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1、数学建模实验题目解答题目一:慢跑者与狗一个慢跑者在平面上沿椭圆以恒定的常速v=1跑步,设椭圆方程为:x=10+20cost,y=20+5sint.突然有一只狗攻击他.这只狗从原点出发,以恒定速率w跑向慢跑者.狗的运动方向始终指向慢跑者.分别求出w=20,w=5时狗的运动轨迹,并分析狗是攻击到慢跑者.一,建立模型.设时刻t慢跑者的坐标为(X(t),Y(t)),狗的坐标为(x(t),y(t)),又X=10+20cost,Y=20+15sint.由于狗的运动方向始终指向慢跑者,故此时狗与人的坐标连线就是此时狗的轨迹曲线弧处的切
2、线,即dy/dx=(Y-y)/(X-x),y’=(dy/dt)/(dx/dt)又运动时间相同:ïïîïïíì==-+-++-+=-+-++-+=0)0(,0)0()sin1520()sin1520()cos2010()cos2010()sin1520()cos2010(2222yxytytxtwdtdyxtytxtwdtdx,解得可得参数方程为:二,求解模型w=20时,建立m-文件xy1.m如下:functiondy=xy1(t,y)dy=zeros(2,1);dy(1)=20*(10+20*cos(t)-y(1))/s
3、qrt((10+20*cos(t)-y(1))^2+(20+15*sin(t)-y(2))^2);constructionqualityacceptanceandassessmentRegulation(ProfessionalEdition)(DL/T5210.2-2009~DL/T5210.8-2009);1.9thequalitycheckoutandevaluationofelectricequipmentinstallationengineeringcode(DL/T5161.1-2002~5161.17-20
4、02);1.10thenormsofconstructionsupervision,theelectricpowerconstructionsupervisionregulationsdy(2)=20*(20+15*sin(t)-y(2))/sqrt((10+20*cos(t)-y(1))^2+(20+15*sin(t)-y(2))^2);取t0=0,tf=6.0,建立主程序fangcheng1.m如下:t0=0;tf=6.0;[t,y]=ode45('eq3',[t0tf],[00]);T=0:0.1:2*pi;X=1
5、0+20*cos(T);Y=20+15*sin(T);plot(X,Y,'-')holdonplot(y(:,1),y(:,2),'*')轨迹线如下图:发现狗没有攻击到慢跑者,constructionqualityacceptanceandassessmentRegulation(ProfessionalEdition)(DL/T5210.2-2009~DL/T5210.8-2009);1.9thequalitycheckoutandevaluationofelectricequipmentinstallationeng
6、ineeringcode(DL/T5161.1-2002~5161.17-2002);1.10thenormsofconstructionsupervision,theelectricpowerconstructionsupervisionregulations于是,从4.0开始,不断的更改tf的值,发现当tf=3.15时,刚好追上慢跑者.其轨迹线如下图所示:W=5时,建立m-文件xy2.m如下:functiondy=xy2(t,y)dy=zeros(2,1);dy(1)=5*(10+20*cos(t)-y(1))/sq
7、rt((10+20*cos(t)-y(1))^2+(20+15*sin(t)-y(2))^2);dy(2)=5*(20+15*sin(t)-y(2))/sqrt((10+20*cos(t)-y(1))^2+(20+15*sin(t)-y(2))^2);取t0=0,tf=30立主程序fangcheng2.m如下:constructionqualityacceptanceandassessmentRegulation(ProfessionalEdition)(DL/T5210.2-2009~DL/T5210.8-2009);
8、1.9thequalitycheckoutandevaluationofelectricequipmentinstallationengineeringcode(DL/T5161.1-2002~5161.17-2002);1.10thenormsofconstructionsupervision,theelectri