美赛数模论文之公式写作

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由假设得到公式
+ m' v6 |1 e: i5 A6 c4 `9 Q9 ^1．We assume laminar flow and use Bernoulli's equation:（由假设得到的公式）
7 a6 U3 Y! Z) F
' q" L. S2 @; {" }- X公式

$ E, T; x9 e  {+ ]1 M# T' wWhere

符号解释

According to the assumptions, at every junction we have （由于假设）
6 h' k) O: h  s* e  R6 L
公式
) }: O. t1 Z2 n- Y3 U/ r
由原因得到公式
2．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）；
8 f! y$ P/ d1 Z8 N
8 x" y( R  ~5 |, C4 m' _; |% C. [公式

Since the fluid is incompressible(由于液体是不可压缩的), we have
8 }( Q3 L4 Z% Y7 Q6 ^7 D
! X. v2 P+ Y9 m公式

$ @3 W/ e" f- `* I! nWhere
0 N  f8 Q# W- I' |3 z6 s8 w8 V5 g
# Q* b" }. E: M5 ?; j1 G8 c5 Z8 q公式

* o/ M" K& L4 J: {' A8 F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）
# h* r+ u( o6 t$ W$ K( v
8 E6 u& N" x8 Q# P$ s7 w4 b公式

5 x+ O( f* a) Q8 l$ e( ]) J11．Putting these together(把公式放在一起), because of the law of conservation of energy, yields：

公式
( C: k& q4 K5 ?/ k, g* d9 p) i
+ ~5 P' a. w0 F12．Therefore, from (2),(3),(5), we have the ith junction（由前几个公式得）
- k+ q: B' o4 ^9 V' O8 b7 H  r
公式
4 W, e8 O2 q- ]/ R% t6 s9 M
+ ]7 _2 s6 w7 V+ _9 G# \Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have
; l  z2 n9 j5 J
: D( l9 g7 Y, T4 ^7 S; F公式

Putting these into (1) ,we get（把这些公式代入1中）
8 D+ v; b' x9 F
  `6 }  x8 R! M2 a5 l公式

, K! d* r2 _! qWhich means that the

9 j( B' [/ e) f8 j6 Z5 yCommonly, h is about

From these equations, （从这个公式中我们知道）we know that ………

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引出约束条件
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
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公式

2 d& M8 u8 o" p1 d# C' m计算结果
6．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）
/ G0 K% \% P4 K  h/ y
( o: b0 d& X1 l  K( K6 g公式
5 t( ?- U, H2 g
Where

() is ;;

7．Solving for VN we obtain (公式的解)

公式

Where n is the …..

 
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8．We have the following differential equations for speeds in the x- and y- directions:
7 c: I& h- o/ z' u1 s
- v9 b. C$ j7 u$ g公式

; Q% r2 `: [$ C' x, N! l% b* }Whose solutions are （解）
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公式
. h- m" R& L. E* l% ?! ~% P' b
& C' l2 X4 \3 u+ N7 g& I9．We use the following initial conditions ( 使用初值 ) to determine the drag constant:

! ]- x, O8 A; ]/ g公式

0 t  l  C0 H/ Z& ~- U根据原有公式
& T/ C! ]- p2 |& h10．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

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

公式
% [" ~! G4 u7 D8 W, l
The increase in kinetic energy is (动能的增加)
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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)

$ x5 ^. z( d& y, c9 UWhere a is the acceleration vector and m is mass

 

Using the Newton's Second Law, we have that F/m=a and
1 _2 t" a9 o, O8 {! [
+ s8 `( N* v' B: U9 z4 J公式

$ k9 d2 r, J1 s6 o! ASo that
' f8 C( E6 [) d# e0 _2 P
公式
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6 J# l! \$ f& s5 Q& n/ k' z' u/ RSetting the two expressions for t1/t2 equal and cross-multiplying gives

* U& P! Q1 y/ [公式

! j( j! }& ^6 u22．We approximate the binomial distribution of contenders with a normal distribution:
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公式

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

公式
+ S7 T! x  d. v& O/ {0 S
As an analytic approximation to . for k=1, we get B=c
, ]7 D& b% ]9 _1 L; i3 r5 k! a
 
: O1 _4 w1 w# e& o
4 A) _8 o( o$ {: @2 P6 k  Q26．Integrating, （使结合）we get PVT=constant, where
) a$ C6 b+ W/ c) |+ l
公式
9 u& K$ c2 C8 q% P
The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so

! e" Z- L! @4 Q; v6 R 

6 z0 M* _+ ^8 }5 t23．According to First Law of Thermodynamics, we get

公式

' Z; `! N, v2 X! m$ O( nWhere ( ) . we also then have

# ]. K$ @' L7 G公式

/ u- X" E% q7 p, g9 `6 X$ ~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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公式

Where

 

对公式变形
13．Define A=nlw to be the ( )（定义）; rearranging (1) produces （将公式变形得到）
9 b4 ]9 n* ^9 [0 a
+ d# ?6 z  P4 q6 p' I* O) ^* O3 T  b公式
. U! V+ J8 W. T4 Q% |
: O6 x  }" ~6 `! f4 Y4 ~* j7 pWe 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 Q% B# w0 J& F5 ~  Q7 ?+ v- n
: @' E. D% t& E: u8 {使服从约束条件
2 e$ F& x; E- c! [% X14．Subject to the constraint （使服从约束条件）

公式

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

And thus f depends only on h , the function f is minimized at （求最小值）

公式

! e1 y' x0 }' z1 K! L7 K; ?At this value of h, the constraint reduces to

公式

结果说明
15．This implies（暗示） that the harmonic mean of l and w should be

公式

1 C- D7 j# B+ w1 O' D$ }0 M, {So , in the optimal situation. ………

5．This value shows very little loss due to friction.（结果说明） The escape speed with friction is
* w) ~6 c' w2 `, l' `6 i; i1 l
公式
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16． We use a similar process to find the position of the droplet, resulting in
' ~" C/ y! f" o& D) T9 ]/ [
公式
3 P+ s2 }6 C9 J) c
With t=0.0001 s, error from the approximation is virtually zero.
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; I  x# s0 W1 n; q& c! s7 G 

/ c. J$ T" T5 o- I9 a3 o+ M7 z17．We calculated its trajectory(轨道) using

, O+ Y! i2 Y* t5 ]0 Y* C( L公式
: c, D. n5 {4 ^5 w0 e4 f
  b) g0 G2 W# u. H+ A/ i18．For that case, using the same expansion for e as above,

公式
" z( D+ H5 F9 |8 |. s
: p& k7 o( f/ d7 m8 Y2 P) l1 I19．Solving for t and equating it to the earlier expression for t, we get

+ X4 h# j" J" |5 L! s: c7 }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
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+ m, l7 M% @5 x- [% u公式
8 ?6 K( L* T; y% G
As v=…, this equation becomes singular (单数的).
) W  ~; a3 Z1 @3 R" f9 z& m) N
 

由语句得到公式
: g2 Z4 h7 ~8 z! O21．The revenue generated by the flight is
3 L. ~, z+ l  Z- X; ^2 Y" G, J. w# X5 n
! G& B! Z# r! i$ \' M公式
7 L! o1 q8 a2 e0 r9 d$ Z9 A7 R, Z
 

1 W7 l7 n* |  g; b; u" M) X24．Then we have
" l4 l: {; N( P: Z% g% g
公式

We differentiate the ideal-gas state equation

/ y9 y. S- V7 I# j8 Q+ Y0 g公式

3 {9 L! c0 O) v# A0 VGetting
4 e# B" m1 \/ `& X) i4 ^/ I0 p
4 Y% A, s1 w8 c( Z公式
+ A4 |5 w9 F$ p/ Y1 H
! g( C: I& W6 v! A" T- e) t25．We eliminate dT from the last two equations to get （排除因素得到）
4 ?9 ]7 `2 J- }' R5 r
公式
' A& h- O8 Z$ J# V
 
# d- x4 M7 r$ W# M7 f% Z, z
22．We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations
* B+ ?% A; z" s" P3 u! w4 p5 Q
公式
& r$ U* K$ M0 S  n! Q4 ^8 X3 p0 p
Where P is the relative pressure. We must first find the speed v1 of water at our source: （找初值）

* a9 y6 K, [2 q  W" d: u5 H) M* F公式
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: O3 @/ y8 R& _( R版权声明：本文为CSDN博主「闪闪亮亮」的原创文章。
8 R* D0 |% y+ v! u7 K原文链接：https://blog.csdn.net/u011692048/article/details/77474386
: N) i* {( J) b$ x

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