美赛数模论文之公式写作

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由假设得到公式
, j/ B6 Q9 J, W! K1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）

& Y9 J. ?5 I/ K- i( |9 G公式
0 ^: o' i6 j. j) C5 m
& u1 A: W  Y3 s: H0 j* cWhere

符号解释
" P& r6 [5 M3 d2 B
9 g5 A! r  P, S0 H, aAccording to the assumptions, at every junction we have （由于假设）
/ C8 P% j$ \2 k7 W% d
* r7 i% v# K$ z2 o8 s公式
6 r! \9 P6 q8 N2 a# D" \' U
/ W( g  f! i! k3 g. M由原因得到公式
9 M: a" w# }& }/ S% D2．Because our field is flat, we have公式, so the height of our source relative to our sprinklers does not affect the exit speed v2 （由原因得到的公式）；

公式

& B: Q# k+ F% c, {, y# N) n/ kSince the fluid is incompressible(由于液体是不可压缩的), we have
! W. V  i4 s3 m9 j8 ]( \: V/ r
# ^/ M" J& a( `公式

( G6 d$ E5 j4 I9 k9 C3 KWhere
  I- n& _1 e- x: e5 i% ^, _5 [
, g3 R) `) U$ Y( ?公式

用原来的公式推出公式
# |% _0 m7 D+ D; Z3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）
- v7 a  i+ ]' H  F0 Z  U: e/ q
! I# ]  s% @( a公式

: H% ~/ x" T' n' \' T11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：

公式
/ M% f5 [, f' X6 n5 f
12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）

, K  d) q7 ~8 T+ q* U1 P4 W' @0 }公式

- D- ]) S2 e0 @" iPutting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have

公式
6 Z! _7 ?" t# N4 }
Putting these into (1) ,we get（把这些公式代入1中）
. N, p! Q' z9 `
: n1 K& s) Q5 M! @) n公式

& o  q) [2 H- L% a" D% v# p1 _. CWhich means that the

6 K3 k% k( q. I% M3 ACommonly, h is about
  `/ R$ o  ]$ _; d/ y
From these equations, （从这个公式中我们知道）we know that ………
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4 T% O( @' A$ i% p9 v, Y2 u: S引出约束条件
4．Using pressure and discharge data from Rain Bird 结果,
  |; h, d0 |! W2 P0 t% Q
We find the attenuation factor （得到衰减因子，常数，系数） to be

公式

计算结果
! O8 `# r( i0 l6．To find the new pressure ,we use the ( 0 0),which states that the volume of water flowing in equals the volume of water flowing out : （为了找到新值，我们用什么方程）
  t! P0 c5 h3 s! Z5 K3 z
  \! t" o) t9 F公式

0 N2 H+ T3 u. q& r1 k7 O- W/ uWhere

' @/ {) ~! N/ r! l1 }, E7 L& ~() is ;;

7．Solving for VN we obtain (公式的解)
! g1 T  G; A3 a- c! U+ c" e
公式

5 C& T; e' o$ d8 `8 p% Y8 zWhere n is the …..
6 M4 |' g0 e2 f, _* S& B. a
0 T% A2 ?4 h( v. m6 ~- @ 
  q; M/ u8 a0 D3 i9 v1 T
8．We have the following differential equations for speeds in the x- and y- directions:
3 e: d: `2 Q+ J+ E% D, `
0 E4 q# s6 F2 ^公式
/ U9 K( h+ v& K; O3 b4 s3 h0 D
0 ~: i" D$ |6 |0 `1 p6 gWhose solutions are （解）

4 f" D( D2 t% b" l& ~& p公式

9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:
  x' B  _; h% ~% ?! w$ v5 |% i
公式

. |/ ^# L; }1 Z- `' S6 I根据原有公式
. j2 ~. b, B8 L  E10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is
- a- l7 z: }- `! P  O2 f- D
公式
$ z1 V# c, U( }9 H& i4 B
The decrease in potential energy is （势能的减少）
  `4 W7 \5 b& [, y. y# ?& g
; Y, I! i5 B/ I+ [8 y) V3 @; {公式

The increase in kinetic energy is (动能的增加)

. r4 b- X0 b2 O* p公式

Drug acts directly against velocity, so the acceleration vector from drag can be found Newton's law F=ma as : (牛顿第二定律)

) x/ R& M( D) j6 K3 [5 L3 WWhere a is the acceleration vector and m is mass

 
: r& X, s* ?' |
$ i/ e# x  o: g$ ~- P3 `+ fUsing the Newton's Second Law, we have that F/m=a and

公式
6 u9 R) P/ ?/ u$ n
So that

公式
* m# j( I0 U; I3 z( M8 c$ v" c) `
5 d7 h$ A9 A: d( g) cSetting the two expressions for t1/t2 equal and cross-multiplying gives
0 R& {) |; M: P5 r# }# y/ L$ a* g# G  c3 s! Y
3 U* h; h4 u" [  d, D0 Q6 E公式
0 ]# n0 g+ o( l/ g# \
22．We approximate the binomial distribution of contenders with a normal distribution:

公式
9 a; s) P) o: G7 p0 I- i
& k8 |; w: f1 q6 a1 c- VWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives

公式

As an analytic approximation to . for k=1, we get B=c
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/ j' }1 q: B+ X' ^
26．Integrating, （使结合）we get PVT=constant, where
/ ?: l+ R# e# K) Y2 }
. N; p6 D" ]( H  ~/ y公式
7 f  A$ |# w) J2 {6 B! }
* Q7 K* f* {4 e& ^  rThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
/ T- e4 V$ y( `; U+ b* [, }
 

23．According to First Law of Thermodynamics, we get
" S' A4 i9 P4 o: e, y- [. n& l
公式
, O& D( K7 ]7 A7 k% g& q
Where ( ) . we also then have

  z/ g( q$ f$ V3 c( F+ K. h$ U6 Q公式
; `) o6 V, o4 e% k( G: H0 [
& R0 D4 n" D/ N4 j0 ^8 I+ AWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:

公式
* H5 G9 }. w: N. l* l
7 p" H) B$ |7 D& J1 l4 g2 lWhere

 
$ \3 U% ?/ y  C, {1 h& r9 x6 q, |
对公式变形
9 Y3 O- R$ f4 G13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）

; ?' Y$ I: ], j( z2 D5 Q公式
3 ?9 |/ L. N4 [
We maximize E for each layer, subject to the constraint (2). The calculations are easier if we minimize 1/E.（为了得到最大值，求他倒数的最小值） Neglecting constant factors （忽略常数）, we minimize

公式
4 d1 B& }7 w2 K! p$ P% M
使服从约束条件
" o' t2 M" B$ T14．Subject to the constraint （使服从约束条件）

公式

6 }. p/ |2 v$ o; I4 z+ h" ~Where B is constant defined in (2). However, as long as we are obeying this constraint, we can write （根据约束条件我们得到）
+ a  p: a( I) W) p8 n! |
公式
, D5 `, ]- j# W6 ~& w
: y; R; v7 c0 L+ ^# r7 JAnd thus f depends only on h , the function f is minimized at （求最小值）
! L! w8 J! z+ n
公式

At this value of h, the constraint reduces to

公式
0 j1 q4 Q/ e& g( Y8 g/ Y& ]
2 ?7 S9 h  T; Z, Y结果说明
15．This implies（暗示） that the harmonic mean of l and w should be
* h# K- |* N2 p9 I4 t
公式
9 m+ q& J5 O& n; O
So , in the optimal situation. ………
  |3 C; ~2 v' C" X# D9 p
0 T; c! x; G" a3 v" N! E5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is

公式

" n5 F5 I9 [) L# x, q0 S16． We use a similar process to find the position of the droplet, resulting in

公式

  Q  G' i& \& R- `4 t; b$ V  ~With t=0.0001 s, error from the approximation is virtually zero.
4 b$ E- e  U8 u% ^# M$ r- a  c
 

5 E$ h; x7 d4 O: Q2 G17．We calculated its trajectory(轨道) using
7 P! M6 S2 n4 h( z  H- p- X
/ c/ l' k( ]: J& d( L) ]公式

+ r) l# M6 C# b+ E* m! I, [1 B18．For that case, using the same expansion for e as above,

1 M* }: K: u6 e1 Y' w公式
- R% E! d1 k  `* s8 H6 `# o# j
; Z7 F# y5 V/ p6 _. u8 `5 m7 @  c/ b# A19．Solving for t and equating it to the earlier expression for t, we get

; a- M/ ?# P, j# u$ D公式

% V+ _, z$ A8 q) N20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is
/ s( I% l/ `2 z" z  b/ H
公式

: M6 Y* U1 {; b: B& b2 B% ZAs v=…, this equation becomes singular (单数的).

2 Y4 Y/ b# c4 ] 

由语句得到公式
! {7 ^4 D0 x2 V& p0 r9 G21．The revenue generated by the flight is

5 J/ `2 P: Q' A, ^9 J2 I/ I公式
/ w" P, _/ N  F6 s2 T
 
$ j. A0 P$ N; w9 S
24．Then we have
) U) \" o. ?4 m+ C7 T2 e5 N- G
公式
' N+ N1 y* X# t. |
We differentiate the ideal-gas state equation

6 I: S0 G+ l2 M! [. w公式

Getting
2 f& G1 f" c2 V, ]/ @4 C
" \' E* O3 e) x3 Z: s$ v' Z公式
& i+ A' {4 ]1 @( Y
25．We eliminate dT from the last two equations to get （排除因素得到）

+ W4 m. g. t- V2 J0 B公式
; G' k/ {  F+ k4 g% w- e' p' T
 
7 i2 r  y* M6 W+ u
3 H; e8 }! @6 K: j  Y8 x. Z' P22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations

: B8 u# W8 V$ F2 e公式
/ m4 J4 E: s4 e% `: e2 e, h# ]1 C
Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）

公式
  O8 J2 y- U# b8 F; ]& [% ~————————————————
2 u* i0 k" v6 e+ K$ z. n版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
! G, O8 [/ T) Z4 s原文链接：https://blog.csdn.net/u011692048/article/details/77474386
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