美赛数模论文之公式写作

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由假设得到公式
1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）
% J4 P0 y; q, a# {* x5 j' l  a
公式
0 X! C0 b& W3 A, e# Q2 R# D
Where
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9 H% b7 k: @1 C" n( I3 f8 u  d5 Q符号解释
% D" n  f2 j( h! b" S. |' @5 e5 X
4 S3 p+ D9 `( J' n# \3 ]/ @& MAccording to the assumptions, at every junction we have （由于假设）

公式

由原因得到公式
, b* F4 `, B* b  @! y. h0 O2．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）；
9 F( t1 Q* Q  O2 R+ X
公式

Since the fluid is incompressible(由于液体是不可压缩的), we have

4 x- z6 \- s% }$ y* G: f公式

Where

公式

用原来的公式推出公式
3．Plugging v1 into the equation for v2 ,we obtain （将公式1代入公式2中得到）
. B1 g' k) }* H5 j. O
, W$ r# t& O0 [2 \& f, O/ O公式

: Y3 L  R* @5 X5 I" x. e+ Y5 X11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：

公式

& L: }- h5 \: W12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）

' ^" ~( M1 [2 N公式
8 c: I; `* e- A) s5 o: g& Y4 Y
/ b3 A) D+ D- b1 m/ gPutting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have

公式

Putting these into (1) ,we get（把这些公式代入1中）
  R9 t' V: {1 g8 D5 C- ~
公式
, F  u, m( A0 `0 x3 m
4 |, v1 }3 u/ c! UWhich means that the

/ v' N, B4 n- U. T5 `! S0 dCommonly, h is about

8 n8 [/ d0 H- Q( j5 J& F% gFrom these equations, （从这个公式中我们知道）we know that ………

 

引出约束条件
4．Using pressure and discharge data from Rain Bird 结果,
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- Y* f3 u; Q% l5 \) JWe find the attenuation factor （得到衰减因子，常数，系数） to be
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公式
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计算结果
7 n! G4 Z" X; g, m5 ~5 N0 G) S6．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）

公式

Where
0 I$ J) r3 K6 t1 n, a$ K; h
() is ;;

7．Solving for VN we obtain (公式的解)

公式
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4 u0 W# A3 |6 B, R; N, ?Where n is the …..
- Z! I+ |1 H* o! \6 X/ p
4 T: [: h3 M- b3 V5 {' \2 M9 Z" f 

8．We have the following differential equations for speeds in the x- and y- directions:
: N2 T7 P4 d% N, }8 n- {
公式

) I) O" t- ]( v" I3 }9 pWhose solutions are （解）

公式

9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:
4 Z' O  n' C( L7 g. h* X" I
4 \. x6 L+ f- K4 l* \公式
- a4 q% m+ x- ]) V1 M" r
根据原有公式
10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is

公式
* _, i% h2 U/ I5 r/ p0 f( m
The decrease in potential energy is （势能的减少）
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公式

, C: _5 J' M+ c4 S1 L8 mThe increase in kinetic energy is (动能的增加)

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)
6 L- a4 e+ G# P, b  F
Where a is the acceleration vector and m is mass
% S2 N" |! K* {: z: N: B
3 K  B& U: U' O6 A" r 

Using the Newton's Second Law, we have that F/m=a and
6 s9 X$ v9 j2 K: u& e+ g: n3 r+ {8 ]3 \
公式
( Z- @- ~5 S1 O: r5 s
) ]3 A6 m4 j4 FSo that

公式

Setting the two expressions for t1/t2 equal and cross-multiplying gives
9 S3 [1 q& U" P
- V& b& V' U: Q6 j& Q% F4 c" e公式
& D, e% E  t1 r$ R2 D9 b
22．We approximate the binomial distribution of contenders with a normal distribution:
: F7 [' O2 }% B) l- z; _
3 Q( O) V& U  @! o4 n# p公式

- q) W. \% t0 x! bWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives

5 I4 I- r3 O7 g, _. R, e7 f公式

6 v, x$ H3 Z0 i) h$ tAs an analytic approximation to . for k=1, we get B=c
. a( h3 a% r) ?+ Z% v0 g
3 J5 P% ?: |# ]9 B7 K 
6 ?/ I/ R  s" x3 X" {
( S) m( S- u8 g8 R* ?8 m  l26．Integrating, （使结合）we get PVT=constant, where
8 T! Y8 N- t( g7 r  h$ R
公式
; W/ ^9 |% T1 a) Y( Q3 M
' i5 @/ \$ v' ]9 U9 WThe main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so

 

3 X; O/ B! j& e23．According to First Law of Thermodynamics, we get

3 J! u1 N$ ~  M: r$ |: j' |* A公式

Where ( ) . we also then have

/ M* l0 V* G0 x. @. r6 W公式

Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
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: p1 H# O" [. a' ~; X( |公式

Where
) u, b) `& j- r# p
 
7 H! o* L/ Y6 p8 B
对公式变形
! h! X' f: b% e  `% o. p- k; _- w13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）

! u# J& a' f$ w5 j# P* U公式
0 q" u! w3 }+ [  z  N6 }
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
2 L5 m# K1 B6 k5 d* H/ l) U
使服从约束条件
14．Subject to the constraint （使服从约束条件）
( |7 }3 f( g0 ^* `! k
+ K- [$ g$ e- R0 c% @0 z! J公式
( ^6 X2 v. q7 x+ w3 w1 @5 g
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）
4 Z1 \, f# f: u6 l# k! q
公式

" P0 i% J" |1 c9 jAnd thus f depends only on h , the function f is minimized at （求最小值）
4 C% m0 T4 {; r2 k( J
) O( G! l$ F; Z( _公式

. y3 q7 ]  V! ~At this value of h, the constraint reduces to

公式
5 s9 R4 x0 W5 D* ~. K+ o! q
结果说明
/ z4 U- T0 o# A. N. _; U9 M5 Q15．This implies（暗示） that the harmonic mean of l and w should be

; ^' u3 t- ~4 `' \% f6 A4 Y. M3 d公式

So , in the optimal situation. ………
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5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
: T- ?4 m0 n! P5 e
公式

4 B0 P8 ^, W  I* m16． We use a similar process to find the position of the droplet, resulting in
! ?* V" }' |: X0 ?! \- l
公式
/ i( ]8 A1 ?$ L5 |
3 ?+ N, w. I( z* c3 L) p* ^With t=0.0001 s, error from the approximation is virtually zero.

3 v3 R1 G' y/ |6 O) v- N 
! `( S  O, l1 M+ ~6 f) Y& N* g9 W
5 E) y& c. @( W' F5 P17．We calculated its trajectory(轨道) using
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0 ]  B. b7 n$ h/ H公式

18．For that case, using the same expansion for e as above,
6 o/ O3 j; K/ ]3 U4 o
7 a, v2 |) W5 r- W: d3 o% \  G$ x- s公式
3 `5 U* v, Z. \! ~; o, E0 W9 g
19．Solving for t and equating it to the earlier expression for t, we get
3 @" C0 o( L8 }: q
公式
' b" e1 q( }" w" X2 ^) R, R4 b
. A1 m% q% D9 c( o7 {' M8 b20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is
7 o" w8 g5 R; F6 R; n) ?
+ e7 j% P% Y; c3 z. O$ `8 ]0 y公式
' l* ]8 Q$ C! I, I' v+ E9 n% g
" J4 S( V0 w5 Z3 i4 AAs v=…, this equation becomes singular (单数的).
' b. t! }5 n: r1 f. x1 X
 
4 Y3 o' n4 x( R# l8 b! r2 B
/ s# x1 U7 N' Z3 x, L) B+ `; F* p由语句得到公式
) B0 i8 k, _/ E+ h( F21．The revenue generated by the flight is
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- Y; I& F* E* l' `/ }0 O9 J公式

 

24．Then we have

公式

We differentiate the ideal-gas state equation

6 @+ v2 e1 t' m  R, R+ Q公式
# K" K: z4 q9 r1 [3 W7 e
: S( X+ o7 z" O- HGetting

公式

25．We eliminate dT from the last two equations to get （排除因素得到）
3 j2 c( W9 n' v  J1 S# ^. G# t
7 J, y, H- y/ R/ X8 D7 u公式

% }8 @) o0 M- } 

22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
+ U( T% n7 {  B/ v2 h
0 ?& P5 e$ o+ I6 G( ~公式
+ M' m& Y8 ?# O1 {' _
7 }/ ]$ q! }1 sWhere P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）
. }5 o- ?+ g; k7 Z* o9 v/ d5 E$ y" {
. s, \/ {+ w" {公式
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版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
原文链接：https://blog.csdn.net/u011692048/article/details/77474386
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