1、精選優質文檔-傾情為你奉上問題：緝私艇問題續。（1）在本問題的求解過程中，假定了走私艇的逃跑方向是正北方向，而初始緝私艇的位置在x軸正向。如果放寬這個假定，也就是當這個夾角是任意角度時，如何建立方程進行求解。以下面數值為例進行求解：b=40,a=20,c=15,其中坐標系如課件上所述，走私船的方向為45°。 （2）如果有多個走私艇在一個位置上進行交易，而緝私艇向該方向追趕。這些走私艇向不同方向四散逃走，問如何安排追趕路線？ （假定緝私艇追上一個立刻掉頭追趕另外一個，中間沒有時間停留）。以下面數值為例進行求解：b=40,a1=20, a2=25, a3=30,c=15,三個角度分別為4

2、5°，90°和-60°。C(x,y)Q(c+atcos,atsin)R(c+ycot,y)  yx0走私船緝私艇（1）緝私艇速度為b，走私船速度為a，初始距離為c。設走私船的速度方向與緝私艇初始速度方向呈角，因為緝私艇速度方向始終指向走私船方向，故兩者大致軌跡如圖所示。根據x與y的速度關系可列出以下微分方程：即： 由dsolve列方程無法得到x(t), y(t)的解析解，通過變換消去t可得到以下微分函數關系：再通過dsolve函數求解，仍無法得到y(x)的解析解。因此只能用數值解法求其解。給定初值：a=20，b=40，c=15，=45°

3、;。使用MATLAB求解可得：模型的數值解tx（t）（海里）y（t）（海里）x1（t）（海里）y1（t）（海里）（其中x(t),y(t)表示緝私艇的坐標，x1(t),y1(t)表示走私船的坐標。）00015.000000.500019.04715.182722.07117.07111.000029.123214.130229.142114.14211.500036.092621.186536.213221.21322.000043.251328.248643.284328.28432.500050.112335.310950.355335.35533.000057.275442.523757.

4、426442.42643.500064.495249.195264.497549.49754.000071.430756.486071.568556.56854.500078.688663.518178.639663.63965.000085.767370.653485.710770.71075.500092.439577.594592.781777.78176.000099.623284.802599.852884.85286.5000106.717791.6557106.923991.92397.0000113.825198.8215113.994998.99497.5000120.964

5、7105.9589121.0660106.06608.0000127.5442112.8577128.1371113.13718.5000134.9007119.1727135.2082120.20829.0000142.2377127.1158142.2792127.27929.5000149.5407134.1205149.3503134.350310.0000156.2282141.2136156.4214141.4214軌跡圖由圖和數據初步判斷大約在t=0.5到t=1之間緝私艇追上走私船。將t的范圍縮小到7-8之間并提高精度使用matlab求出其數值解如下：模型的數值解（0.5-0.7

6、精確）t（h）x（t）海里）y（t）（海里）x1（t）（海里）y1（t）（海里）0.521019.75815.631322.36817.36810.546020.58486.195322.72167.72160.571021.38436.794823.07528.07520.596022.16627.417423.42878.42870.607522.52457.706423.59148.59140.619022.87738.002023.75418.75410.630523.22278.305823.91688.91680.642023.56408.614324.07959.07950.64

7、5623.67058.710924.13039.13030.649223.77668.808224.18129.18120.652823.88218.906024.23219.23210.656423.98709.004324.28299.28290.660024.09269.102224.33389.33380.663624.19689.201524.38469.38460.667224.29829.303324.43559.43550.670824.40019.404624.48639.48630.672324.43349.457824.50849.50840.673924.48589.5

8、03024.53049.53040.675524.56289.526424.55259.55250.677024.59159.553924.57459.57450.678624.53449.617824.59659.59650.680124.50489.648124.61869.61860.681724.53989.620224.64069.64060.683324.60269.606824.66279.66270.684624.66379.646124.68139.68130.685924.68439.683224.69999.69990.687224.63309.709424.71869.

9、71860.688524.60769.721824.73729.73720.689824.66229.721624.75539.75530.691124.71059.741124.77339.77330.692424.74639.787924.79139.79130.693624.78479.815224.80949.80940.694824.84049.782124.82629.82620.696024.86509.760824.84309.84300.697224.84139.787424.85989.85980.698424.83049.834424.87669.87660.699624

10、.89429.876624.89459.89450.700924.91449.903824.91239.91230.702224.85339.901824.93029.93020.703424.81719.899324.94819.94810.704624.86329.916324.96499.96490.705824.90649.937824.98179.98170.707024.94389.965024.99859.99850.708224.98049.993125.015310.01530.709725.05519.984725.036710.0367可以判斷出：大約在第0.6755h時

11、，緝私艇追上走私船。（在計算中通過改變函數ode45中t的取值區間來提高其結果的精度，在0.6755h時，x已略微超過x1，y也十分接近y1，所以判定為在這個時刻緝私艇追上走私船。）45°60° yx0V1=20V2=25V3=30(15,0)Vj=40(2)在這個問題中，三艘走私船各自以恒定的速度、恒定的方向行進。從第一問中能夠知道緝私艇在追一艘走私船時，走私船的速度和方向都對時間有所影響。因為三艘船都不停地在各自所在的直線上勻速運動，所以初步猜測為：在追第一艘船時，選擇一艘在最短時間內能夠追上的船，追上后剩下的一艘最短時間內能夠追上的船，最后再追第三艘走私船，這樣所用時

12、間會最短。簡單的理論依據：三艘走私船在從一個點向四周勻速擴散開來，以不變的速度追趕，要在最短時間內追到，則要使他們的相對距離在每被追上一個時保持最小，則每次追趕應該從最好追的一個點開始追趕。從以上思想出發，計算（標記第i艘走私船為Zi,緝私艇為J）從零點開始分別追上Z1，Z2，Z3的時間： t1在第一問已求出t1=0.6755h。對于t2，改變a=25，=90°，通過數值法求出：t（h）x（t）海里）y（t）（海里）x1（t）（海里）y1（t）（海里）0.504914.725010.990115.000012.62180.518014.805611.511115.000012.951

13、10.531214.866212.034615.000013.28040.544414.912912.559615.000013.60970.553214.941012.910815.000013.82980.562014.962213.262515.000014.04990.570814.974813.614315.000014.27000.579614.983913.966215.000014.49010.585514.996814.202015.000014.63690.591415.002514.437315.000014.78380.597214.993414.670715.0000

14、14.93060.603114.990114.904015.000015.07740.605014.996414.979615.000015.12440.606914.999615.054915.000015.17140.608714.996915.129315.000015.21840.610614.996015.203615.000015.26540.612515.033615.297015.000015.31240.614415.023115.355715.000015.35940.616314.924315.348915.000015.40640.618114.871215.34631

15、5.000015.45340.619614.919715.386815.000015.49110.621214.952415.437515.000015.52890.622714.962815.499015.000015.56660.624214.975215.559415.000015.60440.625815.040915.619815.000015.64450.627415.063715.662915.000015.68470.629014.995415.680615.000015.72490.630614.948715.705215.000015.76500.632015.001515

16、.750315.000015.79890.633315.010815.796515.000015.83270.634714.951215.835815.000015.86660.636014.921515.864315.000015.90050.637214.977515.885515.000015.92980.638414.998015.921215.000015.95910.639514.962615.969715.000015.98850.640714.947816.004115.000016.01780.641815.005116.019815.000016.04480.642915.

17、028416.045215.000016.07170.643914.992916.076915.000016.09860.645014.972216.106515.000016.12560.646015.022516.126915.000016.1505可以判斷出：大約在第0.6125h時，緝私艇追上走私船。（在計算中通過改變函數ode45中t的取值區間來提高其結果的精度，在0.6125h時，x剛略微超過x1，y也十分接近y1，所以判定為在這個時刻緝私艇追上走私船。）即t2=0.6125h同理，對于t3，改變a=30，=-60°，通過數值法求出：t（h）x（t）海里）y（t）（海里）

18、x1（t）（海里）y1（t）（海里） 0.529918.5227 8.939822.9492 13.7685 0.579919.8364 10.451023.6992 15.0675 0.629921.0848 12.016124.4492 16.3666 0.679922.2931 13.611925.1992 17.6656 0.729923.4452 15.249525.9492 18.9646 0.779924.5339 16.930226.6992 20.2637 0.829925.5943 18.627927.4492 21.5627 0.863026.2976 19.746627

19、.9447 22.4209 0.896026.9883 20.873128.4402 23.2790 0.929027.6642 22.008528.9356 24.1372 0.962128.3348 23.147029.4311 24.9953 0.986528.8351 23.988129.7980 25.6309 1.011029.3309 24.831830.1650 26.2665 1.035529.8195 25.679630.5319 26.9021 1.059930.3082 26.527230.8989 27.5377 1.074530.6036 27.030131.117

20、6 27.9164 1.089130.8970 27.534131.3362 28.2952 1.103731.1869 28.040031.5549 28.6740 1.118231.4776 28.545531.7736 29.0527 1.128531.7075 28.889731.9273 29.3190 1.138731.9154 29.244532.0811 29.5853 1.149032.0760 29.618632.2348 29.8516 1.159232.2683 29.971632.3886 30.1179 1.162832.3931 30.055032.4421 30

21、.2105 1.166432.4546 30.184232.4956 30.3032 1.169932.4091 30.381732.5491 30.3959 1.173532.4185 30.501432.6026 30.4885 1.177132.6081 30.492932.6561 30.5812 1.180632.7070 30.561532.7095 30.6738 1.184232.6445 30.716632.7630 30.7665 1.187832.6260 30.832332.8165 30.8592可以判斷出：大約在第1.1699h時，緝私艇追上走私船。（在計算中通過改

22、變函數ode45中t的取值區間來提高其結果的精度，在1.1699h時，y剛略微超過y1，x也十分接近x1且有略微下降，所以判定為在這個時刻緝私艇追上走私船。）即t3=1.1699h因為t1=0.6755h，t2=0.6125h，可判斷：t2<t1<t3所以首先追趕Z2用時最短T1=0.6125h。由此，可以知道：走私船與緝私艇速度方向夾角越小，追趕時間越長；走私船速度越大，追趕時間越長。此時，已追趕上Z2，各點位置已發生變化，在原坐標系中，緝私艇坐標J(15,15.3125),Z1(23.6621,8.6621),Z3(24.1875,-15.9132),如圖所示： 1yx0Z1緝

23、私艇Jx145°60° yx0V1=20V3=30(15,0)Vj=4031x1x2以緝私艇為原點建立坐標系如下圖：因為v1與v2夾角為45°，v2與v3夾角為60+90=150度，所以當緝私艇追上Z2后調轉方向時，（若指向Z1，與其方向夾角為1，距離為x1；若指向Z3，與其夾角為3，距離為x3），很容易判斷：x3>x1，1>3。前面已推出：走私船與緝私艇速度方向夾角越小，追趕時間越長；走私船速度越大，追趕時間越長。因為v3>v1，1>3，x3>x1，所以要在最短時間內追上第二個目標的話，應選擇追趕Z1。通過坐標J(15,15.312

24、5)與Z1(23.6621,8.6621)容易求出：海里通過向量法求出。對于追趕Z1所用時間T2，改變a=20，=82.52°，c=10.9206通過數值法求出：t（h）x（t）海里）y（t）（海里）x1（t）（海里）y1（t）（海里）0.284810.28913.974111.66215.64720.297310.59434.369211.69455.89460.309710.87214.783711.72706.14200.322211.12285.215211.75956.38940.330311.27135.501211.78056.54910.338311.40655.79

25、3711.80146.70870.346411.52626.092611.82246.86830.354411.63276.396411.84347.02800.359611.69886.594011.85697.13080.364811.75666.793611.87047.23360.370011.80176.995211.88397.33650.375211.84297.197811.89747.43930.376811.85707.261211.90167.47150.378411.87027.324811.90587.50370.380011.88247.388711.91007.5

26、3590.381711.89387.452611.91437.56810.383311.89457.499011.91857.60030.384911.90487.573611.92277.63250.386511.93437.684811.92707.66480.388211.94367.725411.93127.69700.389511.91497.662711.93487.72450.390911.89827.627511.93847.75190.392311.90777.652711.94207.77940.393711.92697.712711.94567.80690.394911.

27、92547.753611.94897.83170.396211.93597.809611.95217.85650.397411.96697.882411.95547.88130.398711.96927.912611.95877.90610.399911.90927.892611.96197.93090.401211.87417.887911.96527.95570.402411.89817.901911.96847.98050.403711.94107.936711.97178.00530.405011.98497.989911.97518.03120.406311.98778.033211

28、.97858.05720.407611.91958.054811.98198.08310.408911.88398.068011.98538.1091可以判斷出：大約在第0.3865h時，緝私艇追上走私船Z1。（在計算中通過改變函數ode45中t的取值區間來提高其結果的精度，在0.3865h時，x剛略微超過x1，y也十分接近y1，所以判定為在這個時刻緝私艇追上走私船。）即T1=0.6125h T2=0.3865h。45°60° yx0V3=30(15,0)Vj=40X33此時，緝私艇已追上Z1，所有點位置已改變，此時在最初的坐標系中J(29.1280，14.1280),Z3

29、(29.9850,-25.9548),最后，緝私艇開始追趕Z3。通過坐標J(29.1280，14.1280),Z3(29.9850,-25.9548)易求得：海里。通過向量法可求得。以緝私艇為原點建立坐標系如下圖：1yx0Z3緝私艇JX3對于追趕Z3所用時間T3，改變a=30，=28.78°，c=11.8578通過數值法求出：t（h）x（t）海里）y（t）（海里）x1（t）（海里）y1（t）（海里）0.291611.47961.810419.52424.21110.331412.99842.288120.57084.78610.371214.50362.807221.61745.36

30、100.415916.18153.433922.79486.00770.460717.84324.102523.97226.65440.505519.48924.809225.14967.30120.550321.12155.547026.32707.94790.596422.79326.330827.54068.61450.642624.45277.140228.75419.28120.688726.09997.974729.96779.94780.734927.73928.824531.181310.61440.778829.30049.631932.336211.24880.822730

31、.853010.455033.491111.88320.866732.391711.302134.646012.51760.910633.930412.148935.800913.15200.931534.668212.545036.351413.45430.952535.404812.943336.901813.75670.973436.139913.344337.452314.05910.994336.874613.746138.002714.36141.015337.618514.133138.553214.66381.036238.354214.533839.103614.96611.

32、057139.071614.963539.654115.26851.078139.799615.373240.204615.57091.083039.976015.458940.333315.64161.087940.150715.547940.462115.71241.092840.322715.641540.590915.78311.097740.494615.735340.719715.85381.102540.688115.832640.848515.92461.107440.869515.926040.977215.99531.112341.011116.007741.106016.

33、06611.117241.152516.091441.234816.13681.120341.291816.191041.314916.18081.123341.382316.209841.395016.22481.126441.371516.092641.475116.26881.129441.364416.035141.555216.3128可以判斷出：大約在第1.1203h時，緝私艇追上走私船Z3。（在計算中通過改變函數ode45中t的取值區間來提高其結果的精度，在1.1203h時，y剛略微超過y1，x也十分接近x1，所以判定為在這個時刻緝私艇追上走私船。）即T1=0.6125h T2=0.3865h T3=1.1203hT=T1+T2+T3=2.1193h綜上所述，最佳追趕順序：Z2Z1Z3，所用總