• 2022-06-05 问题

    一空间曲线由参数方程x=t y=sin(2t) , -3<t<3z=cos(3t*t)表示,绘制这段曲线可以由下列哪组语句完成。 A: t=-3:0.1:3;x=t;y=sin(2*t);z=cos(3*t.*t);plot3(x, y, z, t) B: t=-3:0.1:3;x=t;y=sin(2*t);z=cos(3*t*t);plot3(x, y, z) C: t=-3:0.1:3;y=sin(2*t);z=cos(3*t.*t);plot3(x, y, z) D: t=-3:0.1:3;x=t;y=sin(2*t);z=cos(3*t.*t);plot3(x, y, z) E: x=-3:0.1:3;y=sin(2*x);z=cos(3*x.*x);plot3(x, y, z)

    一空间曲线由参数方程x=t y=sin(2t) , -3<t<3z=cos(3t*t)表示,绘制这段曲线可以由下列哪组语句完成。 A: t=-3:0.1:3;x=t;y=sin(2*t);z=cos(3*t.*t);plot3(x, y, z, t) B: t=-3:0.1:3;x=t;y=sin(2*t);z=cos(3*t*t);plot3(x, y, z) C: t=-3:0.1:3;y=sin(2*t);z=cos(3*t.*t);plot3(x, y, z) D: t=-3:0.1:3;x=t;y=sin(2*t);z=cos(3*t.*t);plot3(x, y, z) E: x=-3:0.1:3;y=sin(2*x);z=cos(3*x.*x);plot3(x, y, z)

  • 2021-04-14 问题

    一空间曲线由参数方程x=ty=sin(2t) , -3&lt;t&lt;3z=cos(3t*t)表示,绘制这段曲线可以由下列哪组语句完成。? t=-3:0.1:3;x=t;y=sin (2*t);z=cos (3*t.*t);plot3(x, y, z)|t=-3:0.1:3;x=t;y=sin (2*t);z=cos (3*t*t);plot3(x, y, z)|t=-3:0.1:3;y=sin (2*t);z=cos (3*t.*t);plot3 (x, y, z)|t=-3:0.1:3;x=t;y=sin (2*t);z=cos (3*t.*t);plot3(x, y, z, t)

    一空间曲线由参数方程x=ty=sin(2t) , -3&lt;t&lt;3z=cos(3t*t)表示,绘制这段曲线可以由下列哪组语句完成。? t=-3:0.1:3;x=t;y=sin (2*t);z=cos (3*t.*t);plot3(x, y, z)|t=-3:0.1:3;x=t;y=sin (2*t);z=cos (3*t*t);plot3(x, y, z)|t=-3:0.1:3;y=sin (2*t);z=cos (3*t.*t);plot3 (x, y, z)|t=-3:0.1:3;x=t;y=sin (2*t);z=cos (3*t.*t);plot3(x, y, z, t)

  • 2022-06-14 问题

    曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$

    曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$

  • 2022-06-06 问题

    9. 已知函数$z=z(x,y)$由${{z}^{3}}-3xyz={{a}^{3}}$确定,则$\frac{{{\partial }^{2}}z}{\partial x\partial y}=$( ) A: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ B: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-xy)}{{{({{z}^{2}}-xy)}^{2}}}$ C: $\frac{z({{z}^{3}}-2xyz-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ D: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$

    9. 已知函数$z=z(x,y)$由${{z}^{3}}-3xyz={{a}^{3}}$确定,则$\frac{{{\partial }^{2}}z}{\partial x\partial y}=$( ) A: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ B: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-xy)}{{{({{z}^{2}}-xy)}^{2}}}$ C: $\frac{z({{z}^{3}}-2xyz-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ D: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$

  • 2022-06-08 问题

    设方程\(z^2+y^2+z^2 = 4z\)确定函数\(z=z(x,y)\),则\( { { {\partial ^2}z} \over {\partial {x^2}}} =\) A: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2+ z)}^3}}}\) B: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\) C: \( { { { { (2 - z)}^2} -{x^2}} \over { { {(2 - z)}^3}}}\) D: \( { { { { (2 + z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\)

    设方程\(z^2+y^2+z^2 = 4z\)确定函数\(z=z(x,y)\),则\( { { {\partial ^2}z} \over {\partial {x^2}}} =\) A: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2+ z)}^3}}}\) B: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\) C: \( { { { { (2 - z)}^2} -{x^2}} \over { { {(2 - z)}^3}}}\) D: \( { { { { (2 + z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\)

  • 2022-06-09 问题

    设\(z = {e^{x - 2y}}\),而\(x = \sin t,\;y = {t^3},\)则\( { { dz} \over {dt}} = \)( ) A: \({e^{\sin t - 2{t^3}}}\) B: \({e^{\sin t - 2{t^3}}}\left( {\cos t - 6{t^2}} \right)\) C: \({e^{\sin t - 2{t^3}}}\ {\sin t } \) D: \({e^{\sin t - 2{t^3}}}\,{t^3}\)

    设\(z = {e^{x - 2y}}\),而\(x = \sin t,\;y = {t^3},\)则\( { { dz} \over {dt}} = \)( ) A: \({e^{\sin t - 2{t^3}}}\) B: \({e^{\sin t - 2{t^3}}}\left( {\cos t - 6{t^2}} \right)\) C: \({e^{\sin t - 2{t^3}}}\ {\sin t } \) D: \({e^{\sin t - 2{t^3}}}\,{t^3}\)

  • 2022-06-04 问题

    设\(z = {e^{x - 2y}}\),而\(x = \sin t\),\(y = {t^3}\),则全导数\( { { dz} \over {dt}} = \) A: \({e^{\sin t - {t^3}}}(\cos t - 6{t^2})\) B: \({e^{\sin t - 2{t^3}}}(\sin t - 6{t^2})\) C: \({e^{\cos t - 2{t^3}}}(\cos t - 6{t^2})\) D: \({e^{\sin t - 2{t^3}}}(\cos t - 6{t^2})\)

    设\(z = {e^{x - 2y}}\),而\(x = \sin t\),\(y = {t^3}\),则全导数\( { { dz} \over {dt}} = \) A: \({e^{\sin t - {t^3}}}(\cos t - 6{t^2})\) B: \({e^{\sin t - 2{t^3}}}(\sin t - 6{t^2})\) C: \({e^{\cos t - 2{t^3}}}(\cos t - 6{t^2})\) D: \({e^{\sin t - 2{t^3}}}(\cos t - 6{t^2})\)

  • 2022-06-03 问题

    已知函数定义:def demo(**p): print(p)调用该函数demo(x=1,y=2,z=3),下面输出正确的是: A: {'x': 1, 'y': 2, 'z': 3} B: 1,2,3 C: (1, 2, 3) D: [1, 2, 3]

    已知函数定义:def demo(**p): print(p)调用该函数demo(x=1,y=2,z=3),下面输出正确的是: A: {'x': 1, 'y': 2, 'z': 3} B: 1,2,3 C: (1, 2, 3) D: [1, 2, 3]

  • 2022-06-16 问题

    ()为声压反射率的公式。 A: r<sub>p</sub>=(Z<sub>2</sub>-Z<sub>1</sub>)/Z<sub>1</sub>+Z<sub>2</sub> B: t<sub>p</sub>=1+r<sub>p</sub> C: R=r<sub>p</sub><sup>2</sup> D: D=1-r<sub>p</sub><sup>2</sup>

    ()为声压反射率的公式。 A: r<sub>p</sub>=(Z<sub>2</sub>-Z<sub>1</sub>)/Z<sub>1</sub>+Z<sub>2</sub> B: t<sub>p</sub>=1+r<sub>p</sub> C: R=r<sub>p</sub><sup>2</sup> D: D=1-r<sub>p</sub><sup>2</sup>

  • 2021-04-14 问题

    【填空题】曲线 x = t , y = t 2 , z = t 3 上的点 , , 在该点的切线平行于平面 x + 2 y + z = 4 .

    【填空题】曲线 x = t , y = t 2 , z = t 3 上的点 , , 在该点的切线平行于平面 x + 2 y + z = 4 .

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