Question

u=f(2x-3y,3y-4z,4z-2x) then show that 6ux+4uy=-3uz

Answer 1

$\textbf{Given:}$

$U=f(2x-3y,3y-4z,4z-2x)$

$\text{Let,}\;p=2x-3y,\;q=3y-4z,\;r=4z-2x$

$\text{Consider,}$

$U_x=\frac{\partial{U}}{\partial{x}}$

$U_x=\frac{\partial{U}}{\partial{p}}\,\frac{\partial{p}}{\partial{x}}+\frac{\partial{U}}{\partial{q}}\,\frac{\partial{q}}{\partial{x}}+\frac{\partial{U}}{\partial{r}}\,\frac{\partial{r}}{\partial{x}}$

$U_x=\frac{\partial{U}}{\partial{p}}(2)+\frac{\partial{U}}{\partial{q}}(0)+\frac{\partial{U}}{\partial{r}}(-2)$

$\implies\bf\;U_x=2\,\frac{\partial{U}}{\partial{p}}-2\,\frac{\partial{U}}{\partial{r}}$

$U_y=\frac{\partial{U}}{\partial{y}}$

$U_y=\frac{\partial{U}}{\partial{p}}\,\frac{\partial{p}}{\partial{y}}+\frac{\partial{U}}{\partial{q}}\,\frac{\partial{q}}{\partial{y}}+\frac{\partial{U}}{\partial{r}}\,\frac{\partial{r}}{\partial{y}}$

$U_y=\frac{\partial{U}}{\partial{p}}(-3)+\frac{\partial{U}}{\partial{q}}(3)+\frac{\partial{U}}{\partial{r}}(0)$

$\implies\bf\;U_y=-3\,\frac{\partial{U}}{\partial{p}}+3\,\frac{\partial{U}}{\partial{q}}$

$U_z=\frac{\partial{U}}{\partial{z}}$

$U_z=\frac{\partial{U}}{\partial{p}}\,\frac{\partial{p}}{\partial{z}}+\frac{\partial{U}}{\partial{q}}\,\frac{\partial{q}}{\partial{z}}+\frac{\partial{U}}{\partial{r}}\,\frac{\partial{r}}{\partial{z}}$

$U_z=\frac{\partial{U}}{\partial{p}}(0)+\frac{\partial{U}}{\partial{q}}(-4)+\frac{\partial{U}}{\partial{r}}(4)$

$\implies\bf\;U_z=-4\,\frac{\partial{U}}{\partial{q}}+4\,\frac{\partial{U}}{\partial{r}}$

$\text{Now,}$

$\bf\;6\;U_x+4\;U_y$

$=6(2\,\frac{\partial{U}}{\partial{p}}-2\,\frac{\partial{U}}{\partial{r}})+4(-3\,\frac{\partial{U}}{\partial{p}}+3\,\frac{\partial{U}}{\partial{q}})$

$=12\,\frac{\partial{U}}{\partial{p}}-12\,\frac{\partial{U}}{\partial{r}}-12\,\frac{\partial{U}}{\partial{p}}+12\,\frac{\partial{U}}{\partial{q}}$

$=-12\,\frac{\partial{U}}{\partial{r}}+12\,\frac{\partial{U}}{\partial{q}})$

$=-3(4\,\frac{\partial{U}}{\partial{r}}-4\,\frac{\partial{U}}{\partial{q}})$

$=-3\;U_z$

$\implies\boxed{\bf\;6\;U_x+4\;U_y=-3\;U_z}$

$\textbf{Hence preoved}$