w w w x x x f a b , lim xx xy yx yy w w w fz - Penn Mathwziller/math114s14/ch14-5.pdf · h f a h b...
Transcript of w w w x x x f a b , lim xx xy yx yy w w w fz - Penn Mathwziller/math114s14/ch14-5.pdf · h f a h b...
h
bafbhafbaf
hx
,,lim,
0
baxf ,at respect to with of derivative Partial
Regard as a constant and differentiate , with respect to y f x y x
0
, ,, limy
h
f a b h f a bf a b
h
, x
f zf x y D f
x x x
, y
f zf x y D f
y y y
second partial derivatives , , , , , ,xx xy yx yyf x y f x y f x y f x y
yxfz ,dz z dx z dy
dt x dt y dt
, , ,x g s t y h s t s
y
y
z
s
x
x
z
s
z
, x g t y h t
t
y
y
z
t
x
x
z
t
z
Chain Rule:
Partial derivative of with respect to at ,f y a b
Review:
, 0 implicity defines as a function of :F x y y x
Set , yz F x
2Consider the function 2. Find at 1,4dy
xy x ydx
2 2F x y xy
1
22
xF xy yxy
2 1
2yF x x
xy
2
22
2
yxy
xydy
xdxx
xy
1,4
42 4
2 21
12 2
dy
dx
1 8 28
3 3
4
Implicit differentiation:
solve for : ( )y y f x
Differentiate with respect to : , y , f(x) 0x F x F x
0z F dx F dy
x x dx y dx
Example:
x
y
Fdy
dx F
F F dy
x y dx
x y
dyF F
dx
, , 0w F x y z
yxgz ,
x
z
Fz
x F
y
z
Fz
y F
2 2 2 is defined as a function of and by: 23 . z x y x y y z xz
2 2 2 23 0F x y y z xz
22xF xy z
2 2zF y xz
x
z
Fz
x F
2
2
2
2
xy z
y xz
2
2
1,2,3
2 1 2 3
2 2 1 3
z
x
13
10
Another implicit differentiation:
implicity defines as a function of and z x y
0 x y
w F dx F dy F z F F z zF F
x x dx y dx z x x y x x
similarly ,
Example:
Find at 1,2,3z
x
(Notice 1, 2,3 0 !)F
14.5
Directional DerivativesAnd Gradient Vectors
,z f x y
0 0, is the rate of change of in the direction parallel to the axis.xf x y z x
0 0, is the rate of change of in the direction parallel to the axis.yf x y z y
What about the rate of change of in other directions?z
slope 1,1xf
slope 1,1yf
0 0We want to find the rate of change of at ,
in the direction of an arbitrary vector , .
z x y
unit a bu
The slope of the tangent line to at the point
is the rate of change of in the direction of .
T C P
z u
Project and onto the
plane to get and .
P Q
xy P Q
Surface with equation ,S z f x y 0 0 0,z f x y
0 0 0, ,P x y zThe vertical plane that passes through
in the direction of intersects in a curve .
P
S Cu
Let , , be another point on .Q x y z C
,P Q h ha hb u
0 0 and ha x x hb y y
0 0 and x x ha y y hb
0 0, ,z f x y f x ha y hb
0 0 0 0( , ) ( , )slope of secant line:
f x sa y sb f x y
s
0 0Go along a straight line in the - plane, starting at ( , )
in the direction of at unit speed (arc length).
x y x y
u
2 2, with 1 a b a b u
0 0( ) ( , )t x y s r u 0 0( , ) ,x y s a b
0 0or , yx x sa y sb
0 0 0 0change in : ( , ) ( , ) z z f x sa y sb f x y
0 0 0 0
0
( , ) ( , )rate of change in the direction of lim
s
f x sa y sb f x y
s
u :
0 0
0 0
(Directional Derivative)
The directional derivative of , in the direction of at ,
is denoted by , :
f x y x y
D f x yu
Definition :
u
0 0 0 00 0
0
( , ) ( , ), lim
s
f x sa y sb f x yD f x y
s
u
or cos( ), sin( )a b
0 0,x sa y sb a
a b
b
0 0 0 00 0
0
( , ) ( , ), lim
s
f x sa y sb f x yD f x y
s
u
, with 1a b u u
cos( ),sin( )
a b i j
angle with the axisx
A picture of the directional derivative:
Special cases:
a) ( 1, 0) :a b u i
0 0 0 00 0
0
( s, ) ( , ), lim
s
f x y f x yD f x y
s
u
0 0,xf x y
b) ( 0, 1) :a b ju
0 0 0 00 0
0
( , s) ( , ), lim
s
f x y f x yD f x y
s
u
0 0,yf x y(here )h s
0 0 0 00 0
0
( , ) ( , )A simpler way to compute , lim
s
f x sa y sb f x yD f x y
s
u
0 0Set ( ) ( , )g s f x sa y sb
0 0Then ,D f x y u'(0)g
By the chain rule: '(0)g
0 0 0 0 0 0, ( , ) ( , ) where ,x yD f x y f x y a f x y b a b u u
Example:
If then u i, 0 0 0 0 , ( , )xD f x y f x yu
If then u j, 0 0 0 0 , ( , )yD f x y f x yu
This also shows that it is important that is a unit vector. u
0
( ) (0)lims
g s g
s
0 0 0 0( , ) ( , )x yf x y a f x y b
Example:2 2
0 0
Compute the directional derivative of ( , ) 310 10
at ( , ) (3, 1) in the direction of 1,-1 .
x xyD f f x y
x y
u
2
(3, 1)
23, 1
10 10x
x yf
6 1 5
10 10 10
(3, 1)
23, 1
10y
xyf
6
10
5 1 6 1
3, 110 102 2
D f
u
2
20
0 0 0 0 0 0, ( , ) ( , ) x yD f x y f x y a f x y b u
unit vector u1 1
,2 2
, with 1a b u u
Although decreases in the direction of the axis and axis,
it increases in the direction of !
f x y
u
The gradient vector:
0 0 0 0 0 0, ( , ) ( , ) where ,x yD f x y f x y a f x y b a b u u
0 0 0 0( , ), ( , ) ,x yf x y f x y a b
0 0 0 0 0 0 0 0The gradient vector at ( , ) is f ( , ) ( , ), ( , )x yx y x y f x y f x y
for short: f ,x yf f
If ( , , ), then the directional derivative is:w f x y z
0 0 0 0 0 0 0 0 0 0 0 0, , ( , , ) ( , , ) ( , , )
where , , is a unit vector
x y zD f x y z f x y z a f x y z b f x y z c
a b c
u
u
0 0 0, , where , ,x y zD f x y z f f f f f u u
0 0,D f x y f u u
0 0 0 0( , ), ( , )x yf x y f x y u
Find the directional derivative of , ,
at the point 1,0, 3 in the direction of 6 3 2
z xf x y z
z y
P
a i j k
6 3 2 36 9 4 49 7 a i j k a
, ,z x
f x y zz y
2 2
1, , , ,
x z y xf x y z
z y z y z y
1,0, 3 1,0, 3D f f u
u
6 3 2, ,
7 7 7
u
, ,xf x y z
2
z x
z y
1
or , ,f x y z z x z y
2
z y z x
z y
2
y x
z y
2
x z
z y
1 4 1
1,0, 3 , ,3 9 9
f
1 4 1 6 3 2
1,0, 3 , , , ,3 9 9 7 7 7
D f f
u u 1 6 4 3 1 2
3 7 9 7 9 7
18 12 2
63
8
63
Example:
1
z y
, ,yf x y z
, ,zf x y z
0 0What is the maximum value of the directional derivative at , ?x y
0 0 0 0, ,D f x y f x y u
u 0 0, cosf x y u
0 0, cosf x y
but 1u
This will be maximized when cos 1 and 0
0 0 0 0The maximum value of , is ,D f x y f x y u
0 0This value is obtained when =0, i.e. in the direction of , .f x y
0 0 0 0
0 0
The minimum value of the directional derivative at , is ,
and occurs when has the direction as , .
x y f x y
opposite f x y
u
0 0Fastest increase is , in the direction of f
f x yf
u
0 0Fastest decrease is , in the direction of f
f x yf
u
Example:2 2
Let ( , ) be the temparature of a hot plate 1 3,1 4.
An ant is sitting on the hot plate at ( , ) (2,3).
In what direction should he run to get off the plate without getting burned?
y xT x y x y
x y
x y
xT yT
3 2 2 3
2 1 1 2,
y xT
x y x y
6 1 1 4(2,3) ,
8 9 4 27T
4882(2,3)
108T
2 1By running in the direction of or about , ,
3 10
the ant can decrease the heat at a rate of about 50%.
T
plot3d({y/x^2+x/y^2,0},x=1..3,y=1..4,axes=boxed);
23 11,
36 108
3 2
2 1y
x y
2 3
1 2x
x y
A geometric interpretation of the gradient vector: f ,x yf f
Consider the level curves ( , ) where is some constant.f x y k k
Assume that ( ) ( ), ( ) is a parametrization of this level curvet x t y tr
Then ( ) '( ), '( ) is tangent to the level curvet x t y tr'
Along this curve we have ( ( ), ( ))f x t y t k
in other words: '( ), '( )f x t y t
is orthogonal to the level curves.f
Hence by the chain rule = df
dt' ' 0 x yf x f y
'( ) 0 f t r
0 0
0 0
, is perpendicular
to the level curve
, that passes
through the point ,
and points in the direction
of fastest increase
f x y
f x y k
P x y
On a topographical map, if , represents the height above sea level
at a point with coordinates , , the path of steepest ascent is perpendicular
to all the contour lines.
f x y
x y