求微分方程[img=269x55]17da6536a9fba07.png[/img]的通解；（） A: (C15*sin(2*t))/exp(3*t) + (C16*sin(2*t))/exp(3*t) B: (C15*cos(2*t))/exp(3*t) - (C16*sin(2*t))/exp(3*t) C: (C15*cos(2*t))/exp(3*t) + (C16*cos(2*t))/exp(3*t) D: (C15*cos(2*t))/exp(3*t) + (C16*sin(2*t))/exp(3*t)

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2022-06-19 问题

求微分方程[img=269x55]17da6536a9fba07.png[/img]的通解；（） A: (C15sin(2t))/exp(3t) + (C16sin(2t))/exp(3t) B: (C15cos(2t))/exp(3t) - (C16sin(2t))/exp(3t) C: (C15cos(2t))/exp(3t) + (C16cos(2t))/exp(3t) D: (C15cos(2t))/exp(3t) + (C16sin(2t))/exp(3t)

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: ${e^{\sin t - 2{t^3}}}(\cos t - 6{t^2})$

2021-04-14 问题

已知u(t)=2 cos (2t-90°)V，i(t)= cos (2t+150°)mA，则（）。

2022-06-09 问题

设 $z=e^{x-2y}$, 而 $x=\sin t, y=t^3$, 则 $\displaystyle\frac{\mathrm{d}z}{\mathrm{d}t}=$ A: $\displaystyle\cos t+3t^2$ B: $e^{\sin t-2t^3}(\cos t+3t^2)$ C: $\displaystyle\cos t-6t^2$ D: $e^{\sin t-2t^3}(\cos t-6t^2)$

设 $z=e^{x-2y}$, 而 $x=\sin t, y=t^3$, 则 $\displaystyle\frac{\mathrm{d}z}{\mathrm{d}t}=$ A: $\displaystyle\cos t+3t^2$ B: $e^{\sin t-2t^3}(\cos t+3t^2)$ C: $\displaystyle\cos t-6t^2$ D: $e^{\sin t-2t^3}(\cos t-6t^2)$

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: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$

2022-06-04 问题

下列命令执行后得到的图形是 ______ 。x=@(t) 2sin(t);y=@(t) 2cos(t)

下列命令执行后得到的图形是 ______ 。x=@(t) 2*sin(t);y=@(t) 2*cos(t)

2022-05-26 问题

设函数$$y=y(x)$$由$$\left\{ \begin{matrix} x=a(t-\sin t), \\ y=a(1-\cos t) \\ \end{matrix} \right.$$确定，则$${y}''(x)=$$()． A: $$-\frac{1}{a(1-\cos t)}$$ B: $$-\frac{1}{a{{(1-\cos t)}^{2}}}$$ C: $$\frac{1}{a(1-\cos t)}$$ D: $$\frac{1}{a{{(1-\cos t)}^{2}}}$$

设函数$$y=y(x)$$由$$\left\{ \begin{matrix} x=a(t-\sin t), \\ y=a(1-\cos t) \\ \end{matrix} \right.$$确定，则$${y}''(x)=$$()． A: $$-\frac{1}{a(1-\cos t)}$$ B: $$-\frac{1}{a{{(1-\cos t)}^{2}}}$$ C: $$\frac{1}{a(1-\cos t)}$$ D: $$\frac{1}{a{{(1-\cos t)}^{2}}}$$

2022-06-09 问题

设$z = f(x,y)$，$x = \sin t$，$y = {t^3}$，则全导数$ { { dz} \over {dt}} = $ A: ${f'_x} \sin t+ 3{t^2}{f'_y}$ B: ${f'_x} \cos t+ {t^2}{f'_y}$ C: ${f'_x} \cos t+ 3{t^2}{f'_y}$ D: ${f'_y} \cos t+ 3{t^2}{f'_x}$

设$z = f(x,y)$，$x = \sin t$，$y = {t^3}$，则全导数$ { { dz} \over {dt}} = $ A: ${f'_x} \sin t+ 3{t^2}{f'_y}$ B: ${f'_x} \cos t+ {t^2}{f'_y}$ C: ${f'_x} \cos t+ 3{t^2}{f'_y}$ D: ${f'_y} \cos t+ 3{t^2}{f'_x}$

2022-06-06 问题

cos(2α)=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)

2022-06-27 问题

曲线$x={{\sin }^{2}}t, y=\sin t\cos t, z={{\cos }^{2}}t$在$t=\frac{\text{ }\!\!\pi\!\!\text{ }}{2}$所对应的点处的切向向量为 A: $(0,-1,1)$ B: $(1,-1,0)$ C: $(0,1,1)$ D: $(0,-1,0)$

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