# Mathematics-I (3110014)

BE | Semester-1   Winter-2019 | 17-01-2020

## Q5) (c)

#### If then show that $\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=0$.

$\mathrm{Let},\mathrm{x}-\mathrm{y}=\mathrm{l}$

$\frac{\partial \mathrm{l}}{\partial \mathrm{x}}=1$

$\frac{\partial \mathrm{l}}{\partial \mathrm{y}}=-1$

$\mathrm{Let},\mathrm{y}-\mathrm{z}=\mathrm{m}$

$\frac{\partial \mathrm{m}}{\partial \mathrm{y}}=1$

$\frac{\partial \mathrm{m}}{\partial \mathrm{z}}=-1$

$\mathrm{Let},\mathrm{z}-\mathrm{x}=\mathrm{n}$

$\frac{\partial \mathrm{n}}{\partial \mathrm{x}}=-1$

$\frac{\partial \mathrm{n}}{\partial \mathrm{z}}=1$

$\frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{\partial \mathrm{u}}{\partial \mathrm{l}}-\frac{\partial \mathrm{u}}{\partial \mathrm{n}}$

$\frac{\partial \mathrm{u}}{\partial \mathrm{z}}=-\frac{\partial \mathrm{u}}{\partial \mathrm{l}}+\frac{\partial \mathrm{u}}{\partial \mathrm{m}}$

$\frac{\partial \mathrm{u}}{\partial \mathrm{z}}=-\frac{\partial \mathrm{u}}{\partial \mathrm{m}}+\frac{\partial \mathrm{u}}{\partial \mathrm{n}}$

$\mathrm{Now},\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}$

${\mathrm{No}}=\frac{\partial \mathrm{u}}{\partial \mathrm{l}}-\frac{\partial \mathrm{u}}{\partial \mathrm{n}}-\frac{\partial \mathrm{u}}{\partial \mathrm{l}}+\frac{\partial \mathrm{u}}{\partial \mathrm{m}}-\frac{\partial \mathrm{u}}{\partial \mathrm{m}}+\frac{\partial \mathrm{u}}{\partial \mathrm{n}}$

${\mathrm{No}}=0$