Recall from last week:

Transcription

1 Recall fom last week: Length of a cuve '( t) dt b Ac length s( t) a a Ac length paametization ( s) with '( s) 1 '( t) Unit tangent vecto T '(s) '( t) dt Cuvatue: s ds T t t t t t 3 t ds u du '( t) dt Pincipal unit nomal: s s t t T T N Binomal B T N Fenet fame (o TNB fame) T,N,B Tosion : db N ds db ds N d ds ( t) ( t), y(t), z( t) ' t '' t ''' t B t t

2 Recall: Decompose the acceleation vecto a ''( t) a Ta T N N d s d tangential acceleation: a = ( T ( t) ) dt dt av at v ' '' ' ds nomal acceleation: a N ( t) dt va ' '' an = a a T v ' d constant velocity: a ( T ) 0 dt a N If you tavel at constant speed a on a cicle of adius : a 0 T a N 1 a

3 13.6 Acceleation in Pola Coodinates

4 Newton s law of gavitation (1687): F GmM Invese squae law GM F = ma a '' = d dt ' M m G is the vecto fom the cente of the sun to the planet is the mass of the sun is the mass of the planet is the gavitational constant ' ' '' 11 G = N m kg (fom 1798) '' 0 since '' is paallel to by Newton's law hence ' is a constant vecto C in paticula C 0 the planet moves in a plane othogonal to C!

5 Recall pola coodinates: cos( ) y sin( ) y actan( y ) 0 0 fom Calc I, aea element: da 1 d eplace i, j by (pependicula unit vectos) u = cos( ) i sin( ) j, u sin( ) i cos( ) j cos( ) i sin( ) j cos( ) i sin( ) j u u ' ( cos( ) i sin( ) j)' 'cos( ) i 'sin( ) j 'sin( ) i 'cos( ) j ' u + ' u ' ' u ' u

6 Keple s second law of planetay motion (1609): The planet sweeps out equal aea in equal time Lets show this is tue: ecall: C ' constant i.e., the planet tavels in a plane othogonal to C we can assume planet tavels in y plane ecall u and ' ' u ' u _laws_of_planetay_motion hence C ' u ' u ' u ( ') u u but u and u ae unit vectos in the y plane, hence u u = k (o k) thus C ( ') u u ( ') k = ' k since C and k ae constant vectos, ' is constant as well ecall: da 1 d hence da 1 d ' Aea swept out at constant speed! dt dt

7 14.1 Functions of Seveal Vaiables

8 Recall: Functions of one vaiable: y f ( ) Domain D: set of all whee f ( ) is defined. Range: set of all f ( ) fo in the domain Gaph of f : the set in the plane (, f ( )) D Functions of two (o moe) vaiables: z f (, y) (o w f (, y, z) ) Domain D: egion in the plane whee f (, y) is defined. Range: f (, y) D Eample: Find and sketch the domain of the function f, y 1 y. 1 y 0 o y 1

9 Gaph of f : the set in 3-space (, y, f (, y)) (, y) D

10 Level cuves ae cuves in the y plane given by f (, y) c fo vaious constants c. When lifted to the suface, they ae sometimes called contou cuves.

11 Aveage Januay sea-level tempeatues measued in degees Celsius The level cuves ae called isothemals, they join aeas with the same tempeatue

12 The gaph of z f (, y) 4 y is fomed by lifting the level cuves: (a) Contou map (b) Hoizontal taces ae aised level cuves

13 Level cuves and gaph of y z ye Maple command: plot d y ep y y aes boed 3 ( * * ( ), 5..5, 5..5, );

14 I II III IV V VI VI I II ( a) f (, y) y (b) f (, y) y (c) f (, y) 1 1 y ( e) f (, y) y ( d) f (, y) y (f ) f (, y) sin y V IV III

15 A B C 53. z sin y B, III D E F 54. z y e y C, II 55. z 1 4y FV, I II III 56. z 3 3y A, VI IV V VI 57. z sin sin y D, IV 58. z sin 1 4 y EI,

16 14. Continuity

17 Recall: lim f ( ) L a the values f ( ) appoach L moe and moe as you get close to a. Notice: a does not have to lie in the domain of definition! ( f( a) may be undefined) We can appoach a fom two sides: both limits lim f ( ) and lim f ( ) Eamples: a) lim 0 9 b) lim 3 3 does not eists since 6 sin( ) c) lim since 3 3 a a must eists and be the same. 0 0 lim 1 but lim 1 since sin( ) (sandwich theoem) f is continous at a if lim f ( ) f ( a ) a f is continous, if it is continuous fo all D (domain of definition)

18 Definition: lim f (, y ) L y a b (, ) (, ) If the values of f(, y) appoach the numbe L as the point (, y) appoaches the point (a, b) along any path that stays within the domain of f. Eamples: a) lim ( y, ) (1,1) sin( y ) y sin() b) lim ( y, ) (0,0) sin( y ) y lim 0 sin( ) sin(z) lim 1 z0 z

19 c) lim ( y, ) (0,0) sin( y ) y If 0 then sin( y ) sin( ) lim lim 1 y (, y) (0,0) 0 If y 0 then lim sin( y ) sin( y ) lim y y (, y) (0,0) 0 1 Limit does not eist! We can let (, y) appoach (a, b) fom an infinite numbe of diections in any manne whatsoeve as long as (, y) stays within the domain of f. Fo all of these the limit must be the same.

20 Eamples: a) lim ( y, ) (0,0) y y If y then If 0 o y 0 then y lim 0 y ( y, ) (0,0) y 1 lim lim y (, y) (0,0) 0 b) lim ( y, ) (0,0) y y 4 (To be discussed Thusday) Limit does not eist