Answer to Question #109174 in Calculus for Ivan Wafula

Question

given that v²=x²+y²+z² prove that d²v/dx² +d²v/dy² +d²v/dz² =2/v

Answer 1

$v^2=x^2+y^2+z^2$

$v$ is symmetrical for $x,y,z$ . Let's found $\partial v/\partial x$ :

$v=\sqrt{x^2+y^2+z^2}$

${\frac {\partial v} {\partial x}}={\frac {2x} {2\sqrt{x^2+y^2+z^2}}}={\frac {x} {v}}$

Then ${\frac {\partial v} {\partial y}}={\frac {y} {v}}$ and ${\frac {\partial v} {\partial z}}={\frac {z} {v}}$

Let's found second order derivatives

${\frac {\partial^2 v} {\partial x^2}}={\frac {v-x{\frac {\partial v} {\partial x}}} {v^2}}={\frac {v-{\frac {x^2} {v}}} {v^2}}={\frac {v^2-x^2} {v^3}}$

Then ${\frac {\partial^2 v} {\partial y^2}}={\frac {v^2-y^2} {v^3}}$ and ${\frac {\partial^2 v} {\partial z^2}}={\frac {v^2-z^2} {v^3}}$

Let's prove our statement

${\frac {\partial^2 v} {\partial x^2}}+{\frac {\partial^2 v} {\partial y^2}}+{\frac {\partial^2 v} {\partial z^2}}={\frac {v^2-x^2} {v^3}}+{\frac {v^2-y^2} {v^3}}+{\frac {v^2-z^2} {v^3}}={\frac {3v^2-x^2-y^2-z^2} {v^3}}={\frac {2v^2} {v^3}}={\frac {2} {v}}$

Proved