数学实验期末作业(缉私艇问题).docx

精选优质文档-倾情为你奉上问题：缉私艇问题续。（1）在本问题的求解过程中，假定了走私艇的逃跑方向是正北方向，而初始缉私艇的位置在x轴正向。如果放宽这个假定，也就是当这个夹角是任意角度时，如何建立方程进行求解。以下面数值为例进行求解：b=40,a=20,c=15,其中坐标系如课件上所述，走私船的方向为45°。 （2）如果有多个走私艇在一个位置上进行交易，而缉私艇向该方向追赶。这些走私艇向不同方向四散逃走，问如何安排追赶路线？ （假定缉私艇追上一个立刻掉头追赶另外一个，中间没有时间停留）。以下面数值为例进行求解：b=40,a1=20, a2=25, a3=30,c=15,三个角度分别为45°，90°和-60°。C(x,y)Q(c+atcos,atsin)R(c+ycot,y)  yx0走私船缉私艇（1）缉私艇速度为b，走私船速度为a，初始距离为c。设走私船的速度方向与缉私艇初始速度方向呈角，因为缉私艇速度方向始终指向走私船方向，故两者大致轨迹如图所示。根据x与y的速度关系可列出以下微分方程：即： 由dsolve列方程无法得到x(t), y(t)的解析解，通过变换消去t可得到以下微分函数关系：再通过dsolve函数求解，仍无法得到y(x)的解析解。因此只能用数值解法求其解。给定初值：a=20，b=40，c=15，=45°。使用MATLAB求解可得：模型的数值解tx（t）（海里）y（t）（海里）x1（t）（海里）y1（t）（海里）（其中x(t),y(t)表示缉私艇的坐标，x1(t),y1(t)表示走私船的坐标。）00015.000000.500019.04715.182722.07117.07111.000029.123214.130229.142114.14211.500036.092621.186536.213221.21322.000043.251328.248643.284328.28432.500050.112335.310950.355335.35533.000057.275442.523757.426442.42643.500064.495249.195264.497549.49754.000071.430756.486071.568556.56854.500078.688663.518178.639663.63965.000085.767370.653485.710770.71075.500092.439577.594592.781777.78176.000099.623284.802599.852884.85286.5000106.717791.6557106.923991.92397.0000113.825198.8215113.994998.99497.5000120.9647105.9589121.0660106.06608.0000127.5442112.8577128.1371113.13718.5000134.9007119.1727135.2082120.20829.0000142.2377127.1158142.2792127.27929.5000149.5407134.1205149.3503134.350310.0000156.2282141.2136156.4214141.4214轨迹图由图和数据初步判断大约在t=0.5到t=1之间缉私艇追上走私船。将t的范围缩小到7-8之间并提高精度使用matlab求出其数值解如下：模型的数值解（0.5-0.7精确）t（h）x（t）海里）y（t）（海里）x1（t）（海里）y1（t）（海里）0.521019.75815.631322.36817.36810.546020.58486.195322.72167.72160.571021.38436.794823.07528.07520.596022.16627.417423.42878.42870.607522.52457.706423.59148.59140.619022.87738.002023.75418.75410.630523.22278.305823.91688.91680.642023.56408.614324.07959.07950.645623.67058.710924.13039.13030.649223.77668.808224.18129.18120.652823.88218.906024.23219.23210.656423.98709.004324.28299.28290.660024.09269.102224.33389.33380.663624.19689.201524.38469.38460.667224.29829.303324.43559.43550.670824.40019.404624.48639.48630.672324.43349.457824.50849.50840.673924.48589.503024.53049.53040.675524.56289.526424.55259.55250.677024.59159.553924.57459.57450.678624.53449.617824.59659.59650.680124.50489.648124.61869.61860.681724.53989.620224.64069.64060.683324.60269.606824.66279.66270.684624.66379.646124.68139.68130.685924.68439.683224.69999.69990.687224.63309.709424.71869.71860.688524.60769.721824.73729.73720.689824.66229.721624.75539.75530.691124.71059.741124.77339.77330.692424.74639.787924.79139.79130.693624.78479.815224.80949.80940.694824.84049.782124.82629.82620.696024.86509.760824.84309.84300.697224.84139.787424.85989.85980.698424.83049.834424.87669.87660.699624.89429.876624.89459.89450.700924.91449.903824.91239.91230.702224.85339.901824.93029.93020.703424.81719.899324.94819.94810.704624.86329.916324.96499.96490.705824.90649.937824.98179.98170.707024.94389.965024.99859.99850.708224.98049.993125.015310.01530.709725.05519.984725.036710.0367可以判断出：大约在第0.6755h时，缉私艇追上走私船。（在计算中通过改变函数ode45中t的取值区间来提高其结果的精度，在0.6755h时，x已略微超过x1，y也十分接近y1，所以判定为在这个时刻缉私艇追上走私船。）45°60° yx0V1=20V2=25V3=30(15,0)Vj=40(2)在这个问题中，三艘走私船各自以恒定的速度、恒定的方向行进。从第一问中能够知道缉私艇在追一艘走私船时，走私船的速度和方向都对时间有所影响。因为三艘船都不停地在各自所在的直线上匀速运动，所以初步猜测为：在追第一艘船时，选择一艘在最短时间内能够追上的船，追上后剩下的一艘最短时间内能够追上的船，最后再追第三艘走私船，这样所用时间会最短。简单的理论依据：三艘走私船在从一个点向四周匀速扩散开来，以不变的速度追赶，要在最短时间内追到，则要使他们的相对距离在每被追上一个时保持最小，则每次追赶应该从最好追的一个点开始追赶。从以上思想出发，计算（标记第i艘走私船为Zi,缉私艇为J）从零点开始分别追上Z1，Z2，Z3的时间： t1在第一问已求出t1=0.6755h。对于t2，改变a=25，=90°，通过数值法求出：t（h）x（t）海里）y（t）（海里）x1（t）（海里）y1（t）（海里）0.504914.725010.990115.000012.62180.518014.805611.511115.000012.95110.531214.866212.034615.000013.28040.544414.912912.559615.000013.60970.553214.941012.910815.000013.82980.562014.962213.262515.000014.04990.570814.974813.614315.000014.27000.579614.983913.966215.000014.49010.585514.996814.202015.000014.63690.591415.002514.437315.000014.78380.597214.993414.670715.000014.93060.603114.990114.904015.000015.07740.605014.996414.979615.000015.12440.606914.999615.054915.000015.17140.608714.996915.129315.000015.21840.610614.996015.203615.000015.26540.612515.033615.297015.000015.31240.614415.023115.355715.000015.35940.616314.924315.348915.000015.40640.618114.871215.346315.000015.45340.619614.919715.386815.000015.49110.621214.952415.437515.000015.52890.622714.962815.499015.000015.56660.624214.975215.559415.000015.60440.625815.040915.619815.000015.64450.627415.063715.662915.000015.68470.629014.995415.680615.000015.72490.630614.948715.705215.000015.76500.632015.001515.750315.000015.79890.633315.010815.796515.000015.83270.634714.951215.835815.000015.86660.636014.921515.864315.000015.90050.637214.977515.885515.000015.92980.638414.998015.921215.000015.95910.639514.962615.969715.000015.98850.640714.947816.004115.000016.01780.641815.005116.019815.000016.04480.642915.028416.045215.000016.07170.643914.992916.076915.000016.09860.645014.972216.106515.000016.12560.646015.022516.126915.000016.1505可以判断出：大约在第0.6125h时，缉私艇追上走私船。（在计算中通过改变函数ode45中t的取值区间来提高其结果的精度，在0.6125h时，x刚略微超过x1，y也十分接近y1，所以判定为在这个时刻缉私艇追上走私船。）即t2=0.6125h同理，对于t3，改变a=30，=-60°，通过数值法求出：t（h）x（t）海里）y（t）（海里）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 19.746627.9447 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 27.030131.1176 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 30.2105 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可以判断出：大约在第1.1699h时，缉私艇追上走私船。（在计算中通过改变函数ode45中t的取值区间来提高其结果的精度，在1.1699h时，y刚略微超过y1，x也十分接近x1且有略微下降，所以判定为在这个时刻缉私艇追上走私船。）即t3=1.1699h因为t1=0.6755h，t2=0.6125h，可判断：t2<t1<t3所以首先追赶Z2用时最短T1=0.6125h。由此，可以知道：走私船与缉私艇速度方向夹角越小，追赶时间越长；走私船速度越大，追赶时间越长。此时，已追赶上Z2，各点位置已发生变化，在原坐标系中，缉私艇坐标J(15,15.3125),Z1(23.6621,8.6621),Z3(24.1875,-15.9132),如图所示： 1yx0Z1缉私艇Jx145°60° yx0V1=20V3=30(15,0)Vj=4031x1x2以缉私艇为原点建立坐标系如下图：因为v1与v2夹角为45°，v2与v3夹角为60+90=150度，所以当缉私艇追上Z2后调转方向时，（若指向Z1，与其方向夹角为1，距离为x1；若指向Z3，与其夹角为3，距离为x3），很容易判断：x3>x1，1>3。前面已推出：走私船与缉私艇速度方向夹角越小，追赶时间越长；走私船速度越大，追赶时间越长。因为v3>v1，1>3，x3>x1，所以要在最短时间内追上第二个目标的话，应选择追赶Z1。通过坐标J(15,15.3125)与Z1(23.6621,8.6621)容易求出：海里通过向量法求出。对于追赶Z1所用时间T2，改变a=20，=82.52°，c=10.9206通过数值法求出：t（h）x（t）海里）y（t）（海里）x1（t）（海里）y1（t）（海里）0.284810.28913.974111.66215.64720.297310.59434.369211.69455.89460.309710.87214.783711.72706.14200.322211.12285.215211.75956.38940.330311.27135.501211.78056.54910.338311.40655.793711.80146.70870.346411.52626.092611.82246.86830.354411.63276.396411.84347.02800.359611.69886.594011.85697.13080.364811.75666.793611.87047.23360.370011.80176.995211.88397.33650.375211.84297.197811.89747.43930.376811.85707.261211.90167.47150.378411.87027.324811.90587.50370.380011.88247.388711.91007.53590.381711.89387.452611.91437.56810.383311.89457.499011.91857.60030.384911.90487.573611.92277.63250.386511.93437.684811.92707.66480.388211.94367.725411.93127.69700.389511.91497.662711.93487.72450.390911.89827.627511.93847.75190.392311.90777.652711.94207.77940.393711.92697.712711.94567.80690.394911.92547.753611.94897.83170.396211.93597.809611.95217.85650.397411.96697.882411.95547.88130.398711.96927.912611.95877.90610.399911.90927.892611.96197.93090.401211.87417.887911.96527.95570.402411.89817.901911.96847.98050.403711.94107.936711.97178.00530.405011.98497.989911.97518.03120.406311.98778.033211.97858.05720.407611.91958.054811.98198.08310.408911.88398.068011.98538.1091可以判断出：大约在第0.3865h时，缉私艇追上走私船Z1。（在计算中通过改变函数ode45中t的取值区间来提高其结果的精度，在0.3865h时，x刚略微超过x1，y也十分接近y1，所以判定为在这个时刻缉私艇追上走私船。）即T1=0.6125h T2=0.3865h。45°60° yx0V3=30(15,0)Vj=40X33此时，缉私艇已追上Z1，所有点位置已改变，此时在最初的坐标系中J(29.1280，14.1280),Z3(29.9850,-25.9548),最后，缉私艇开始追赶Z3。通过坐标J(29.1280，14.1280),Z3(29.9850,-25.9548)易求得：海里。通过向量法可求得。以缉私艇为原点建立坐标系如下图：1yx0Z3缉私艇JX3对于追赶Z3所用时间T3，改变a=30，=28.78°，c=11.8578通过数值法求出：t（h）x（t）海里）y（t）（海里）x1（t）（海里）y1（t）（海里）0.291611.47961.810419.52424.21110.331412.99842.288120.57084.78610.371214.50362.807221.61745.36100.415916.18153.433922.79486.00770.460717.84324.102523.97226.65440.505519.48924.809225.14967.30120.550321.12155.547026.32707.94790.596422.79326.330827.54068.61450.642624.45277.140228.75419.28120.688726.09997.974729.96779.94780.734927.73928.824531.181310.61440.778829.30049.631932.336211.24880.822730.853010.455033.491111.88320.866732.391711.302134.646012.51760.910633.930412.148935.800913.15200.931534.668212.545036.351413.45430.952535.404812.943336.901813.75670.973436.139913.344337.452314.05910.994336.874613.746138.002714.36141.015337.618514.133138.553214.66381.036238.354214.533839.103614.96611.057139.071614.963539.654115.26851.078139.799615.373240.204615.57091.083039.976015.458940.333315.64161.087940.150715.547940.462115.71241.092840.322715.641540.590915.78311.097740.494615.735340.719715.85381.102540.688115.832640.848515.92461.107440.869515.926040.977215.99531.112341.011116.007741.106016.06611.117241.152516.091441.234816.13681.120341.291816.191041.314916.18081.123341.382316.209841.395016.22481.126441.371516.092641.475116.26881.129441.364416.035141.555216.3128可以判断出：大约在第1.1203h时，缉私艇追上走私船Z3。（在计算中通过改变函数ode45中t的取值区间来提高其结果的精度，在1.1203h时，y刚略微超过y1，x也十分接近x1，所以判定为在这个时刻缉私艇追上走私船。）即T1=0.6125h T2=0.3865h T3=1.1203hT=T1+T2+T3=2.1193h综上所述，最佳追赶顺序：Z2Z1Z3，所用总时间为T=2.1193h。附：代码：function dx=jisi(t,x,a,b,c,e)s=sqrt(c+a*t*cos(e)-x(1)2+(a*t*sin(e)-x(2)2);dx=b*(c+a*t*cos(e)-x(1)/s;b*(a*t*sin(e)-x(2)/s;x0=0 0;a=20;b=40;c=15;e=pi/3;t,x=ode45(jisi,0 2,x0,a,b,c,e);x1=c+a*t*cos(e);y1=a*t*sin(e);t,x,x1,y1plot(x(:,1),x(:,2),'r*',x1,y1,'g*'),gridxlabel('x','FontSize',16),ylabel('y','FontSize',16)axis(0 100 0 100);x0=0 0;a=20;b=40;c=15;e=pi/4;opt=odeset('RelTol',1e-6,'AbsTol',1e-9);t,x=ode45(jisi,0 1,x0,opt,a,b,c,e);x1=c+a*t*cos(e);y1=a*t*sin(e);t,x,x1,y1;plot(x(:,1),x(:,2),'r',x1,y1,'g'),gridxlabel('x','FontSize',16),ylabel('y','FontSize',16)axis(0 200 0 200);（提高精度）y=dsolve('D2y=(sqrt(1+(Dy)2)*(sin(e)-cos(e)*Dy)2)*a/b/(c-x)*sin(e)+y*cos(e)','y(0)=0','Dy(0)=0','x')（求解y(x)的微分方程，无法解出。）x,y=dsolve('Dx=b*(c+a*t*cos(e)-x)/sqrt(c+a*t*cos(e)-x(1)2+(a*t*sin(e)-x(2)2)','Dy=b*(a*t*sin(e)-y)/sqrt(c+a*t*cos(e)-x(1)2+(a*t*sin(e)-x(2)2)','Dy(0)=0','Dx(0)=b','t')（求解y(t)，x(t)的微分方程，无法解出。）专心-专注-专业