美赛数模论文之公式写作

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由假设得到公式
, F" u6 |4 n" ]3 m! |1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）

公式
8 }8 D- w8 }6 X9 W  T2 v: P
; a$ X* f6 Q2 p9 N- T) L4 YWhere
) c/ d; x9 Q) U4 m2 f/ l5 \; g8 B
符号解释
- ?4 y* ]% _- b8 Y1 n/ Q5 `
According to the assumptions, at every junction we have （由于假设）
- ^, M# c4 R- Y; \4 o3 Z- M
8 ^- R% C! q* V$ ?* L公式
' _9 s3 ~: a( P
由原因得到公式
+ U: l0 x, H8 j' e6 W2．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）；
/ |* X6 m2 Q1 z. D
公式

9 E! t* I" ^1 u) E3 B( [Since the fluid is incompressible(由于液体是不可压缩的), we have

; d, j8 c8 i  I$ E1 W( j公式

Where

公式
5 D# a! @( \3 s4 S
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）

( E- F  e: G0 f* G6 a4 k" o& T2 V公式

7 R+ l# U/ t  t4 j& F11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：
; ?! M/ ]5 W) [4 Q$ j4 `
公式
) C$ E6 I2 E! R3 \1 q  R
12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）

公式

Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have

. g8 @, u& o' C公式
# J1 K# d) n, t/ c" L6 v
* O% G3 ]/ m% k3 M1 J7 ^Putting these into (1) ,we get（把这些公式代入1中）
( w4 P8 K1 t# p/ X
; S7 \9 s1 q) D公式
% ?2 ?. J$ n3 R. c3 `: A) O% N
! [; ~* }' ^. t  V5 F5 nWhich means that the
0 o- X6 t, R0 O6 O( [" R0 f
Commonly, h is about
4 k% j8 N8 W0 \) v$ b
From these equations, （从这个公式中我们知道）we know that ………

* Q0 U7 a8 T8 r; ?& Q5 v 

' q8 P; u1 Y) d+ M& T6 q6 T引出约束条件
  t/ [1 t" ]" o! Y9 d0 z+ y9 @4．Using pressure and discharge data from Rain Bird 结果,

We find the attenuation factor （得到衰减因子，常数，系数） to be
6 r( J+ ?' [: I& v) i
: S3 S. F9 `; u9 v公式

计算结果
0 q3 }: V" E1 g5 i6．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）
/ M) p/ F& p( ], i; u( s
7 B8 O# K% H6 q1 C9 P公式

Where

() is ;;

* l8 J8 j9 J& }# [  S7．Solving for VN we obtain (公式的解)
  }8 N  h5 z  [
5 |. q5 {  G# Z4 n7 ?. {  C公式

2 @8 f+ D7 \) Z* @5 G" L0 q7 I4 AWhere n is the …..
8 \% D& n2 ?; h! Q
 

8．We have the following differential equations for speeds in the x- and y- directions:
  @8 k' D7 k8 f, w6 d* ~8 D
6 ~5 H4 J" a: y; l公式
" x( v8 B- Q+ l! o( p
+ Z# m: n1 K# ]% x; yWhose solutions are （解）

5 c5 I5 L4 R2 k$ o- J# V: C公式
: D4 N2 b/ U3 K
5 ^6 O+ L6 p8 q( [" [* v9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:

公式
  j  V$ h5 ^& U; K9 C' d+ c) p
- b" Y5 C( v( k9 o7 k根据原有公式
% G% W2 a# g) v- \, ^; }10．We apply the law of conservation of energy（根据能量守恒定律）. The work done by the forces is

/ k9 r% M7 N2 P: G. R公式

The decrease in potential energy is （势能的减少）

' v" Z2 I$ G/ I! M4 u9 `公式
7 [* d6 g7 Z, q3 ^1 P
5 V3 `7 C1 J2 h  t2 R  d/ {& \0 IThe increase in kinetic energy is (动能的增加)
8 x( H0 S' m/ W
" b; C# Z1 A+ Z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)

' M. A/ H! z4 g% d8 v0 fWhere a is the acceleration vector and m is mass
  @9 G  x/ h" H* j$ @, Y
! `8 \0 P3 ~, k) s3 F# } 
' J* _6 Q- Q; [( P. ~9 y
& `' K% p3 e5 ]0 L. RUsing the Newton's Second Law, we have that F/m=a and
' s8 W) i- M: s& d3 y
公式

8 G) X- ~% J2 t! P/ ^& T/ mSo that
7 y% v' n" z* ]! |8 `2 X
1 y: y% [2 s* `6 K, k& b公式
9 A# Y( p3 w: l2 ?
# T& G- W$ z) K9 R0 a& R4 y! @Setting the two expressions for t1/t2 equal and cross-multiplying gives

# N8 r% \- q  t' r0 [1 |公式
, b* d  Q6 }6 L8 [0 z) @
22．We approximate the binomial distribution of contenders with a normal distribution:

公式
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Where x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives

公式
1 N5 s: s$ M: l# B$ D" t" j
6 _  I7 v  {0 _9 sAs an analytic approximation to . for k=1, we get B=c
- U6 g! `: x  A( H; ~! R7 A6 i: }
 

, c2 D" e2 ~3 U& U+ e( w) @& c26．Integrating, （使结合）we get PVT=constant, where
+ D8 M' t" R6 @% S" z
2 ~* [- ]* G( L* }公式

) b" X' t0 j$ |! `/ p: e" S7 d* `The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so
( [3 N+ t0 h: l: p; V% F8 F
 

23．According to First Law of Thermodynamics, we get

5 A8 G; ~* ^; }# b  b公式

Where ( ) . we also then have
8 Z! Q. k5 y) _
9 k0 S9 a! t& ?" L公式
' u' H% i2 D1 x' T7 j8 m, d
' b$ T( ?; c* g: f/ F+ ^" _6 Q; ?Where P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:
3 F2 A- {4 q0 n+ a7 }
公式
4 `5 R; O  v+ d# W! ]3 f- P2 i+ [
. }0 H3 N; n' D4 s, @* V1 M4 UWhere
* z3 k) Y( k2 C  G) a9 `
 
2 V% i/ ~) j9 Z0 t
: E& k* b2 `5 {6 B$ f对公式变形
- e! z3 {7 N! b" Z7 t& u13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）
. V' O1 |% T9 y  J7 O
; J- e8 J! [( a公式
+ H8 P; f/ j' ~+ ]
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

! U5 J  t" }8 B5 v使服从约束条件
# u" D) L! N  E% D8 O: t' s& s% \14．Subject to the constraint （使服从约束条件）

( e+ A1 `+ G/ _1 [公式
2 P4 w! {/ H- T  A$ T. m. S
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）

公式
/ k6 S6 k. f* d5 p7 ~
& T- A. ]) E# b6 HAnd thus f depends only on h , the function f is minimized at （求最小值）
! B( |" e; y" [$ X" F0 X" v: |
公式
( W' x, `; H& X$ P+ N
At this value of h, the constraint reduces to
0 o2 C) b# ?9 B) O7 [  P
* A+ q% S# U; M; p" t% A. \公式
4 @0 w& j1 [' l. J" Z! T
3 b  B% [% G% P7 a# T5 x$ x结果说明
15．This implies（暗示） that the harmonic mean of l and w should be

公式
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6 Z9 v6 \- t5 s2 |So , in the optimal situation. ………
# x, h9 z5 V$ v8 ]" @0 t
5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is

6 x, X! U3 G! d; r0 T6 X/ a6 _公式

. _- I8 S) i# K; p16． We use a similar process to find the position of the droplet, resulting in

公式

With t=0.0001 s, error from the approximation is virtually zero.
* I! w! j+ z  c9 U5 \' W- `- \
 

17．We calculated its trajectory(轨道) using

  m" j3 j: n# X- _" k! c4 V公式

3 v1 |7 Z+ w% w18．For that case, using the same expansion for e as above,

公式

19．Solving for t and equating it to the earlier expression for t, we get
) n2 d3 d. u2 W! ^/ g5 A
公式

20．Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is

  [; q. }- N6 [& \! g公式
9 h( \- t' j7 k- g
As v=…, this equation becomes singular (单数的).
6 e8 a, D' c# w6 ~
 
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由语句得到公式
21．The revenue generated by the flight is

公式

 
0 L! _3 @$ ?( S  _1 J5 l+ M+ P0 [# s
! \: y  ]2 i! n9 O24．Then we have

公式
2 ]( A) @5 |3 Z
! K- `" s- T4 M4 c- m+ R/ xWe differentiate the ideal-gas state equation

公式
/ s1 c0 w8 C) @' z  d- O) `6 m
Getting
) m5 |! h$ s; z
3 t5 w  L0 C- V$ u公式
5 O) C* |- i: D6 T& n
+ @& G% Z# @1 w8 e- `5 e25．We eliminate dT from the last two equations to get （排除因素得到）

7 }) F3 A' L7 \# X; f  L公式

$ {# p0 F, }: e( G6 s  ]- [+ U 

22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations

公式
- S$ W: D: ]9 J& r3 M; B! P. Z+ d
Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）

4 s5 f0 s6 J3 e' J/ c8 S公式
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5 a8 P6 [4 Q5 U; o# H' y版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
( n. U6 m. u( o3 ~8 {2 \4 D; K$ s原文链接：https://blog.csdn.net/u011692048/article/details/77474386
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