2) quiz tuesdayadwiggin/teachingfiles... · so in three dimensions f(x,y ÷), the graph is in three...
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Announcements.
I ) HW #4 due Thursday
2) Quiz Tuesday next week
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Lagrange Multiplier Example.
Arithmetic - Geometric Mean
Inequality :
If xi ,X2 ,.
. . ,×n are positive
real numbers ,
Y*III?lkt÷O
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For n=3
zj%< x+±z3
why is this true ?
Maximize ( Xyz )"
. same
as maximizing XYZ .
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want to maximize subject
to the constraint
×ty¥ =K,
same
as Xtytz =3 k .
Using calculus
Maximize FCX ,y,z)=×yz
Subject to 3k=×tytz=gG . ,⇒
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Lagrange Multipliers
÷< yz ,XZ
, xy )
0g = ( l,
I,I )
÷f = 7 pg
( yz , xz , xy ) = Xm- ( x , 7,1 )
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we get
D= Xy = yz = XZ
Ww×=Z ×=y
if YFO . since Z > 0 .
but yso !
Then ×=y= Z .
Use constraint :
xtytz = 3k
3×234 .
×=k ,y=k ,
2- =L ,
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We found that
f( × , y ,z ) = Xyz , subject
to ×ty¥-k ,
has
a maximum when
× =y=Z=K .
In this
case f ( × , y ,z ) = k3
( xytl"
=HM÷/ tytz
-
3
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This says :
(×yzY3=xt±z3when ×=y=Z ;
Otherwise ,we have
strict inequality !
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Integration.
( Chapter 15 )
we 've Seen two heads of the
Calculus Cerberus in two
or three variables - limits
and differentiation .
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In One Dimension.
Two types of regions :
intervals ( }f(x)dx
when a < b- the interval
is [ a ,bJ ) or points
( §fCx)d×=o )
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Q : what geometric concept
does the definite integral
capture ?
A : AREA ( under a
Curve with non- negative
y - values )
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Definition : ( one variable integral )
ha Ddx÷idle { fixes )where xi is a point in the
interval [ athilnbtatitbif
provided the limit exists !
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The definite integral
exists for all continuous
functions f.
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Picturepftxit
'y
- fcx )
¥#tEEje±¥
:* "
n÷telexes¥¥¥E
Approximate using rectangles ,
make the width go to zero .
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So in three dimensions
÷f(x,y ) ,the graph
is in three dimensions .
If fiyy ) > 0,the
integral over a region
Should give volume -
provided the region is
non. degenerate ! So
no points or one - dimensional
regions ( lines ,curves ,
etc. )
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Goodwife
EE'sIE
#weird 2 - D region
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Bad Regions of Integration.
.
•
point line segment
in"¥÷in.
NO.Curve
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The Definite Integral
÷2
,rectangle )
A solid rectangle in. 1122 is
denoted by
[ a ,b]x[c,d ]
={ ( x ,y ) / aexeb ,C Eyed }
Example [ 0,1 ]x[ 0,1 ]
istie " "
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Subdivide [ a ,b] into
n intervals of size b÷9 .
Subdivide [ Gd ] into
m intervals of sizeDIPick a point ( xit, ;
,y*gj)
in the rectangle
f.at#nh,atiEjaDxfttiIagctHf' ]
for each pair C i ,j ) with
I Ej< m,
lei En .
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Picture
:In ;) 7×3*2,95's ' cxiipiss ;)
explainKinlin cxziiysiis
[ 0,3 ]X[ 1,2 ]
m=2,
n =3
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Horrid Definition-
For
z=f(x,y ) defined on a
rectangle [ a ,bJt[c,d ] in 1123,
the definite integral off
over [ a ,b]×[ c,d ] is denoted
BT S f(×,y,d@Afor "area "
[a,Di[ Gd ]
and defined as
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ask.us#@=nnsnmphbTttnE.IfcxIjnIn)
provided the limit exists !
Again , if f is continuous ,
the limit will exist .
How
do you find it ?
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Fubinis Theorem.
If z=f(x,y ) is continuous
on [ a ,b]x [ C,d) ,
then
a!a¥gBBYv*@= § ( §bf( xiyldx )dy
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where
§f(×,y)dx means treat
f( yy ) as a function of × with
y constant , integrate with
respect to x. For
§f( x ,y)dy ,interchange the
roles of × and y.
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Examine : S ye*dA[ o ,1n(aDx[ 1,4 ]
Use Fubinis'
Theorem :
S ye⇒dA[0,1^121]×[1,41
= §4 § ye "dy)d×w
integration by parts!
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The integral is also equal to
§ ( §"
ye"dx ) dy
=§ly§ne*d×)ds( since y is a constant writ .
x )
= § ( us ( Elon "D⇒= § y1y\ ( end 'Ll ) dy
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= § (emihld "
=e÷Ynh"
=h¥i±ut¥¥s= h1#
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Easy Scenario.
If z=f(x,y)=g(x)h(y )
on [ a ,bJx[ Sd ] and f is
Continuous y )dA
C ["קYnd×.§n@
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Exampled S ×3
arctanclncusly)l)dA
G, ]Ji[ 0,731
By Fubinis'
theorem , this integral
is equal to
-
§X3dx' §3arctan( lnccusly ) ))dy
x0 since ×3 is odd and
the domain is symmetric .
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So the value of the
integral is Zero !
Note : I have no idea
how to integrate
arctanclnlcoscy ) ) ) .
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Boring Properties of the Integral.
Let R be the region
[ a ,b]x[9d ] in 1122,let
f, g be continuous real .
valued
functions on R .
Then
1) § ( flay ) tgiyy ) )dA
= § f(x,y)dAt§gGsy)dA
(integrals distribute over addition )
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21 If C is any constant ,
§ ( flay )dA=C§fCx,y)dA
( can pull constants out of integrals )
3) If D is a one or zero
dimensional region in IR ? then
§ fcxiylda = 0
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Fake Fubini Justification.
§§f(x,y)dy)dx
=§k÷¥{ fans ;D .
think one { Iiftxnisdx
=mli zd֤,
§fKy%d×
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anti;od÷{ ti;xb÷&fain ;D
antis "Ind÷±n'
,I Iftxitiit
= S f(x , y ) DA
[a ,D×[ c , d ]
Where did I lie to you ?