數(shù)學實驗課后習題解答

實驗一曲線繪圖y=3*a*t.A3J(l+t.A2);

【練習與思考】plot(x,y)

畫出下列常見曲線的圖形。8.擺線x=a(t-sinf),y=。(1-cost)

以直角坐標方程表示的曲線：clear;clc;

_3a=l;b=l;

1.立方曲線y=xt=0:pi/50:6*pi;

clear;x=a*(t-sin⑴)；

x=-2:0.1:2;y=b*(l-cos(t));

y=x.A3;plot(x,y);

plot(x,y)axisequal

2.立方拋物線y=融gridon

9.內(nèi)擺線(星形線)

clear;222

y=-2:0.1:2;x=flcos3y=tzsin3r(x3+y3=a3)

x=y.A3;

plot(x,y)

gridonclear;

a=l;

__-x2

3.高斯曲線y=et=0:pi/50:2*pi;

x=a*cos(t).A3;

clear;

y=a*sin(t).A3;

x=-3:0.1:3;

plot(xy)

y=exp(-x.A2);9

10.圓的漸伸線(漸開線)

plot(x,y);

gridonx=Q(COSr+/sinr),y=Q(sint-tcost)

%axisequal

以參數(shù)方程表示的曲線

2t=0:pi/50:6*pi;

4.奈爾拋物線x=/，y=廣(y=x3)x=a*(cos(t)+t.*sin(t));

y=a*(sin(t)+t.*cos(t));

clear;plot(x,y)

t=-3:0.05:3;gridon

x=t.A3;y=t.A2;11.空間螺線

plot(x,y)x=acost.y=bsint.z=ct

axisequal

gridonclear

a=3;b=2;c=l;

5.半立方拋物線x=t2,y=t3(y2=x3)

t=0:pi/50:6*pi;

clear;x=a*cos(t);

t=-3:0.05:3;y=b*sin⑴；

x=t.A2;y=t.A3;z=c*t;

plot(x,y)plot3(x,y,z)

%axisequalgridon

gridon以極坐標方程表示的曲線：

6.迪卡爾曲線12.阿基米德線r二。0,r2()

（3.30）clear;

小需叼=a=l;

phy=0:pi/50:6*pi;

rho=a*phy;

clear;

polar(phy,rho,*r-*')

a=3;t=-6:0.1:6;

x=3*a*tJ(l+t.A2);13.對數(shù)螺線「=6"。

y=3*a*t.A2J(l+t.A2);clear;

piot(x,y)a=0.1;

7.蔓葉線phy=0:pi/50:6*pi;

233rho=exp(a*phy);

atat2x、

~~(zy=—)polar(phy,rho)

l+r1+ra-x14.雙紐線

clear;r2=a2cos2^((x2+y2)2=a2(x2-y2))

a=3;t=-6:0.1:6;

x=3*a*t.A2./(l+t.A2);

clear;16.四葉玫瑰線r=?sin2(p,r>0

a=l;

clear;close

phy=-pi/4:pi/50:pi/4;

a=l;

rho=a*sqrt(cos(2*phy));

phy=0:pi/50:2*pi;

polar(phy,rho)

rho=a*sin(2*phy);

holdon

polar(phy,rho)

polar(phy,-rho)

15.雙紐線17.三葉玫瑰線r=Qsin3e，rN0

r2=a2sin((x24-y2)2=2a2xy)clear;close

a=l;

phy=0:pi/50:2*pi;

clear;rho=a*sin(3*phy);

a=l;polar(phy,rho)

phy=0:pi/50:pi/2;

18.三葉玫瑰線r=Qcos3o，rNO

rho=a*sqrt(sin(2*phy));

polar(phy,rho)clear;close

holdon

polar(phy,-rho)

a=l;

phy=0:pi/50:2*pi;

rho=a*cos(3*phy);

polar(phy,rho)

實驗二極限與導數(shù)

【練習與思考】

1.求下列各極限

(1)lim(l——)"(2)lim-"+3"(3)lim(J〃+2-2j.+l+6)

〃一>8〃〃一>8n—>00

dear;

symsn

yl=limit((l-l/n)An,n,inf)symsxm

y2=limit((nA3+3An)A(l/n),nunf)y7=!imit(cos(m/x),x,inf)

y3=Iimit(sqrt(n+2)-2*sqrt(n+l)+sqrt(n),n,inf)y8=limit(1/x-l/(exp(x)-1),x,l)

y9=limit(((l+x)A(l/3)-l)/x,x,0)

yl=l/exp(1)

y2=3y7=1

y3=0y8=(exp(1)-2)/(exp(1)-1)

y9=1/3

(4)lim(-----------)(5)limxcot2x(6)2.考慮函數(shù)

xT廠一1x-\XT。

f(x)=3x2sin，)，-2<尤<2

lim(7x2+3x-x)

作出圖形，并說出大致單調區(qū)間；使用diff求/'(X),并求

clear;/(X)確切的單調區(qū)間。

symsx;

clear;close;

j4=limit(2/(xA2-l)-l/(x-l),x,l)

symsx;

y5=limit(x*cot(2*x),x,0)

f=3*xA2*sin(xA3);

y6=liniit(sqrt(xA2+3*x)-x,x,inf)

czplot(f,[-2,2J)

gridon

y4=-1/2

大致的單調增區(qū)間：

y5=1/2[-2,-1.7],[-1.3,1.2],[1.7,2];

y6=3/2大致的單點減區(qū)間：[-1.7,?1.3"1.21.7]

n=din(f,x,i)

ezplot(n,[-2,2])

Z11、

(7)lim(cos—⑻*K)(9)Iine([-53],[0,0])

?reXgridon

axis([-2.1,2.1,-60,120J)

V1+x—1

lim---------

fl=

6*x*sin(XA3)+9*x^4*cos(xA3)

clear;

用fzero函數(shù)找/'(X)的零點，即原函數(shù)/(X)的駐點2*x*sin(x^2-x-2)+x^2*cos(x^2-x-

2)*(2*x-1)

xl=fzeroC6*x*sin(xA3)+9*xA4*cos(xA3),,[-2,-1.7])

用函數(shù)找)的零點，即原函數(shù))的駐點

x2=fzero(,6*x*sin(xA3)+9*xA4*cos(xA3),,[-1.7,-1.5])fzerof'(Xf(X

x3=fzeroC6*x*sin(xA3)+9*xA4*cos(xA3),,[-1.5,-l.l])x1=fzero(*2*x*sin(xA2-x-2)+xA2*cos(xA2-x-2)*(2*x-l)\[-2,-

x4=fzero(,6*x*sin(xA3)+9*xA4*cos(xA3)\0)1.2])

x5=fzero(*6*x*sin(xA3)+9*xA4*cos(xA3)\|1,1,5])x2=fzero('2*x*sin(xA2-x-2)+xA2*cos(xA2-x-2)*(2*x-l)*,[-1.2,-

x6=fzero(*6*x*sin(xA3)+9*xA4*cos(xA3),,[1.5,1.7J)0,5])

x7=fzero(*6*x*sin(xA3)+9*xA4*cos(xA3)*,|1.7,2|)x3=fzero(,2*x*sin(xA2-x-2)+xA2*cos(xA2-x-2)*(2*x-l),,[-

xl=0.5,1.2])

-1.9948x4=fzcro(,2*x*sin(xA2-x-2)+xA2*cos(xA2-x-2)*(2*x-l)',[1.2,2J)

x2=

-1.6926xl=

x3=-1.5326

-1.2401x2=

x4=-0.7315

0x3=

x5=-3.2754e-027

1.2401x4=

x6=1.5951

1.6926ff=@(x)x.A2.*sin(x.人2-X-2)

x7=ff(-2),fT(xl),n(x2)4T(x3)JT(x4),ff(2)

1.9948

確切的單調增區(qū)間：[-1.9948,-1.6926],[-ff=

1.2401J.24011.6926,1.9948]@(x)x.A2.*sin(x.x2-x-2)

確切的單調減區(qū)間：[-2,-1.9948],[-1.6926,-ans=

1.2401],[1.2401,1.6926]，口.9948,2]-3.0272

3.對于下列函數(shù)完成下列工作，并寫出總結報告，評論極ans=

值與導數(shù)的關系，2.2364

ans=

⑴作出圖形，觀測所有的局部極大、局部極小和全局最大、全局

-0.3582

最小值點的粗略位置；ans=

3)求r(x)所有零點(即/(%)的駐點)；-9.7549e-054

ans=

求出駐點處()的二階導數(shù)值；

(iii)/X-2.2080

(iv)用fmin求各極值點的確切位置；ans=

(v)局部極值點與/'(X),/”(X)有何關系？0

(1)f(x)=x1sin(A：2-x-2),XG[-2,2]實驗三級數(shù)

【練習與思考】

(2)f(x)=3x5-20/+10,XG[-3,3]

I.用taylor命令觀測函數(shù))=/(X)的Maclaurin展開式的前

(3)f(x)=|x3-x2-x-2|,G[0,3]

幾項,然后在同一坐標系里作出函數(shù)＞=/(X)和它的Tayloi"展開

clear;close;式的前幾項構成的多項式函數(shù)的圖形，觀測這些多項式函數(shù)的圖

symsx;

形向))=/(X)的圖形的逼近的情況

f=xA2*sin(xA2-x-2)

ezplot(f,[-2,2])(1)/(x)=arcsinx

gridonclear;

symsx

y=asin(x);

x*2*sin(x^2-x-2)

yl=taylor(y,0,l)

局部極大值點為：-1.6,局部極小值點為為：-0.75,-1.6y2=taylor(y,0,5)

全局最大值點為為：?1.6,全局最小值點為：-3y3=taylor(y,0,10)

n=diff(f,x,l)y4=taylor(y,0,15)

ezplot(fl,[-2^J)x=-l:0.1:l;

line([-53],[0,0])y=subs(y,x);

gridonyl=subs(yl,x);

axis([-2.1,2.1,-6,20])y2=subs(y2,x);

y3=subs(y3,x);

y4=subs(y4,x);

flplot(x,y,x,yl,,:,^,y2,,-.,,x,y3,,-,^,y4,':,,,lincwidth，3)

clear;

yi=symsx

oy=sin(x)A2;

y2=yl=taylor(y,0J)

x人3/6+xy2=taylor(y,0,5)

y3=y3=taylor(y,0J0)

(35*x^9)/1152+(5*XA7)/112+(3*xA5)/40+y4=taylor(y,0,15)

x人3/6+xx=-pi:0.1:pi;

y4=y=subs(y,x);

(231*xA13)/13312+(63*x*ll)/2816+yl=subs(yl,x);

(35*x^9)/1152+(5*xx7)/112+(3*xA5)/40+y2=subs(y2,x);

x人3/6+xy3=subs(y3,x);

y4=subs(y4,x);

,,,,,,

(2)f(x)=arctanxplot(x,y,x,yl,:,x,y2,-.,x,y3/-*,x,y4,':',linewidthr3)

clear;

yl=

symsx

0

y=atan(x);yl=taylor(y,0,3)

y2=

y2=taylor(y,0,5),y3=taylor(y,0,10),y4=taylor(y,0,15)

x人2-XA4/3

y3=

y=subs(y,x);yl=subs(yl,x);y2=subs(y2,x);x

-x人8/315+(2*x6)/45-x人4/3+x入2

y3=subs(y3,x);y4=subs(y4,x);

y4=

plot(x,y,x,yl,,:,,x,y2,,-.,,x,y3,*-,,x,y4,,:,,'linewidth,,3)

(4*x*14)/42567525-(2*x*12)/467775+

(2*x^l0)/14175-x入8/315+(2*x^6)/45-

yi=

X人4/3+X人2

X

y2=

x-xx3/3⑸/(X)=

y3=l-X

x

x人9/9-x7/7+x人5/5-x人3/3+xclear;

y4=symsx

A

x人13/13-xll/ll+x入9/9-XA7/7+x入5/5-y=exp(x)/(l-x);

x人3/3+xyl=taylor(y,0,3)

y2=taylor(y,0,5)

(3)/(X)=J

y3=taylor(y,0,10)

y4=taylor(y,0,15)

clear;x=-l:0.1:0;

symsxy=subs(y,x);

y=exp(xA2);yl=subs(yl,x);

yl=taylor(y,0,3)y2=subs(y2,x);

y2=taylor(y,0,5)y3=subs(y3,x);

y3=taylor(y,0,10)y4=subs(y4,x);

y4=taylor(y,0,15)plot(x,y,x,yl,,:,^,y2,,-/,x,y3,,-,,x,y4,':',Tlinewidth'4)

y=subs(y,x);yl=

yl=subs(yl,x);(5*x*2)/2+2*x+1

y2=subs(y2,x);y2=

xA

y3=subs(y3,x);(65*x*4)/24+(8*x3)/3+(5*x2)/2+2*x+

y4=subs(y4,x);1

plot(x,y,x,yl；:,,x,y2,,-.,,x,y3,,-,,x,y4,':,,,linewidth'3)y3=

(98641*XA9)/36288+(109601*xA8)/40320+

yl=(685*XA7)/252+(1957*xA6)/720+

XA2+1(163*x^5)/60+(65*x^4)/24+(8*x^3)/3+

y2=(5*x*2)/2+2*x+1

x人4/2+XA2+1y4=

x

y3=(47395032961*x14)

A

x人8/24+x人6/6+x入4/2+x人2+1(8463398743*X13)/3113510400+

y4=(260412269*xA12)/95800320+

A

X^14/5040+XA12/720+x人10/120+x8/24+/4989600+

AA

x人6/6+x人4/2+x人2+1(9864101*x10)/3628800+(98641*x9)/36288+

(109601*XA8)/40320+(685*xA7)/252+

(1957*x^6)/720+(163*x^5)/60+(65*x^4)/24

2

(4)f(x)=sinX+(8*x^3)/3+(5*x^2)/2+2*x+1

ans=

(6)/(x)=ln(x+J1+，)

2.718281828459045534884808148490265011787414

clear;55078125

symsxn=

y=Iog(x+sqrt(l+xA2));70

yl=taylor(y,0,3)精確到小數(shù)點后100位，這時應計算到這個無窮級數(shù)的前71項,

y2=taylor(y,0,5)理由是誤差小于1()的負100次方,需要最后一項小于10的負100

y3=taylor(y,0,10)次方，由上述循環(huán)知n=70時坡后一項小于10的負100次方，故

y4=taylor(y,0,15)應計算到這個無窮級數(shù)的前71項.

x=-l:0.1:l;

4.用練習3中所用觀測法判斷下列級數(shù)的斂散性

y=subs(y,x);

81

yl=subs(yl,x);⑴E——T

y2=subs(y2,x);

y3=subs(y3,x);

y4=subs(y4,x);clear;clc;

plot(x,y,x,yl,,:'?x,y2,'-.*,x,y3,*-*,'linewidth',3)epsl=0.000001;

N=50000;p=1000;

yi=symsn

XUn=l/(nA2+nA3);

y2=sl=symsum(Un,l,N);

x-x人3/6s2=symsum(Un,l,N+p);

y3=sa=vpa(s2-sl);

(35*x^9)/1152-(5*XA7)/112+(3*xA5)/40-sa=setstr(sa);

x人3/6+xsa=str2num(sa);

y4=fprintf('級數(shù)')

(231*xA13)/13312-(63*xAll)/2816+disp(Un)

(35*x^9)/1152-(5*xx7)/112+(3*xx5)/40-ifsa<epsl

xx3/6+xdispC收斂，

coi--2kelse

2.求公式EK=—#=1,2,…，中的數(shù)dispC發(fā)散?

〃=inend

m,k=4,5,6,7,8的值.

k級數(shù)1/(n*3+n，2)收斂

k=[45678);clcar;closc

symsnsymsn

A

symsum(lJn.(2*k),1,inf)s=n；

fork=l:100

ans=

s(k)=symsum(l/(nA3+nA2),l,k);

[pi人8/9450,pi-10/93555,

end

(691*pi*12)/638512875,(2*piA14)/18243225

zplot(s,'.')

(3617*pi^l6)/325641566250]

81

001(2)y——

3.利用公式>一二e來計算e的近似值。精確到小數(shù)點后占〃2"

clcar;clc;

100位,這時應計算到這個無窮級數(shù)的前多少項?請說明你的理由.epsl=0.000001;

解：Matlab代碼為N=50000;p=1000;

clear;clc;closesymsn

epsl=1.0e-100;Un=l/(n*2An);

ep=l;fn=l;a=l;n=l;sl=symsum(Un,l,N);

whileep>epsls2=symsum(Un,l,N+p);

a=a+fn;sa=vpa(s2-sl);

n=n+l;sa=setstr(sa);

fn=fn/n;sa=str2nuni(sa);

ep=fn;fprintfC級數(shù)

enddisp(Un)

fn