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18.02Lecture1. Thu,Sept6,2007Handouts:
syllabus;PS1;flashcards.
Goalofmultivariablecalculus:
toolstohandleproblemswithseveralparametersfunctionsofseveralvariables.
Vectors. A vector (notation: A) has a direction, and a length
(|A|). It is represented bya directed line segment. In a coordinate
system its expressed by components: in space, A =a1, a2,
a3=a1+a2j+a3k.
(Recallinspacex-axispointstothelower-left,ytotheright,zup).
ScalarmultiplicationFormula for length? Showedpicture of 3,2,1
andusedflashcards toask for its length. Most
studentsgottherightanswer(14).why |A| = a
12+a
22+a
32 by reducing to the Pythagorean theorem in the
plane(Draw a picture,
showing
A
andits
projection
to
the
xy-plane,
then
derived
|A| from lengthofprojection+Pythagoreantheorem).Vectoraddition:
A+B byhead-to-tailaddition: inaparallelogram(showedhow
thediagonalsareA+B andB A);A =3+2j+k
onthedisplayedexample.Dotproduct.Definition: A B =a1b1+a2b2+a3b3
(ascalar,notavector).Theorem: geometrically, AB
=|A||B|cos.Explainedthetheoremasfollows: first,A A = A 2cos0=|A 2
isconsistentwiththedefinition. | | |
2Next, consider a triangle with sides A, B, C = A B. Then the
law of cosines gives C =|A|
2+|B|22|A||B|cos,whileweget
| |
C =C C= (AB) (AB) =|A 2+| 2A B.||2 | B|2 Hencethetheorem
isavectorformulationofthe lawofcosines.
A BApplications. 1)computing lengthsandangles: cos= .|A||B|
Example: triangle inspacewithverticesP
=(1,0,0),Q=(0,1,0),R=(0,0,2),findangleatP:P Q P R
cos=
= 1,1,01,0,2 = 1 , 71.5.| P R| 25 10P Q||
Note the sign of dot product: positive if angle less than 90,
negative if angle more than 90,zero ifperpendicular.
2)detectingorthogonality.Example: whatisthesetofpointswherex+
2y+ 3z=0? (possibleanswers: emptyset,apoint,
a line,aplane,asphere,noneoftheabove,Idontknow).Answer: plane;
can see by hand, but more geometrically use dot product: call A
=1,2,3,
P = (x,y ,z),thenAOP =x+ 2y+ 3z= 0 | OP|cos= 0=/2 OP. Sowe A||
AgettheplanethroughOwithnormalvectorA.
1
You can explain
addition works componentwise, and it is true that
Draw a picture
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218.02Lecture2.
Fri,Sept7,2007Weveseentwoapplicationsofdotproduct:
findinglengths/angles,anddetectingorthogonality.
Athirdone: findingcomponentsofavector. Ifu isaunitvector,A u
=|A|cosisthecomponentofA alongthedirectionofu. E.g.,A =componentofA
alongx-axis.
Example: pendulum making an angle with vertical, force = weight
of pendulum F pointingdownwards: then the physically important
quantities are the components of F along
tangentialdirection(causespendulumsmotion),andalongnormaldirection(causesstringtension).
Area. E.g. of a polygon in plane: break into triangles. Area of
triangle = 1baseheight=2
1|A||B|sin(=1/2areaofparallelogram).
Couldgetsinusingdotproducttocomputecosand2sin2+cos2
=1,butitgivesanuglyformula. Instead,reducetocomplementaryangle
=/2byconsideringA =A rotated90 counterclockwise(drewapicture).
Thenareaofparallelogram=|A||B|sin=|A||B|cos =A B.
Q: ifA =a1, a2,thenwhat isA? (showedpicture,usedflashcards).
Answer: A =a2, a1.(explainedonpicture). Soareaofparallelogramisb1,
b2a2, a1=a1b2a2b1.
a1
a2
Determinant. Definition: det( B) =A, =a1b2a2b1.b1
b2Geometrically: a1 a2
b1 b2 =areaofparallelogram.
signof2DdeterminanthastodowithwhetherB iscounterclockwiseor
clockwise from A
,withoutdetails.Determinantinspace: det( B,A, C) = a1 a2 a3b1 b2
b3
c1 c2 c3=a1 b2 b3c2 c3 a2
b1 b3c1 c3 +a3
b1 b2c1 c2 .
Geometrically: det( B,A, C) =volumeofparallelepiped.
Referredtothenotesformoreaboutdeterminants.
Cross-product.
(onlyfor2vectorsinspace);givesavector,notascalar(unlikedot-product).
j k
a2 a3 a1 a3 +k a1 a2AB = = jDefinition: a1 a2 a3 .b2 b3 b1 b3 b1
b2b1 b2 b3(the3x3determinant isasymbolicnotation,theactualformula
istheexpansion).Geometrically: |AB| = area of space parallelogram
with sides A, B; direction = normal to
theplanecontainingA
andB.Howtodecidebetweenthetwoperpendiculardirections=right-handrule.
1)extendrighthand
indirectionofA;2)curlfingerstowardsdirectionofB;3)thumbpointsinsamedirectionasAB.
j=?(answer: k,checkedbothbygeometricdescriptionandby
calculation).Triple product: volume of parallelepiped =
area(base)height = |B C|(A n), where n=
B |B | (B C)=det( B,C/ C . Sovolume=A A, C). The latter
identitycanalsobecheckeddirectlyusingcomponents.
The
Flashcard Question:
