e d e f i n i t i o n o f c o n t i n u i t y , gx i s n o t c o n t i n u o u s a t
(0,O).
Theref o r e
t h e r e f ore
*
N o t i c e t h a t we p i c k e d o n e s p e c i a l p a t h t o c o m p u t e 1i m
( x , y ) + ( O , O ) gx ( ~ Y Y ) .
S i n c e a l o n g t h i s p a t h , t h e l i m i t was n o t e q u a l t o g ( 0 , 0 ) , i t
X g u a r a n t e e s t h a t gx i s n o t c o n t i n u o u s a t ( 0 , 0 ) , f o r i f i t w e r e , 1i m then
g ( x , y ) w o u l d e q u a l gx(O,O) a l o n g e v e r y p a t h .
(x,y)+(O,O) x
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.2
continued
and s i n c e
f
( x , y ) = 2x+3x2y and f
X
Y
( x I y ) = x3+ 32 ~
w e see t h a t
fxry)Ay +klAx+k2Ay
Y
f(x+Ax,y+Ay)-f(x,y)=fx(x,y)Ax+f
where kl = A X + ~ X ~ A X + ~2 A
E ~ ++ ~~ XX A
X A ~ + K ~ A ~
-2
k 2 = 3yAy+Ay
Hence, k
k +O as Axl,Ay+O.
1' 2
b. To estimate ( 1 . 0 0 1 ) ~+ (1.001)
w i t h x=y=l, Ax=0.001,
3
(1.002)
+
(1.002)
3
w e may u s e ( a )
and Ay=0.002
f(1,l) = 3
Then so t h a t
Theref o r e f(1.001,
1.002)
2
f (111)+ f x ( l , l ) A x + f
%
3
+
Y
(1,l)Ay
0.013 = 3.013
3.3.3(L)
The a i m o f t h i s e x e r c i s e i s t o h e l p make it c l e a r t h a t f x and
f a r e e s s e n t i a l l y j u s t two s p e c i a l d i r e c t i o n a l d e r i v a t i v e s , b u t
Y
t h a t t h e c o n c e p t of a d i r e c t i o n a l d e r i v a t i v e makes j u s t a s good
sense regardless of t h e direction.
A t t h e same t i m e , w e want t o
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.3 (L) c o n t i n u e d
make i t c l e a r t h a t i f f x and f
p o i n t (al,a2)
e x i s t and a r e continuous a t t h e
Y
t h e n t h e d i r e c t i o n a l d e r i v a t i v e of f a t ( a l I a 2 )
e x i s t s i n e v e r y d i r e c t i o n , and, i n f a c t , t h e d i r e c t i o n a l
d e r i v a t i v e i n e a c h d i r e c t i o n can be q u i c k l y o b t a i n e d by knowing
.
t h e v a l u e s of f x (a ) and f
(a)
Y Before c o n t i n u i n g f u r t h e r , perhaps a p i c t u r e o r two w i l l be
helpful.
a=(1,2)
"(314)
s 3 2
G f ( x , y ) = x +x y +y Theref o r e
P=[1,2,f
1
4AwK
a
-
and b
determrne a
straight line
W e t h e n draw
the plane
which
passes
through
this line and i s perpendi
ular t o th
xy-plane
11
.'
(3) T h i s p l a n e i n t e r s e c t s
5 3 2 4
t h e s u r f a c e w=x +x y +y
i n some curve C which
p a s s e s through P.
( 4 ) This plane i s a s "logical"
t o c o n s i d e r i n any d i r e c t i o n a s it i s i n t h e
s p e c i a l c a s e s where t h e
l i n e j o i n i n g a and b
happens t o b e p a r a l i e l
t o either the
x-axis o r t h e
+Y
y-axis.
c-vx
.
r(
X ( F i g u r e 1)
-
a. I f w e now l a b e l by s t h e d i r e c t i o n o f t h e l i n e from a t o b, t h e
c u r v e C l i e s i n t h e wls-plane.
F i g u r e 1 may be viewed a s
T h a t i s , o u r shaded p o r t i o n of
Solutions
.
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.3 (L) c o n t i n u e d
The s l o p e o f
W1
( 3 ) T h e r e f o r e , w e may
t h i n k of t h e wl-axis
C a t P is
a s being s h i f t e d t o
coincide with the
w-axis, and t h e sa x i s a s being s h i f t e d
parallel t o itself t o
p a s s through 0.
The complication that
e x i s t s here
b u t didn 't
when w e computed
a w and aw
aY
ax
a
b
-
lc .,-;I
( 4 ) I n t h i s way
dwl
dw
s i n c e t h e curve C i s
n o t a f f e c t e d (only i
position i n space).
is t h a t while
t h e x- and y-axes
p a s s through 0 , t h e
s - a x i s need n o t .
( F i g u r e 2)
The i n t e r e s t i n g p o i n t i s t h a t w e can compute
r e f e r t o a 3-dimensional diagram.
dw
w i t h o u t having t o
Namely, w e know t h a t viewed
s i s t h e l i n e determined by ( 1 , 2 ) and ( 3 , 4 ) ;
hence, i t s s l o p e i s 1. Its e q u a t i o n , t h e r e f o r e , i s x-1 = 1, o r
y = x+l.
i n t h e xy-plane,
Thus, w h i l e w i s a f u n c t i o n of t h e two independent v a r i a b l e s x and
y , once w e r e s t r i c t o u r c o n s i d e r a t i o n t o t h o s e p o i n t s (x,y)
*N o t i c e
t h a t a = ( a l , a 2) r e f e r s t o c o o r d i n a t e s i n t h e x y - p l a n e .
particular
I n t h e w l s - p l a n e a would h a v e t h e form ( ~ ~ ~ 0I n ) our
.
That
i
s
:
example, a = ( 1 , 2 ) + s l =
a.
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.3 (L) c o n t i n u e d
f o r which y = x + l , x and y a r e no l o n g e r independent.
In other
words, i n t h e d i r e c t i o n of s , w can be e x p r e s s e d e n t i r e l y i n terms
of x ( o r of y o r , f o r t h a t m a t t e r , of s ) . I n p a r t i c u l a r , o u r
d e f i n i t i o n o f w l e a d s t o t h e f a c t t h a t i n t h e d i r e c t i o n of s ,
whereupon expanding and c o l l e c t i o n o f t e r m s y i e l d s
5 3 2 4
[Notice t h a t (1) and ( 2 ) a r e s p e c i a l c a s e s of w = x +x y +y
f o r a p a r t i c u l a r path.
I n g e n e r a l , i f s had been any l i n e drawn
from ( 1 , 2 ) b u t n o t p a r a l l e l t o t h e y - a x i s , t h e n t h e l i n e s would
have t h e form y = mx+b, whereupon (1) would have been
s o t h a t i t s h o u l d be c l e a r t h a t how w v a r i e s w i t h x does indeed
depend on t h e d i r e c t i o n of t h e l i n e from ( 1 , 2 ) . I
I f w e d i f f e r e n t i a t e w i n (1) w i t h r e s p e c t t o x ,
(See Remarks i n
F i g u r e 3) w e o b t a i n
and e v a l u a t i n g ( 3 ) f o r x = l [ i . e . ,
a t t h e p o i n t ( 1 , 2 ) 1 we o b t a i n :
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
3.3.3 (L) continued
Given x
0
a'6x 06b'
-
(2) I n t h e s ~ e c i a lc a s e
t h a t ab i s p a r a l l e l
t o thex-axis
d
w
dx happens t o be t h e
same a s a W , b u t we
ax
may t a l k about dw
f o r any d i r e c t i o n s.
A.
w e draw
ab a t c .
)
Y
(Figure 3)
The t r o u b l e w i t h ( 4 ) i s t h a t it i s t h e r i g h t answer t o t h e wrong
dw n o t of dw
problem: We wanted t h e v a l u e of This m a t t e r i s
ds
dx
e a s i l y remedied by our use of t h e chain r u l e (and h e r e i t i s
important t o understand f u l l y t h a t w i s a f u n c t i o n of t h e s i n g l e
-.
r e a l v a r i a b l e s; f o r while s does depend on both x and y, along
any f i x e d l i n e t h e v a l u e of x determines t h e value of y, s o t h a t
we have indeed o n l y one degree of freedom). We have
-2,namely,
and, p i c t o r i a l l y , i t i s easy t o v i s u a l i z e Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.3 (L) c o n t i n u e d
t a n r$ = s l o p e of l i n e s=l o r r$ =
aIT
therefore
Ax - = c ITo s1 ~ = ~ ~ ~
-AS
(Figure 4)
*
a,
Ax i s t h e c o n s t a n t 5
1
d
-x -- l i m A x - 1
Since ds
A s + - O ~ - ~
As
bining t h i s with ( 4 ) , equation (5) y i e l d s
,
and com-
That i s , t h e d e r i v a t i v e o f w i n t h e d i r e c t i o n s which j o i n s ( 1 , 2 )
.
w i t h (3.4) i s
(With r e f e r e n c e t o F i g u r e 1, t h e s l o p e of
5342
t h e c u r v e C a t t h e p o i n t P i s -T-- .)
b. S i n c e , i n o u r p r e s e n t e x e r c i s e , w i s a polynomial i n x and y , i t
i s r e a d i l y v e r i f i e d t h a t w, w
w a l l e x i s t and a r e continuous
x'
Y
at a . Thus, t h e fundamental r e s u l t of t h i s assignment h o l d s ; namely,
*N o t e
t h a t F i g u r e 4 i s i n d e p e n d e n t o f w.
That i s , a l l F i g u r e 3
t a k e s i n t o a c c o u n t i s what i s happening i n t h e x y - p l a n e .
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
3 . 3 . 3 (L) continued
Aw = fx(1,2)Ax
+
f
Y
(1,2)Ay
+
klAx
+
k2Ay
where kl and k2+0 a s Ax and Ay+O.
Therefore,
(This was d e r i v e d i n both t h e t e x t and t h e l e c t u r e , b u t i t i s
important enough s o t h a t you should see it again.)
L e t t i n g A s + O makes both Ax and Ay+O, s o t h a t kl and k2+0
o r since
dw
ds
I
dx
=
ds
cos
= f(l.2)
and
coo (
= sin (
+ i
Y
.
(1,2) s i n (
o r i n d o t product n o t a t i o n
Our f i r s t f a c t o r on t h e r i g h t s i d e of (7) i s $f (1,2) which depends
only on f and t h e p o i n t ( 1 , 2 ) while our second f a c t o r 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 of s which i s independent of f and ( 1 , 2 ) .
Pictorially,
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
- - - --
--
-
3.3.3 (L) c o n t i n u e d
c o s 4)
Thus,
( 7 ) becomes
s i n c e f ( x , y ) = x5+x3y2+y4, w e have f x ( x I y ) = 5x4+3x2y2 and
3
3
Hence, f x ( 1 , 2 ) = 1 7 and f ( x , y ) = 36.
f ( x , y ) = 2x y+4y
Y
Y
Theref o r e NOW,
.
S i n c e t h e v e c t o r from ( 1 , 2 ) t o ( 3 , 4 ) i s
P u t t i n g t h i s r e s u l t and t h e r e s u l t of ( 9 ) i n t o (8) we o b t a i n
Solutions Block 3: Partial Derivatives Unit 3: Differentiability and the Gradient 3.3.3 (L) continued .
which agree& with our result in (a)
c.
From ( 8 ) and (9)
If our direction s is now determined by the vector from (1,2) to (4,6), we have, Hence, (10) becomes Notice how much more convenient this approach is than if we had to resort to the approach of part (a) every time we wanted to find a directional derivative. d.
*
I
Hopefully, we see at a glance from (10) that ds is maximum
(1121
when GS is in the direction of (17.36) , and, in this event,
-P
-P
= (17,36)-u = 11(17,36)11 [since u
S
S
is parallel to (17.361,
Solutions Block 3: Partial Derivatives
Unit 3: Differentiability and the Gradient
3.3.4
a. Since w
=
f(x,y) = eX
X
fx(x,Y) = e
fx(kn 2,
sin y
= e
IT y
+
x2YI
cos y + ~ X Y
X
f (x,y) = -e
Y COS
+
x
cos
2
$+
"'sin 7 +
f (an 2, Z) = -e
Y TT
(2 an 2)
IT
= n an 2
(tn2) = -2+ (an 2)
2
Therefore, d. The direction in which dw
is maximum is in the direction of the
2
Rn 2,-2 + (an 2) I
gradient, which we computed in part (a) to be [TT
*1
max ds .
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
The main aim of t h i s e x e r c i s e i s t o show t h a t t h e d e f i n i t i o n of
t h e g r a d i e n t v e c t o r i s independent of any p a r t i c u l a r coordinate
system, b u t t h a t t h e form of it does depend on t h e coordinate
system.
I n p a r t i c u l a r w e wish t o show t h a t t h e convenient expression
+
fX1 + f
3
which e x p r e s s e s t h e g r a d i e n t of f i n terms of C a r t e s i a n
Y
c o o r d i n a t e s i s a r a t h e r s p e c i a l case. For example, i f one were
merely t o memorize t h e expression f o r t h e g r a d i e n t i n C a r t e s i a n
c o o r d i n a t e s , one might e x p e c t t h a t i n , say, p o l a r c o o r d i n a t e s
:
-f
where t h e b a s i c u n i t v e c t o r s a r e
r and u g , t h e g r a d i e n t would be
-t
+
given by frur + feu8. The aim of t h i s e x e r c i s e i s t o show t h a t
n o t t h e case.
t h i s is a.
Given t h a t r and 8 a r e t h e independent v a r i a b l e s of p o l a r
aw = s i n 8 and
c o o r d i n a t e s , we have from w = r s i n 8 t h a t -f
t
-+
it
- = r cos 8. Since +
ur = cos 81 + s i n 61 and ug = - s i n 8 ~ + c o s 8 ]
ae
[why?). it follows t h a t
= sin 8 (cos e I + s i n
= ( s i n 8 cos 8
Recalling t h a t s i n 8 =
t
t
e$) + r cos 8 ( - s i n 81+cos
83)
-f
2
2 t
-r s i n 8 cos 8) I+ ( s i n 8+r cos 8) 3
$ and
X
cos 8 = F, we o b t a i n
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3 : D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3 . 3 . 5 (L) c o n t i n u e d
b. I f w e c o n v e r t t o C a r t e s i a n c o o r d i n a t e s , we have t h a t w = r s i n 8
implies
whence
Hence,
But w e have a l r e a d y s e e n t h a t
Hence, from ( 2 )
aw
?w = ax
:
, +
aw t
aY 3 .
qw=;.
While p a r t s ( a ) and ( b ) s h o u l d n o t have caused you t o o much
d i f f i c u l t y t o o b t a i n , t h e i m p l i c a t i o n we want t o make i s t h a t w e
must n o t be s p o i l e d by t h e convenience of C a r t e s i a n c o o r d i n a t e s
and conclude t h a t i f w i s e x p r e s s e d i n terms of u and v and i f
+
eU and 2 r e p r e s e n t t h e u n i t v e c t o r s i n t h i s new c o o r d i n a t e system.
aw
+ av 2v r e p r e s e n t s t h e g r a d i e n t of w w i t h r e s p e c t t o
then au u
u and v. To be s u r e w e can compute t h e v e c t o r , b u t i t need n o t
be t h e g r a d i e n t . T h i s i s p r e c i s e l y what w e showed i n t h i s e x e r c i s e .
f
*
Namely, e q u a t i o n (1) made w e l l - d e f i n e d s e n s e , b u t e q u a t i o n (2)
showed t h a t (1) was n o t t h e g r a d i e n t v e c t o r .
With r e s p e c t t o t h i s p a r t i c u l a r e x e r c i s e , w e may u s e t h e f o l l o w i n g
"geo-analytic"
a p o i n t i n terms of
*See
<
If we e l e c t t o write a l l quantities a t
+
and u t h e n
approach.
0
N o t e at the end of this exercise. Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3 : D i f f e r e n t i a b i l i t y and t h e Gradient
3.3.5 (L) continued
+
+
i f w = [ I u r + ( 1 Ue
+
We u t i l i z e t h e f a c t t h a t ur
-t
u
0
= 0. t o o b t a i n from (3) t h a t
while
+ +
But Vw*ur, by t h e d e f i n i t i o n of t h e g r a d i e n t , i s t h e d i r e c t i o n a l
+
+
d e r i v a t i v e of w i n t h e d i r e c t i o n of ur, and s i n c e ur i s measured
r a d i a l l y , i t seems g e o m e t r i c a l l y c l e a r t h a t t h i s d i r e c t i o n a l
aw
d e r i v a t i v e i s simply -
ar
That i s ,
In a similar
+
of uO. Now,
things) t h i s
'
-+
manner, ifw-u0 i s t h e d e r i v a t i v e of w i n t h e d i r e c t i o n
+
u0 i s n o t i n t h e d i r e c t i o n of 0 , s i n c e (among o t h e r
doesn't mean anything. What i s t r u e is t h a t t h e
+
and
d i r e c t i o n of u0 i s t h a t of a 90' p o s i t i v e r o t a t i o n of
f o r small A8, t h e change a t r i g h t a n g l e s t o ur i s measured by rA8.
I n o t h e r words, our geometric i n t u i t i o n might now l e a d us t o
s u s p e c t t h a t t h e d e r i v a t i v e of w i n h e d i r e c t i o n of Go i s
lim
A0+O
AW
3
or,
aw
-1
r ae .
I n o t h e r words,
Without worrying how t o determine t h i s r e s u l t more r i g o r o u s l y
( o r f o r t h a t m a t t e r , more g e n e r a l l y i n t h e case t h a t t h e c o o r d i n a t e
system i s n o t a s simple a s P o l a r c o o r d i n a t e s ) , i t appears t h a t a
reasonable guess f o r w i s
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3 . 3 . 5 ( ~ ) continued
To conclude o u r d i s c u s s i o n , l e t us compute t h e r i g h t s i d e o f
( 6 ) i n o u r p r e s e n t e x e r c i s e and see how i t compares w i t h t h e
g r a d i e n t found i n p a r t ( b ) .
TO t h i s end: W e a l r e a d y know t h a t +
*
t
+
-t
t
-a W- - s i n 0 , a w - r c o s 8, ur
= c o s + s i n 01 , and ue =-sin Bl+cose]
ar
ae
Hence,
= s i n 8 (coo 8:
2
= (sin 8
+
+
s i n 03)
+
EOS
2
t
t
c o s 8) 3 = 3
-t
8 ( - s i n 01
+
COB
0;)
I and t h i s a g r e e s w i t h $w a s determined i n p a r t ( b ) of t h i s e x e r c i s e . -
Note :
Suppose w e a r e g i v e n t h a t
and t h a t w e a l s o know t h a t x and y a r e themselves f u n c t i o n s of two o t h e r v a r i a b l e s , s a y , u and v.
x = g ( u , v ) and y = h ( u , v ) P u t t i n g ( 2 ) i n t o (1) y i e l d s
For example, Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
and from (3) we s e e t h a t w i s now a f u n c t i o n of u and v.
(This
discussion w i l l ultimately generalize i n a l a t e r u n i t t o the
chain r u l e f o r s e v e r a l v a r i a b l e s , b u t t h i s i s n o t important a t
t h i s moment.)
The p o i n t i s t h a t how u and v must be combined t o y i e l d a
p a r t i c u l a r w need n o t be t h e same a s how x and y must be combined
To make t h i s p o i n t more c o n c r e t e l y ,
t o y i e l d t h i s same v a l u e of w.
suppose t h a t
w = f(x,y) = x
+
y
and t h a t
x = u2 while y = -v 2
P u t t i n g ( 2 ' ) i n t o (1') y i e l d s
The p o i n t i s t h a t (1') and ( 3 ' ) d e f i n e e n t i r e l y d i f f e r e n t f u n c t i o n s .
Namely i n (1') we add t h e f i r s t component t o t h e second t o o b t a i n
w, while i n ( 3 ' ) w e s u b t r a c t t h e square of t h e second component
I n o t h e r words, we
from t h e square of t h e f i r s t t o o b t a i n w.
could have r e w r i t t e n ( 3 ' ) a s
where we use 'F t o i n d i c a t e t h a t t h e r u l e f o r combining u and v
t o form w i s d i f f e r e n t from t h e r u l e f o r combining x and y t o
y i e l d w.
Returning t o t h e g e n e r a l c a s e , (3) may be w r i t t e n i n t h e form
I f we now want t o form t h e p a r t i a l of w with r e s p e c t t o u, it i s
c l e a r t h a t w e a r e r e f e r r i n g t o F U ( u , v ) , while i f we wish t o form t h e
p a r t i a l of w w i t h r e s p e c t t o x , it i s c l e a r t h a t we a r e r e f e r r i n g
t o fx(x,y).
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
With r e g a r d t o t h e p r e s e n t e x e r c i s e , when w e w r i t e
ar
or aa w0 we
a r e r e f e r r i n g t o t h e formula w = r s i n €Iw h i l e i f w e w r i t e
aW o r aW
w e a r e r e f e r r i n g t o t h e formula w = y.
ax
ay
3.3.6tL)
T h i s e x e r c i s e i s a c t u a l l y a b u f f e r ( i n terms o f i t s geometric
simplicity) t o help us c l a r i f y a point t h a t i s often d i f f i c u l t
f o r the beginner t o grasp.
Given w = f ( x , y ) , w e have s e e n
g r a p h i c a l l y t h a t t h e g r a d i e n t i s a v e c t o r i n t h e xy-plane which
dw and whose magnitude i s
p o i n t s i n t h e d i r e c t i o n of t h e maximum
t h i s maximum v a l u e .
Now " a l l of a sudden" ( i n p a r t i c u l a r , see
w e a r e t a l k i n g about t h e g r a d i e n t
n o t b e i n g i n t h e xy-plane b u t r a t h e r b e i n g normal t o o u r s u r f a c e .
There i s a s u b t l e ( b u t e x t r e m e l y i n p o r t a n t ) d i f f e r e n c e between
t h e r o l e of $f when w e t h i n k of f i n t h e form, w = f ( x , y , z ) ;
o r i n t h e form f ( x , y , z ) = c.
E x e r c i s e s 3 . 3 . 7 and 3.3.8)
To see t h i s d i f f e r e n c e a s c l e a r l y a s p o s s i b l e , it was o u r f e e l i n g
t h a t w e s h o u l d i n t r o d u c e t h e i d e a w i t h an example i n v o l v i n g an
even lower dimension t h a n t h a t d i s c u s s e d i n t h e t e x t .
2
We have t h a t w = f ( x , y ) = y - x
.
t
From t h i s it i s e a s y t o see t h a t $f i s g i v e n by -2x1
+
1. T h i s
t e l l s u s t h a t i f w e move from t h e p o i n t (xo,yo) t h e maximum
dw
directional derivative, ds
,
is
+.
and t h i s o c c u r s i n t h e d i r e c t i o n -2x 01
So f a r , s o good!
+
t
1.
Now, w e c o n s i d e r a completely d i f f e r e n t q u e s t i o n .
( x , y ) i n t h e xy-plane f o r which
W e look a t a s p e c i a l s e t of p o i n t s
f ( x , y ) = 4; i . e . ,
y
-
x
= 4.
I n t h i s e v e n t , x and y a r e no
l o n g e r independent v a r i a b l e s s i n c e t h e r e s t r i c t i o n t h a t f ( x , y ) = 4 ,
2
o r i n t h i s c a s e , y-x = 4 , f i x e s t h e v a l u e of y , f o r example, once
t h e v a l u e o f x i s known.
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
3.3.6 (L) continued
While it i s c l e a r t h a t y-x2 = 4 i s a curve i n t h e xy-plane,
2
r e l a t i v e t o w = y-x , t h i s curve c o n s i s t s p r e c i s e l y of t h o s e
p o i n t s f o r which w=4.
I n o t h e r words, we say t h a t y-x2 = 4 i s an
I f we l e t C denote
e q u i p o t e n t i a l curve of t h e s u r f a c e , w = y-x 2
d
w
t h e curve y-x2 = 4 , i t follows t h a t
must be i d e n t i c a l l y zero,
.
where s denotes t h e d i r e c t i o n of C ( i . e . ,
s denotes t h e d i r e c t i o n
of t h e l i n e t a n g e n t t o C a t any p o i n t ) , s i n c e w i s c o n s t a n t along
C.
But w e know t h a t
-+
where us i s t h e un it t a n g e n t v e c t o r C a t any given p o i n t .
Since
dw - 0, it must be t h a t e i t h e r $f i s d o r t h a t $f i s perpendicular
ds
+
t o Gs, s i n c e t h e s e a r e t h e only ways i n which $£-us can e q u a l zero
+
( s i n c e u_ i s a u n i t v e c t o r , i t can1t e q u a l 8) Now we have a l r e a d y
A
a
seen t h a t $f = -2x
+
+
1+7;
.
and s i n c e t h e c o e f f i c i e n t of j i s always
1, we see t h a t $f i s never e q u a l t o
-+
Vf i s p e r p e n d i c u l a r t o C.
6.
Hence, i t follows t h a t
AS a quick check, w e may observe t h a t t h e s l o p e of y = x2-4 a t
+ t while t h e s l o p e of -2x1+-~a t t h i s same
(xo,y0) i s given by 2x
0'
Since 2xo and 1/-2xo a r e n e g a t i v e r e c i p o r c a l s ,
p o i n t i s 1/-2xo.
we s e e t h a t t h e g r a d i e n t of f a t (xo,yo) g i v e s t h e s l o p e of the
l i n e normal t o f ( x , y ) = 4 a t t h i s p o i n t .
I n a s i m i l a r way, when we work with w = f ( x ,y, z) , t h e g r a d i e n t of
f i s a v e c t o r i n 3-dimensional space whose d i r e c t i o n and magnitude
dw
t e l l us t h e d i r e c t i o n i n which d s i s a maximum and what t h e value
of t h i s maximum i s . When, i n s t e a d , we work with t h e s u r f a c e
(2 degrees of freedom) f (x,y , z ) = c o n s t a n t , then t h e g r a d i e n t of
f i s a v e c t o r normal t o t h i s s u r f a c e .
between w = f ( x , y , z ) and f (x,y,z) = c.
The key i s t o d i s t i n g u i s h
In t h i s p a r t i c u l a r e x e r c i s e
w e compared w = f ( x ,y ) with f (x,y ) = c o n s t a n t .
Solutions
Block 3 : P a r t i a l D e r i v a t i v e s
U n i t 3: D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.6(L)
continued
A s a f i n a l n o t e on t h i s e x e r c i s e , w e should l i k e t o d i s c u s s t h e
n o t i o n of o r t h o g o n a l t r a j e c t o r i e s .
parameter f a m i l y of c u r v e s ( i . e . ,
arbitrary constant).
Suppose w e a r e g i v e n a 1t h e f a m i l y c o n t a i n s one
The most g e n e r a l r e p r e s e n t a t i o n of such a
family i s
( T h a t i s , t h e r e i s some f u n c t i o n a l r e l a t i o n s h i p between x , y ,
and an a r b i t r a r y c o n s t a n t , c )
.
We c a l l t h e f a m i l y of c u r v e s g ( x , y , c ) = 0 a f a m i l y of o r t h o g o n a l
t r a j e c t o r i e s t o t h e f a m i l y f ( x , y , c ) = 0 i f a t e v e r y p o i n t of
i n t e r s e c t i o n between a member of g ( x , y , c ) = 0 and a member of
f ( x , y , c ) = 0 , t h e a n g l e o f i n t e r s e c t i o n i s 90" (hence t h e name
orthogonal).
R e c a l l t h a t a n g l e s between c u r v e s a r e d e f i n e d t o
be t h e a n g l e s between t h e i r t a n g e n t l i n e s a t t h e p o i n t of i n t e r section.
Computationally, t h e way w e f i n d a f a m i l y of o r t h o g o n a l t r a j e c t o r i e s
( i f i n d e e d such a f a m i l y e x i s t s ) , i s t h a t w e compute
f ( x , y , c ) = 0 i n a way t h a t e l i m i n a t e s c .
9
in
dx t h e form
= h (x. y )
.
from
That i s , we e x p r e s s
W e then solve the d i f f e r e n t i a l
equation t h e i d e a b e i n g t h a t t h e r e s u l t i n g c u r v e s must be o r t h o g o n a l t o
t h o s e of t h e f i r s t f a m i l y a t t h e p o i n t s of i n t e r s e c t i o n by v i r t u e
of t h e f a c t t h a t t h e i r s l o p e s a r e negative r e c i p r o c a l s .
I n o u r p a r t i c u l a r example, w e have t h a t t h e 1-parameter f a m i l y
o f c u r v e s y-x2 = c (which a r e t h e e q u i p o t e n t i a l c u r v e s f o r
w = y-x 2 ) s a t i s f i e s t h e d i f f e r e n t i a l e q u a t i o n
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
3.3.6 (L) continued
Hence t h e family of orthogonal t r a j e c t o r i e s i s given by t h e
d i f f e r e n t i a l equation
Solving t h i s equation y e i l d s
Hopefully, it i s c l e a r t h a t t h e n o t i o n of orthogonal t r a j e c t o r i e s
i s n o t r e s t r i c t e d t o g r a d i e n t s and e q u i p o t e n t i a l curves, b u t
Suppose we
t h e r e i s an i n t e r e s t i n g a p p l i c a t i o n i n t h i s r e s p e c t .
a r e given t h e s u r f a c e w = f ( x , y ) and we wish t o s t a r t a t t h e p o i n t
(xo,yo) and move i n such a way i n t h e xy-plane t h a t t h e h e i g h t of
t h e s u r f a c e i s always changing a s r a p i d l y a s p o s s i b l e .
What we do i s s k e t c h t h e e q u i p o t e n t i a l curves of w.
W e then
s t a r t a t (xo,yo) on t h e e q u i p o t e n t i a l curve wo = f ( x , y ) , where
w0 i s t h e value of f ( x o , y o ) , .and we move along t h e orthogonal
t r a j e c t o r y t o t h i s e q u i p o t e n t i a l curve. I n o t h e r words (and t h i s
w
may seem c l e a r from an i n t u i t i v e p o i n t of view), t o make d
-maximum
ds
we must " c u t a c r o s s " t h e family of e q u i p o t e n t i a l curves a s s w i f t l y
a s p o s s i b l e , and t h i s i s done by moving orthogonally t o t h e family.
A t any r a t e ,
i t i s o u r hope t h a t t h i s d i s c u s s i o n has c l e a r e d up
t h e d i s t i n c t i o n between t h e r o l e played by $f when we consider
w = f (x,y) and t h e r o l e of ?f when we c o n s i d e r f ( x , y ) = c. The
n e x t two e x e r c i s e s extend t h i s n o t a t i o n t o w = f ( x , y , z ) and
f ( x , y , z ) = 0.
Solutions
Block 3 : P a r t i a l D e r i v a t i v e s
U n i t 3 : D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.7(L)
I n o u r p r e v i o u s e n c o u n t e r s w i t h f i n d i n g normals t o s u r f a c e s , o u r
s u r f a c e was i n t h e form z = f ( x , y ) . I n t h e p r e s e n t e x e r c i s e ,
i t i s a t b e s t d i f f i c u l t t o s o l v e f o r any v a r i a b l e e x p l i c i t l y i n
t e r m s o f t h e o t h e r two.
Now, g i v e n
w e may t h i n k o f a new f u n c t i o n
I n ( 2 ) , x, y , z a r e independent v a r i a b l e s .
Again, s i n c e g i s a
polynomial i n x , y , and z , w e may compute t h e g r a d i e n t of g t o
obtain
therefore,
Now, e q u a t i o n (1) may be viewed a s a s p e c i a l s e t of p o i n t s , S,
f o r which g ( x , y , z ) = 3 f o r a l l ( x , y , z ) S.
I n o t h e r words,
r e s t r i c t e d t o S, g i s a c o n s t a n t .
Hence, on S, Ag = 0 .
+
Ag = $gous a l o n g any c u r v e s i n t h e s u r f a c e S. That i s
But -+
+
-+ AS
% * u s must be z e r o ; y e t , from ( 3 ) bg#b, n o r can u s=d s i n c e u s i s
a u n i t vector.
-+
Hence, $g-us = 0
-+
$gl
Gs .
+
S i n c e $g i s p e r p e n d i c u l a r t o each curve on S a t (1,1,1), Vg i s
normal t o S.
I n o t h e r words
6f
+
57
+
82
5 4 3
i s a v e c t o r normal t o t h e s u r f a c e x +y z +xyz5 = 3 a t t h e p o i n t
( 1 1 ~ 1 1 )
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
A Note on Normal Lines t o Surfaces
Suppose we a r e given t h e s u r f a c e whose equation i s , s a y ,
.
f(x,y,z) = c
There i s a danger t h a t you may confuse two q u i t e d i f f e r e n t p o i n t s
h e r e . For t h e sake of argument, l e t us suppose t h a t equation
(1) can be s o l v e d f o r z, e x p l i c i t l y i n terms of x and y.
That i s ,
l e t u s suppose t h a t t h e s u r f a c e S whose equation i s given by
equation (1), i s a l s o given by
I t i s important t o n o t e t h a t i t i s t h e g r a d i e n t of f , n o t 9, t h a t
i s normal t o t h e s u r f a c e S. I n f a c t , t h e g r a d i e n t of g i s a v e c t o r
i n t h e xy-plane, and it i s h a r d l y l i k e l y t h a t t h e normal t o t h e
s u r f a c e S would always be i n t h e xy-plane ( o r p a r a l l e l t o i t ) .
To see t h i s from a d i f f e r e n t p e r s p e c t i v e ,
equation (2)
.
suppose w e s t a r t with
W e could then d e f i n e a f u n c t i o n f ( x , y, z ) by r e w r i t i n g
( 2 ) as
and l e t
That i s , t h e g r a d i e n t of f i s normal
I n t h i s c a s e , f ( x , y , z ) = 0.
t o t h e s u r f a c e d e f i n e d by ( 2 ) s i n c e t h i s s u r f a c e i s an e q u i p o t e n t i a l
s u r f a c e of f ( x , y , z ) .
A s a check, n o t i c e t h a t t h e g r a d i e n t of f i n t h i s case i s t h e
vector
and t h i s checks with o u r e a r l i e r r e s u l t t h a t t h e t a n g e n t plane t o
t h e s u r f a c e z = g ( x , y ) h a s the equation:
2-zo = gx(x0,yo) (x-xo)+g (x ,yo) (y-yo)
Y
I
O
from which we can see t h a t t h e normal i s indeed a s given by ( 4 ) .
Solutions
Block 3 : P a r t i a l D e r i v a t i v e s
U n i t 3 : D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.8
W e have
4 6
5
f ( x , y , z ) = x +y z+xyz
Hence,
$f = (4x3 +yz5 , 6y 5 z+xz5 , y 6 +5xyz 4 )
theref ore
(1
~ f ( l , l , l )= ( 5 , 7 , 6 )
4 6
S i n c e x +y z+xyz5 = 3 d e f i n e s an a q u i p o t e n t i a l s u r f a c e of f ( i . e . ,
Af = 0 on S ) , w e have from (1) t h a t 5:+75+6$
i s normal t o S a t
(1,111).
T h e r e f o r e t h e e q u a t i o n of t h e t a n g e n t p l a n e i s
( N o t i c e t h a t t h e u s e o f t h e normal and t h e e q u a t i o n of a p l a n e a r e
t h e same a s always; a l l t h a t ' s new i n t h e l a s t two problems i s t h e
t e c h n i q u e f o r f i n d i n g t h e normal t o a s u r f a c e . )
3.3.9
a.
W e simply invoke t h e t e c h n i q u e of w r i t i n g ab+cd a s ( a ,c) ( b, d )
This y i e l d s
I f w e now l e t I f = (kl , k 2 ) ,
Ax =
[ f x (al,a2) , f
(alla2) I i s
1
X2
Vf
(Axl ,Ax2) and observe t h a t
(a), w e
obtain
.
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
3.3.9
continued
The v a l i d i t y of
( 2 ) required t h a t f x
and f x e x i s t and be
1
2
continuous a t x = a . Once t h e s e requirements a r e m e t we observe
t h a t equation ( 2 ) b e a r s a s t r o n g resemblance t o t h e 1-dimensional
r e s u l t t h a t i f f i s d i f f e r e n t i a b l e a t x = a , then
lim k = 0
Af = f ' (a)Ax+kAx, where Ax+O
The i n t e r e s t i n g p o i n t i s t h a t (3) i s a s p e c i a l case of ( 2 ) .
That
a ) with t h e n-dimensional analog of f ' ( a )
i s , i f we i d e n t i f y Vf (and r e a l i z e t h a t i n 1-dimensional space t h e d o t product and t h e
"ordinary p r o d u c t w a r e t h e same, we s e e t h a t when t r a n s l a t e d i n t o
1-space equation ( 2 ) becomes equation ( 3 )
.
More g e n e r a l l y t h e most important a s p e c t of ( 2 ) i s t h a t i t a p p l i e s
i n every dimension. Granted t h a t o u r t e x t proved it i n t h e 2dimensional case and i n d i c a t e d t h a t t h e proof a p p l i e d almost
verbatim i n h i g h e r dimensions, t h e f a c t i s t h a t t h e t e x t was conducive
t o our concluding t h a t t h e r e were geometrical reasons f o r (2)
Our aim now i s t o show t h a t t h e technique of t h e t e x t
does apply v i r t u a l l y word-for-word t o h i g h e r dimensions i n which
t h e r e i s no i n t u i t i v e appeal t o geometry. We have chosen t h e case
being t r u e .
The p o i n t i s t h a t our proof reviews t h e one i n t h e t e x t
a s we show t h a t i t a p p l i e s t o a l l dimensions. Admittedly, t h e
n = 4.
proof i s a b s t r a c t b u t w e f e e l t h a t it i s important f o r you t o
s u f f e r through it i n t h e hope t h a t you w i l l once and f o r a l l g e t
b.
t h e t r u e f e e l i n g of what t h e r e s u l t means.
4
4
W e have t h a t f:E +E i s continuous and t h a t a = (al,a2 , a 3 , a 4 ) ~ E
.
,f
,f
, and f a l l e x i s t and a r e X4 X1 X2 X3
W e want t o show t h a t W
e a r e a l s o t o l d t h a t f
continuous a t
2.
Solutions
Block 3 : P a r t i a l D e r i v a t i v e s
Unit 3 : D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3 . 3 . 9 continued
The " t r i c k " i n problems o f t h i s t y p e i s t o add and s u b t r a c t t e r m s
i n such a way t h a t i n each d i f f e r e n c e a l l b u t one o f t h e v a r i a b l e s
a r e being held constant.
A l l t h a t i s r e q u i r e d t o make s u r e t h a t
w e can do t h i s i s t h a t t h e v a r i a b l e s be independent.
The o n l y o t h e r " t r i c k " i s t h a t t h e r e a r e many d i f f e r e n t ways t h a t
w e can add and s u b t r a c t t h e terms. The key p o i n t i s t h a t by
c o n t i n u i t y , w e need n o t worry a b o u t more t h a n one p a r t i c u l a r p a t h ,
s i n c e c o n t i n u i t y g u a r a n t e e s u s t h a t whatever answer w e g e t along
one p a t h w e g e t a l o n g any o t h e r p a t h a s w e l l .
Thus, f o r example, w e may w r i t e
*In
n-tuple n o t a t i o n ,
Af = f(++Ax)-f(2)
where
Ax
=
(Ax1,Ax2,Ax3,Ax4)
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
3.3.9
continued
Equation ( 4 ) may then be r e w r i t t e n s o t h a t t h e v a r i o u s p a r t i a l
d e r i v a t i v e s a r e emphasized (and t h i s i s why we assume t h a t
, f , f , and f x e x i s t i n a neighborhood of a , f o r we
X1 X2
X3
4
c a n ' t r e a l l y t a k e advantage of t h e p a r t i a l d e r i v a t i v e s i f they
f
don't exist!)
We then o b t a i n f ram ( 4 ) t h a t
The f i r s t bracketed expression i n (5) i s r e l a t e d t o f x i n t h e
1
sense that
lim 1
f (al+Axl, a2+Ax2,a3+Ax3,a4+Ax4 -f (aYa2+Ax2,a3+Ax3 ,a4+hx4
Thus f o r a f i x e d Axl#O,
t h e d e f i n i t i o n of
lim i m p l i e s t h a t
Axl+O
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3 : D i f f e r e n t i a b i l i t y and the G r a d i e n t
3 . 3 . 9 continued
where
lim
Ax1-0
'1
= 0.
Using s i m i l a r r e a s o n i n g w e may a l s o o b t a i n
~ h u s( 5 ) becomes
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
Unit 3: D i f f e r e n t i a b i l i t y and t h e Gradient
- - -
3.3.9
continued
where
E ~ , c2
,
c3 , c4'f0 a s
Ax2 , Ax3, Ax44
Axl,
A l l t h a t keeps (6) from being t h e c o r r e c t answer a r e t h e p o i n t s a t
, f , f a r e computed. To o b t a i n t h e d e s i r e d r e s u l t , we
1
X2
X3
f
, and f a r e continuous a t
need only assume t h a t f x , f
1 X2
X3
X4
[Actually a s we a r e doing t h e problem i t
a = (al ,a 2 , a 3 ,a 4 )
which f x
,
-
.
s i n c e i t is being
appears t h a t we have no w o r r i e s about f
X4
evaluated a t t h e desired point
5.
However, a s w e mentioned we
could have chosen o t h e r paths i n which c a s e f x might have been
e v a l u a t e d a t a d i f f e r e n t p o i n t . The assumptiod, t h e r e f o r e , t h a t
is continuous a t
X4
a d i f f e r e n t path.]
f
5 p r o t e c t s us i n t h e event w e wish t o use
i s continuous a t a means
Namely, f o r example, t h e f a c t t h a t f x
1
Solutions Block 3: Partial Derivatives Unit 3: Differentiability and the Gradient 3.3.9 continued That is, where c +O as AxZ, Ax3, Ax4+0
1 .
Similarly, where c2+0 as Ax3, Ax4+0 f (al,a2,a3,a4+Ax4) = fx (alIa2,a3,a4)+ c3 X3
3 where c3+0 as Ax4+0 whereupon ( 6 becomes
Letting kl = E 1+c1' k2 = E ~ + c k3=
~ , E ~ + c k4
~ ,= E ~ and
, abbreviating
(alra2,a3 ,a4) by 5, ( 7 ) becomes
Solutions
Block 3: P a r t i a l D e r i v a t i v e s
U n i t 3 : D i f f e r e n t i a b i l i t y and t h e G r a d i e n t
3.3.9
continued
W e are now home f r e e i f w e i n t r o d u c e t h e n o t a t i o n
+
AZ = ( A x 1 , A ~ 2 , A ~ 3 , A ~ 4 )V
, f (a)
=
Ifx
(a),f
1X2
(511
X3
4
and
k
- = (kltk2
for, then,
tk3
,k4)
( 8 ) becomes
lim k
0
Af = ?£(a)-Ax+k=Ax
- - - - , where Ax+O
-= a f i n a l n o t e on t h i s e x e r c i s e , n o t i c e t h a t w e can always d e f i n e
t h e g r a d i e n t o f f a t (a ) = ( a l , a 2 , a 3 , a 4 ) t o be t h e 4-tuple
As
a ) ] , b u t i f w e want t h e d i f f e r e n c e
(5),f x (a ) f X4 (( a ) ,f
X1X2
3
between Af and V f (a ) * A-x t o go t o z e r o f a s t e r t h a n Il~xll , t h e n
[f
,
w e must i n s i s t t h a t f x ,f
,f
, and f
a l s o be continuous a t 5.
1 X2 X3
X4
S i n c e we s h a l l o f t e n r e q u i r e t h a t t h e above d i f f e r e n c e does
approach z e r o " s u f f i c i e n t l y f a s t , " w e s h a l l a g r e e t h a t even though
( a ) 1 e x i s t s once t h e p a r t i a l
t h e $ - t u p l e [ f x '(5),f x (5),f x (5),f
1
2
3
X4 derivatives e x i s t a t a , we w i l l n o t c a l l t h i s 4-tuple t h e g r a d i e n t
of f a t
5 u n l e s s t h e p a r t i a l s a l s o happen t o be continuous a t _a.
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Calculus Revisited: Multivariable Calculus
Prof. Herbert Gross
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