第2类拉格朗日方程表达式为____ 。
A: $\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial T}}{{\partial {{\dot q}_k}}}} \right) - \frac{{\partial T}}{{\partial {q_k}}} = 0$
B: $\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial T}}{{\partial {{\dot q}_k}}}} \right) - \frac{{\partial T}}{{\partial {q_k}}} = {Q_k}$
C: $\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial T}}{{\partial {{\dot q}_k}}}} \right) + \frac{{\partial T}}{{\partial {q_k}}} = {Q_k}$
D: 以上都不对
A: $\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial T}}{{\partial {{\dot q}_k}}}} \right) - \frac{{\partial T}}{{\partial {q_k}}} = 0$
B: $\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial T}}{{\partial {{\dot q}_k}}}} \right) - \frac{{\partial T}}{{\partial {q_k}}} = {Q_k}$
C: $\frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial T}}{{\partial {{\dot q}_k}}}} \right) + \frac{{\partial T}}{{\partial {q_k}}} = {Q_k}$
D: 以上都不对
举一反三
- \[设z=z(x,y)是由方程f(cx-az,cy-bz)=0所确定的函数,则a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}=（）\] A: a B: b C: c D: 0
- \(已知二元函数f(x,y)=\sin{x^2y},则\frac{\partial f}{\partial x}(1,\pi)=(\,)\) A: \(\frac{\pi}{2}\) B: \(2\pi\) C: \(-2\pi\) D: \(-\frac{\pi}{2}\)
- 从$\sum\limits_{k = 1}^N {\left( {{Q_k} + Q_k^*} \right)\delta {q_k}} = 0$ 得到${Q_k} + Q_k^* = 0{\rm{ }}(k = 1,2...N)$，需要的条件是____。 A: 理想约束 B: 双面约束 C: 定常约束恒 D: 以上都不对
- 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: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$
- \(已知u=e^{xyz},则\frac{\partial^3u}{\partial x\partial y\partial z}=(\,)\) A: \[(x^2y^2z^2+zyz+1)e^{xyz}\] B: \[(x^2yz^2+3zyz+1)e^{xyz}\] C: \[(x^2y^2z^2+3zyz+1)e^{xyz}\] D: \[(x^2y^2z^2+3zyz)e^{xyz}\]