曲线 x 2 +y 2 +z 2 =6 ,x+y+z=0在点(1,-2,0)处的切线方程为()。
举一反三
- 4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$,并且$z(x,0)=x,z(0,y)={{y}^{2}}$,则$z(x,y)=$( ) A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$
- 曲线\( \left\{ {\matrix{ { { x^2} + {y^2} = {z^2}} \cr { { z^2} = y} \cr } } \right. \)在坐标面\( yoz \) 上的投影曲线方程为( ) A: \( \left\{ {\matrix{ { { x^2} + { { \left( {y - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \) B: \( \left\{ {\matrix{ { { z^2} = y} \cr {x = 0} \cr } } \right. \) C: \( \left\{ {\matrix{ {z = {y^2}} \cr {x = 0} \cr } } \right. \) D: \( \left\{ {\matrix{ { { y^2} + { { \left( {x - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)
- 过点(1, -2, -2)且与平面x -2 y + 3z = 2平行的平面方程为 A: x -2 y + z = 6; B: x -2y + 3z = 0; C: x -2y + 3z = 0; D: 2x - y + 3z = 9.
- 曲线 z=x 2 +y 2 ,y=1,在(1,1)处的切线与x轴的夹角为()度。
- 设\(f\left( {x,y,z} \right) = x{y^2} + y{z^2} + z{x^2}\),则\({f_{yz}}\left( {0,-1,0} \right) = \)( ) A: 1 B: 0 C: -1 D: 2