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1 5H7) mr sinh cosh R ] R mr ] R - -mr sin h ] - R 4 O h R mr sinh cosh P Rsin h msinh h h m mr sinh ] R ] R = ] R
2 (a) (b) Y t= p 4 p P ar p X ar R sinh m (R sinh ) =mr mr sinh cosh F=-msinh + =-msinh +mr sinh cosh 3p 4 sinh F= h sin h =cosh = R h cosh = R C = ] R N=mcosh +mr sin h sin h =mcosh +mr sin h =mr cos h +mr h= h +u sin h =mr F=-msin h+ u +mr sin h+ u cos h+ u7-m(-r )u =-m 8 - Ru=-k + k=m R = R 9 ] k m = ] R - F=-msinh +mr sinh cosh =-(m-mr cosh )sinh = sin h 'h -(m-mr cosh )=+ cosh = R h u sinh =sin h+ u =sinh cosu+sinu cosh 7sin h +u
3 cosh =cos h+ u =cos h cosu-sinh sinu 7cosh -usin h sinh cosh =sin h+ u cos h+ u7sin h cosh +u(cosh -sin h )-u sinh 7sin h cosh +u(cosh -sin h ) F7-m(sinh +u)+mr =-mu+mr =-(mr sin h cosh +mr u(cosh -sin sin h )%u =-mr F7-(mR sin h sin h )%u=-(m h )=-{m-mr u(cos h -sin h ) (cosh -sin h )}%u sin h )%Ru=-kk=m sin h =Ru F = ] k m = U sin -cos h = ] - h =U R 4 = ] R - + ] R = ] - ] R R 4 = ] R + ] R = = ] R =] - = ] e R 8] 9 R - ) ] R 8] 9 R 3 ] R t= p 4 cos p 4 = U sin p 4 = U X= Ru U = Racost = Racost = ar U U t= p cos p =sin p = X=Y=Ru=aRcost=aRcost= t= 3p 4 cos 3p 4 = - U sin 3p 4 = U X=- Ru U =- Racost U = ar t=pcosp=-sinp= Y= Ru U = ar Ru Y= U = Racost = Racost =- ar U U
4 X=-Ru=-aR=-aRcost=aRY= X=Rucost=aRcostcost=aRcos t Y=Rusint=aRcostsint=aRcostsint X + Y =(Ru) ( cos t+ sin t)= a R cos t= X + Y =arx+ X- 8 ar ar + Y 9 = 8 9 ar ar 8 9 cos tcos t= X ar a R, l B = I pa u + =pb Bu - =-p B b i RT u + - RT u + T 6 T 3 T T 3 5T 3 T t lw= pr u + K= M(l ) lk7ml l l= pml R u + l7% -4 rad/s
5 A B y H= I pa y H y H= I pa -a I a I I I AD 6, 6, BC E -a a F (b) D C (a) (a)(b) h=bc DAI y y (b) H y =Hcos6,%=H= I pa l B = B y = l H y = I pa B B y h %B u + =pb Bu - =-pb B ABCD V=- lu lt i= V lu =- R Rlt h=y h= p 3 p 3 G. H (h p 3 p 3 (h p 3 G. H p 3 (h 4p 3 4p 3 (h 5p 3 H. G 5p 3 (h(p
6 h=t= p T tt= h p T p 3 (h p 3 T 6 (t T 3 lt= T 6 lu=u --u + =-u + i=- lu Rlt = RT u + 4p 3 (h 5p 3 4T 6 (t 5T 6 lt= T 6 lu=u +-u - =u + i=- lu =- Rlt RT u + =% lw= i R% T 6 = 8RT9 u + R% T 6 = 4 RT u + = pr u + v=l K= M v = M(l ) K-lK= M l (-l) = M(l ) -Ml l+ M(ll ) 7 M(l ) -Ml l l 8 9 ( M(ll ) lk7ml l lw=lk pr u + =Ml l+ l= pml R u + l= pml R u 7p + = -6 %u + = -6 % % p% %. % p% %. % 7 =996.6% =.9966% -4 7% -4 rad/s u + =pb B = 8 Bpb = l B = B y = l H y = I pa b li a = 4 7 %(.3 ) % 3 % %. 4p 8 9 = 7p 7 % h p pr k e 8 - n n H 9 6.5% 8.% Rch R8 n n H n H -n 9
7 dsinh dlz 8 c -v c 9 k kbt m kbt m k 8 9 c k= h p = h mv h p m v pr k =n e n= n H n - Rch n H - Rch n n= n H n= n - Rch n H Rch n 9 k= R k= R8 n n H n H -n 9 =8 - Rch=hm m= n n H 9 c k n n H - = R8 n n H n H -n 9 n = n H =3k= 36 5R n =3k= R8 3 n H n H -3 9 = 3 =6.5% m R n H 3 =8.% m R n H. * m h dsinh dsinh =kk(k=!!! )
8 sinh 7tan hzsinh 7tan h= z dz =kkz z= kk d lz= k d 4.5% m k 7.% m 4.95 kr n n -. n H n H 9 n = - =.75 n = + k= dlz n =3 3. n =3 n )3 n = =.5 n H =3 3 = n H=3 k=6.5% m n H =4 4 = n H=4 k=4.8% m n H =5 5 =.4-4. n H=5 n H )5 4.5% m k 7.% m ( n H n ) z= kk k=4.8% m d z=4.8% k d % mk=6.5% mz=6.5% k d % m d % m, $4.8, $9.6,!!!, $6.5, $3.!!! dsinh d d h z h
9 m km=c v km=c-v k= c- v = m 8 c -v k c 9 m T k B m v = 3 k B T v k = v + v y + v z =3v v = B T m v v lk=k- k =- c k (lk ) = k 8 9 c v + lk = k 8 9 c k v = B T m k 8 9 c lk=k- k lk
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Gmech08.dvi
51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r
) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)
4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7
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#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =
#A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/
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4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P = 90, = ( ) = X
4 4. 4.. 5 5 0 A P P P X X X X +45 45 0 45 60 70 X 60 X 0 P P 4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P 0 0 + 60 = 90, 0 + 60 = 750 0 + 60 ( ) = 0 90 750 0 90 0
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O r r r, F F r,r r = r r F = F (. ) r = r r 76 77 d r = F d r = F (. ) F + F = 0 d ( ) r + r = 0 (. 3) M = + MR = r + r (. 4) P G P MX = + MY = + MZ = z + z PG / PG = / M d R = 0 (. 5) 78 79 d r = F d
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all.dvi
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1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O
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1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0
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表1-表4_No78_念校.indd
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九州大学学術情報リポジトリ Kyushu University Institutional Repository 物理工科のための数学入門 : 数学の深い理解をめざして 御手洗, 修九州大学応用力学研究所 QUEST : 推進委員 藤本, 邦昭東海大学基盤工学部電気電子情報工学科 : 教授 http://hdl.handle.net/34/500390 出版情報 : バージョン :accepted
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005-10-14 1 1 [1] [] [3] [4] (a + b) = a +ab + b [5] (a + b) 3 a 3 +a b + ab + a b +ab + b 3 a 3 +3a b +3ab + b 3 [6] (a + b) 4 (a + b) 5 [7] technology expand((a+b) n n =?) [8] technology n =6, 7, 8,
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Taro10-名張１審無罪判決.PDF
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[] a x f(x) = ( + a)( x) + ( a)x f(x) = ( a + ) x + a + () x f(x) a a + a > a + () x f(x) a (a + ) a x 4 f (x) = ( + a) ( x) + ( a) x = ( a + a) x + a + = ( a + ) x + a +, () a + a f(x) f(x) = f() = a
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CONTENT S 01 02 04 06 08 10 20 25 25 30 38 49 53 58 60 64 74 8 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 6 1 2 2002,000 200 7 Web 20 20 13 13 4 13 13F 13F 20 20 13 13 4 13 13F 13F 13 6 A B C A B 13 20 13 20
[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s
![[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s](/thumbs/94/118579915.jpg "[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s")
[ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =
2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t) = x 0 + v 0 t 1 2 gt2 v(t) = v 0 gt t x = x 0 + v2 0 2g v2 2g 1.1 (x, v) θ
1 1 1.1 (Isaac Newton, 1642 1727) 1. : 2. ( ) F = ma 3. ; F a 2 t x(t) v(t) = x (t) v (t) = x (t) F 3 3 3 3 3 3 6 1 2 6 12 1 3 1 2 m 2 1 x 1.1: v mg x (t) = v(t) mv (t) = mg 0 x(0) = x 0 v(0) = v 0 x(t)
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18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................
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No. 1 2 3 1 4 310 1 5 311 7 1 6 311 1 7 2 8 2 9 1 10 2 11 2 12 2 13 3 14 3 15 3 16 3 17 2 18 2 19 3 1 No. 20 4 21 4 22 4 23 4 25 4 26 4 27 4 28 4 29 2760 4 30 32 6364 4 36 4 37 4 39 4 42 4 43 4 44 4 46
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2009 I 2 II III 14, 15, α β α β l 0 l l l l γ (1) γ = αβ (2) α β n n cos 2k n n π sin 2k n π k=1 k=1 3. a 0, a 1,..., a n α a
009 I II III 4, 5, 6 4 30. 0 α β α β l 0 l l l l γ ) γ αβ ) α β. n n cos k n n π sin k n π k k 3. a 0, a,..., a n α a 0 + a x + a x + + a n x n 0 ᾱ 4. [a, b] f y fx) y x 5. ) Arcsin 4) Arccos ) ) Arcsin
() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y
![() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y](/thumbs/85/92979039.jpg "() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y")
5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =
di-problem.dvi
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高校生の就職への数学II
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I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg
( 4) ( ) (Poincaré) (Poincaré disk) 1 2 (hyperboloid) [1] [2, 3, 4] 1 [1] 1 y = 0 L (hyperboloid) K (Klein disk) J (hemisphere) I (P
![( 4) ( ) (Poincaré) (Poincaré disk) 1 2 (hyperboloid) [1] [2, 3, 4] 1 [1] 1 y = 0 L (hyperboloid) K (Klein disk) J (hemisphere) I (P](/thumbs/94/121424413.jpg "( 4) ( ) (Poincaré) (Poincaré disk) 1 2 (hyperboloid) [1] [2, 3, 4] 1 [1] 1 y = 0 L (hyperboloid) K (Klein disk) J (hemisphere) I (P")
4) 07.3.7 ) Poincaré) Poincaré disk) hyperboloid) [] [, 3, 4] [] y 0 L hyperboloid) K Klein disk) J hemisphere) I Poincaré disk) : hyperboloid) L Klein disk) K hemisphere) J Poincaré) I y 0 x + y z 0 z
18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (
![18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (](/thumbs/100/143999768.jpg "18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (")
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AVIC-DRZ90 AVIC-DRZ80 2 3 4 5 66 7 88 9 10 10 10 11 12 13 14 15 1 1 0 OPEN ANGLE REMOTE WIDE SET UP AVIC-DRZ90 SOURCE OFF AV CONTROL MIC 2 16 17 1 2 0 0 1 AVIC-DRZ90 2 3 4 OPEN ANGLE REMOTE SOURCE OFF
Contents
Contents 6-1 6-2 780 630 440 385 355 325 295 205 80 1-1 1-2 1-3 1-4 1-5 1-6 1-7 1-8 1-9 1-10 1-11 1-12 1-13 1-14 1-15 1-16 1-17 1-18 1-19 1-20 1-21 1-22 1-23 1-24 1-25 1-26 1-27 1-28 1-29 1-30 MEMO G
lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d
lim 5. 0 A B 5-5- A B lim 0 A B A 5. 5- 0 5-5- 0 0 lim lim 0 0 0 lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d 0 0 5- 5-3 0 5-3 5-3b 5-3c lim lim d 0 0 5-3b 5-3c lim lim lim d 0 0 0 3 3 3 3 3 3
Onsager SOLUTION OF THE EIGENWERT PROBLEM (O-29) V = e H A e H B λ max Z 2 Onsager (O-77) (O-82) (O-83) Kramers-Wannier 1 1 Ons
Onsager 2 9 207.2.7 3 SOLUTION OF THE EIGENWERT PROBLEM O-29 V = e H A e H B λ max Z 2 OnsagerO-77O-82 O-83 2 Kramers-Wannier Onsager * * * * * V self-adjoint V = V /2 V V /2 = V /2 V 2 V /2 = 2 sinh 2H
瀬田唐橋景観検討について
3 3 12,174 /12hr 1,110 /12hr 3,066 /12hr 172.0m 51.75m 23.5+5@25.0+23.5m 3@17.25m 14.0m 15.0m 5 6 2009 12 3 2010 2 25 6/22 8/24 10/18 11/16 1/13 1797 16 15 15 13 6 (S49) 54 7 23-2 24 54 50
一休デイサービスセンター
2770105472 2770105472 2776300135 2776300135 - - 072-274-7219 072-204-7319 http://www.19day.jp/ 072-274-7219 072-204-7319 http://www.19day.jp/ 2007 11 1 45 2007 11 1 45 11 12 4km 10 11 -2- 3-3 - 4 - 5 11
熊本県数学問題正解
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chap1.dvi
1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f
新しい気象情報の利活用状況等に関するアンケート調査
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学習の手順
NAVI 2 MAP 3 m 17 13 19 12 17 24 1 20 18 23 18 12 1 12 17 11 14 16 19 22 m 12 16 A 16 20 B 20 24 24 28 C 20 AC 40 cm AD A 0.20 12 0.300 B 0.200 0.12 12 C D 40 1.000 20 2 2 0 20 30 cm 14 1 1 160 160 16
A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6
1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67
(ii) (iii) z a = z a =2 z a =6 sin z z a dz. cosh z z a dz. e z dz. (, a b > 6.) (z a)(z b) 52.. (a) dz, ( a = /6.), (b) z =6 az (c) z a =2 53. f n (z
B 4 24 7 9 ( ) :,..,,.,. 4 4. f(z): D C: D a C, 2πi C f(z) dz = f(a). z a a C, ( ). (ii), a D, a U a,r D f. f(z) = A n (z a) n, z U a,r, n= A n := 2πi C f(ζ) dζ, n =,,..., (ζ a) n+, C a D. (iii) U a,r
ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d
A 2. x F (t) =f sin ωt x(0) = ẋ(0) = 0 ω θ sin θ θ 3! θ3 v = f mω cos ωt x = f mω (t sin ωt) ω t 0 = f ( cos ωt) mω x ma2-2 t ω x f (t mω ω (ωt ) 6 (ωt)3 = f 6m ωt3 2.2 u ( v w) = v ( w u) = w ( u v) ma22-9
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Gmech08.dvi
145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2
A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π
4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan
さくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1
... 0 60 Q,, = QR PQ = = PR PQ = = QR PR = P 0 0 R 5 6 θ r xy r y y r, x r, y x θ x θ θ (sine) (cosine) (tangent) sin θ, cos θ, tan θ. θ sin θ = = 5 cos θ = = 4 5 tan θ = = 4 θ 5 4 sin θ = y r cos θ =
キラッ都ナゴヤ10月号 _cs5.indd
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(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0
1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45
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76 3 B m n AB P m n AP : PB = m : n A P B P AB m : n m < n n AB Q Q m A B AQ : QB = m : n (m n) m > n m n Q AB m : n A B Q P AB Q AB 3. 3 A(1) B(3) C(
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振動と波動
Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3
d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r
![d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r](/thumbs/85/92098187.jpg "d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r")
2.4 ( ) U(r) ( ) ( ) U F(r) = x, U y, U = U(r) (2.4.1) z 2 1 K = mv 2 /2 dk = d ( ) 1 2 mv2 = mv dv = v (ma) (2.4.2) ( ) U(r(t)) r(t) r(t) + dr(t) du du = U(r(t) + dr(t)) U(r(t)) = U x = U(r(t)) dr(t)
.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,
[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b
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