MA202 Calculus III Fall, 2009 Laboratory Exploration 3: Vector Fields Solution Key

Transcription

1 MA0 Calculus III Fall, 009 Laborator Eloration 3: Vector Fields Solution Ke Introduction: This lab deals with several asects of vector elds. Read the handout on vector elds and electrostatics from Chater 1 of the tet Div, Grad, Curl and All That. You ma ignore the last aragrah on age 8, beginning with "Ver often it is convenient..." and ending with the trile integral in formula I-7. You will not need this for the eercises. Then answer the following questions. You will not need the handout to answer questions Eercise 1 on ages 8-9 of Div, Grad, Curl and All That. I want ou to attemt to sketch these vector elds b hand. If ou get stuck, ou ma use Mathematica. However, to get the most bene t out of this eercise ou should reall force ourselves to do it b hand. Be sure to lot several vectors in each quadrant. a) F (; ) = (; )

2 b) F (; ) = ( 1 ; 1 ) c) F (; ) = (; ) d) F (; ) = (; 0)

3 e) F (; ) = (0; ) 3

4 f) F (; ) = ( ; ) g) F (; ) = (; ) h) F (; ) = (1; )

5 Eercise on age 9 of Div, Grad, Curl and All That. Again, ou should attemt this b hand. The electric eld is the formula given in equation I- on age 7. Notice the r term in the denominator - this is essentiall the "inverse square law" for electric elds. F (; ; z) = 1 q! r u ; where! u is a unit vector ointing radiall out from the origin. Here, we assume q = 1 and assume units are chosen so 1 that = 1 also. This simli es the formula. Finall, note that r is the length of the osition vector! r = (; ; z): Thus, the vector eld is: F (; ; z) = z + + z ; + + z ; z + + z! = 1! r (; ; z) = ( + + z ) 3 r 3 : 5

6 z Here is a -dimensional version of this: F (; ) = ( + ) 3 ; ( + ) Eercise 3 on age 9 of Div, Grad, Curl and All That. (No sketches required for this roblem.) 6

7 a) F (; ) = ; + + b) F (; ) = (+) ; (+) c) F (; ) = ( ; ) d) F (; ; z) = ; + +z ; + +z z + +z z Eercise 5 on age 9 of Div, Grad, Curl and All That. 7

8 Assuming all constants are 1 as above, the eld in question is E(! X r ) = j! r i=1 q i! r i j! u i where! r i = ( i ; i ; z i ) is the location of charge q i ;! r = (; ; z) is an arbitrar osition vector (the roblem asked about restricting this to the - ais) and! u i is a unit vector from! r i to! r ; that is,! u i =! r! r i jj! r! : Since r ijj! r 1 = (1; 0; 0) and q 1 = +1; and! r = ( 1; 0; 0) and q 1 = 1; the eld reduces to: E(! ( 1; ; z) r ) = (( 1) + + z ) 3 ( + 1; ; z) (( + 1) + + z ) 3 Now, if we restrict to the -ais, we have! r = (; ; z) = (0; ; 0); so the eld becomes: ( 1; ; 0) (1 + ) 3 (1; ; 0) (1 + ) 3 1 = ( ; 0; 0) (1 + ) 3 8

9 Here is a lot of the entire eld: 5. Eercise 6 on ages 9-10 of Div, Grad, Curl and All That. a) Since F (; ) = (u(; ); v(; )) is tangent to (); we have that the sloe of () is equal to the ratio of the second coordinate of F to the rst: d v(; ) = d u(; ) 9

10 b) Using the above vector elds: i) F (; ) = (; ) Z d d = d = d Z d = d = + c = C; where C = c: The eld lines are rectangular herbolas. ii)f (; ) = ( 1 ; 1 ) d d = 1 d = d = + C The eld lines are straight lines with sloe 1: iii) F (; ) = (; ) d d = d d = Z Z d d = ln jj = ln + C jj = e ln +C = A 1 where A = e C > 0: But if we dro the absolute values, A is arbitrar. So the eld lines are the curves = A; a famil of herbolas with the aes as asmtotes. iv) F (; ) = (; 0) d d = 0 = C 10

11 The eld lines are horizontal lines. v) F (; ) = (0; ) d d = 0 since this is unde ned, the eld lines must be vertical lines. vi) F (; ) = ( ; ) + + d d = + + This is eactl the same di erential equation as in (i), so the led lines are the same rectangular herbolas. = vii) F (; ) = (; ) d d = = = + C The eld lines are a famil of arabolas. The solu- viii) (1; ) d d = 1 = This is the di erential equation for eonential growth & deca. tions are a famil of eonential curves = Ce : 6. Let F : R 3! R 3 be a vector eld in three dimensions, so that we can write F (; ; z) = (u; v; w); where u; v & w are each functions of ; ; & z: The divergence of a vector eld is a te of derivative. We ll see later what it measures. For now, here is the formula for comuting the divergence algebraicall: div Use this formula to calculate the divergence of the following vector elds: 11

12 a) F (; ; z) = (u; v; w) = ( ; ; z ) @ = + + b) F (; ; z) = (u; v; w) = (z; z; ) = = 0 c) F (; ; z) = (u; v; w) = ( ; ; z 3 ) @ = + 1 d) F (; ; z) = (u; v; w) = (e ; e ; z) = e + e + = e + [Remark: There is a similar de nition for Euclidean n-sace for an n: Notice that the divergence of F is a scalar quantit.] 7. Let F : R 3! R 3 be a vector eld in three dimensions, so that we can write F (; ; z) = (u; v; w); where u; v & w are each functions of ; ; & z: The curl of a vector eld is a te of derivative. We ll see later what it measures. For now, here is the formula for comuting the curl @w @ Use this formula to calculate the curl of the vector elds in roblem 6. [Remark: Unlike the divergence, the curl is de ned onl in 3 dimensions. Notice that the curl is a vector quantit.] a) F (; ; z) = (u; v; w) = ( ; ; @w ; 0; 0 0; 0 0) = (0; 0; 0) b) F (; ; z) = (u; v; w) = (z; @w ; ; ; z z) = (0; 0; 0) c) F (; ; z) = (u; v; w) = ( ; ; z 3 ) 1

13 @w = (0 0; 0 0; 0 0) =) = (0; 0; d) F (; ; z) = (u; v; w) = (e ; e @w ; 0; 0 z; e 0) = (z; z; e ) 8. Given a vector eld F in R 3 ; we see that curl F is another vector eld, so it makes sense to comute its divergence. Comute div(curl F ) ; where F is the vector eld in roblem 6d. d) F (; ; z) = (u; v; w) = (e ; e curl(f @w ; 0; 0 z; e 0) = (z; z; e ) so div (curl F ) = div (z; z; e ( (e ) = z z + 0 = 0 13