A) \[\frac{z}{6}\]
B) \[\frac{z}{3}\]
C) \[\frac{z}{2}\]
D) \[\frac{z}{12}\]
Correct Answer: D
Solution :
\[{{z}^{2}}\] is homogeneous in x, y of degree \[\frac{1}{6}\] \[\therefore \] \[x\,\frac{\partial }{\partial x}({{z}^{2}})+y\frac{\partial }{\partial y}({{z}^{2}})=\frac{1}{6}({{z}^{2}})\] Þ \[2xz\frac{\partial z}{\partial x}+2yz\frac{\partial z}{\partial y}=\frac{1}{6}{{z}^{2}}\]Þ \[x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{1}{12}z\].
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