• 2022-06-15 问题

    计算曲线积分\({\oint_L {({x^2} + {y^2})} ^3}ds\),其中\(L\)为圆周\(x = a\cos t,y = a\sin t(0 \le t \le 2\pi )\)。 A: \(2\pi {a^7}\) B: \(2\pi {a^6}\) C: \(2\pi {a^5}\) D: \(2\pi {a^8}\)

    计算曲线积分\({\oint_L {({x^2} + {y^2})} ^3}ds\),其中\(L\)为圆周\(x = a\cos t,y = a\sin t(0 \le t \le 2\pi )\)。 A: \(2\pi {a^7}\) B: \(2\pi {a^6}\) C: \(2\pi {a^5}\) D: \(2\pi {a^8}\)

  • 2022-06-15 问题

    已知\(L\)为圆周 \(x = a\cos t,y = a\sin t(0 \le t \le 2\pi )\),则\({\oint_L {({x^2} + {y^2})} ^n}ds{\rm{ = }}\) ( ). A: \(2\pi {a^{2n + 1}}\) B: \(2\pi {a^{2n - 1}}\) C: \(\pi {a^{2n + 1}}\) D: \(\pi {a^{2n - 1}}\)

    已知\(L\)为圆周 \(x = a\cos t,y = a\sin t(0 \le t \le 2\pi )\),则\({\oint_L {({x^2} + {y^2})} ^n}ds{\rm{ = }}\) ( ). A: \(2\pi {a^{2n + 1}}\) B: \(2\pi {a^{2n - 1}}\) C: \(\pi {a^{2n + 1}}\) D: \(\pi {a^{2n - 1}}\)

  • 2022-11-03 问题

    产生周期为1的三角波信号,正确的代码是 A: t=0:1/1000:5;y=sawtooth(2*pi*t,0.5);号,正确的代码是 B: t=0:1/1000:5;y=sawtooth(2*pi*10*t,0.5);,正确的代码是 C: t=0:1/1000:5;y=square(2*pi*t,0.5);� D: t=0:1/1000:5;y=square(2*pi*10*t,0.5);

    产生周期为1的三角波信号,正确的代码是 A: t=0:1/1000:5;y=sawtooth(2*pi*t,0.5);号,正确的代码是 B: t=0:1/1000:5;y=sawtooth(2*pi*10*t,0.5);,正确的代码是 C: t=0:1/1000:5;y=square(2*pi*t,0.5);� D: t=0:1/1000:5;y=square(2*pi*10*t,0.5);

  • 2022-06-15 问题

    计算\({\oint_L {({x^2} + {y^2})} ^n}ds\),其中\(L\)为圆周\(x = a\cos t\),\(y=asint\)\((0 \le t \le 2\pi )\)。 A: \(2\pi {a^{n + 1}}\) B: \(2\pi {a^{2n + 1}}\) C: \(\pi {a^{n + 1}}\) D: \(2\pi {a^{n + 1}}\)

    计算\({\oint_L {({x^2} + {y^2})} ^n}ds\),其中\(L\)为圆周\(x = a\cos t\),\(y=asint\)\((0 \le t \le 2\pi )\)。 A: \(2\pi {a^{n + 1}}\) B: \(2\pi {a^{2n + 1}}\) C: \(\pi {a^{n + 1}}\) D: \(2\pi {a^{n + 1}}\)

  • 2022-05-30 问题

    函数 $y=x^ \pi + \pi x + \pi$的导数 A: $y'=\pi x^ {(\pi-1)} + \pi $ B: $y'=x^ {(\pi-1)} + \pi $ C: $y'=x^ \pi $ D: $y'=x^ \pi \ln \pi $

    函数 $y=x^ \pi + \pi x + \pi$的导数 A: $y'=\pi x^ {(\pi-1)} + \pi $ B: $y'=x^ {(\pi-1)} + \pi $ C: $y'=x^ \pi $ D: $y'=x^ \pi \ln \pi $

  • 2021-04-14 问题

    下列程序运行后,看到的图形( ?????)。 ?????t=0:pi/6:2*pi;[x,y,z]=cylinder(t, 6);surf(x,y,z)view(0,90);axis equal

    下列程序运行后,看到的图形( ?????)。 ?????t=0:pi/6:2*pi;[x,y,z]=cylinder(t, 6);surf(x,y,z)view(0,90);axis equal

  • 2022-07-23 问题

    在使用函数square产生方波信号时,下列哪个代码能够产生幅度为1的单极性,占空比为1/2,周期为8的周期方波信号。( ) A: t=-10:0.01:10; y=square(2*pi*t/4,25)/2+0.5; B: t=-10:0.01:10; y=square(2*pi*t/8,25)/2+0.5; C: t=-10:0.01:10; y=square(2*pi*t/8,50)/2+0.5; D: t=-10:0.01:10; y=square(2*pi*t/4,50)/2+0.5;

    在使用函数square产生方波信号时,下列哪个代码能够产生幅度为1的单极性,占空比为1/2,周期为8的周期方波信号。( ) A: t=-10:0.01:10; y=square(2*pi*t/4,25)/2+0.5; B: t=-10:0.01:10; y=square(2*pi*t/8,25)/2+0.5; C: t=-10:0.01:10; y=square(2*pi*t/8,50)/2+0.5; D: t=-10:0.01:10; y=square(2*pi*t/4,50)/2+0.5;

  • 2022-06-05 问题

    一平面简谐波以速度\(u\)沿\(x\)轴正方向传播,在\(t=t'\)时波形曲线如图所示.则坐标原点\(O\)的振动方程为 A: \(y=a\)cos[\(\frac{u}{b}\)\((t-t')\)\(+\frac{\pi}{2}\)] B: \(y=a\)cos[2\(\pi\)\(\frac{u}{b}\)\((t-t')\)\(-\frac{\pi}{2}\)] C: \(y=a\)cos[\(\pi\)\(\frac{u}{b}\)\((t+t')\)\(+\frac{\pi}{2}\)] D: \(y=a\)cos[\(\pi\)\(\frac{u}{b}\)\((t-t')\)\(-\frac{\pi}{2}\)]

    一平面简谐波以速度\(u\)沿\(x\)轴正方向传播,在\(t=t'\)时波形曲线如图所示.则坐标原点\(O\)的振动方程为 A: \(y=a\)cos[\(\frac{u}{b}\)\((t-t')\)\(+\frac{\pi}{2}\)] B: \(y=a\)cos[2\(\pi\)\(\frac{u}{b}\)\((t-t')\)\(-\frac{\pi}{2}\)] C: \(y=a\)cos[\(\pi\)\(\frac{u}{b}\)\((t+t')\)\(+\frac{\pi}{2}\)] D: \(y=a\)cos[\(\pi\)\(\frac{u}{b}\)\((t-t')\)\(-\frac{\pi}{2}\)]

  • 2022-07-23 问题

    在0~4p区间绘制y=5cos(10t+p/3)关系曲线,下述哪个程序正确 A: t=0:4*pi,y=5cos(10t+pi/3) B: t=0:4p,y=5*cos(10*t+p/3) C: t=0:0.1:4*pi,y=5*cos(10*t+pi/3) D: t=0:0.1:4p,y=5*cos(10*t+p/3)

    在0~4p区间绘制y=5cos(10t+p/3)关系曲线,下述哪个程序正确 A: t=0:4*pi,y=5cos(10t+pi/3) B: t=0:4p,y=5*cos(10*t+p/3) C: t=0:0.1:4*pi,y=5*cos(10*t+pi/3) D: t=0:0.1:4p,y=5*cos(10*t+p/3)

  • 2022-07-23 问题

    在0~4π 区间绘制y=5cos(10t+π/3)关系曲线,下述哪个程序正确? A: t=0:4*pi, y=5cos(10t+pi/3) B: t=0:0.1:4π, y=5*cos(10*t+π/3) C: t=0:4π, y=5*cos(10*t+π/3) D: t=0:0.1:4*pi, y=5*cos(10*t+pi/3)

    在0~4π 区间绘制y=5cos(10t+π/3)关系曲线,下述哪个程序正确? A: t=0:4*pi, y=5cos(10t+pi/3) B: t=0:0.1:4π, y=5*cos(10*t+π/3) C: t=0:4π, y=5*cos(10*t+π/3) D: t=0:0.1:4*pi, y=5*cos(10*t+pi/3)

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