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Question Number 59682 by aliesam last updated on 13/May/19

$$if$$$$$$$$H=X^2+Y^2+Z^2$$$$$$$$prove$$$$$$$$\frac{{\partial }^{2}H}{\partial X^2}+\frac{{\partial }^{2}H}{\partial Y^2}+\frac{{\partial }^{2}H}{\partial Z^2}=\frac{2}{H}$$

Commented by mr W last updated on 13/May/19

$$H^2=X^2+Y^2+Z^2$$$$2H\frac{\partial H}{\partial X}=2X$$$$H\frac{\partial H}{\partial X}=X\Rightarrow \frac{\partial H}{\partial X}=\frac{X}{H}$$$$\frac{\partial H}{\partial X}\times \frac{\partial H}{\partial X}+H\frac{{\partial }^{2}H}{\partial X^2}=1$$$$\frac{X^2}{H^2}+H\frac{{\partial }^{2}H}{\partial X^2}=1$$$$\frac{X^2+Y^2+Z^2}{H^2}+H(\frac{{\partial }^{2}H}{\partial X^2}+\frac{{\partial }^{2}H}{\partial Y^2}+\frac{{\partial }^{2}H}{\partial Z^2})=3$$$$1+H(\frac{{\partial }^{2}H}{\partial X^2}+\frac{{\partial }^{2}H}{\partial Y^2}+\frac{{\partial }^{2}H}{\partial Z^2})=3$$$$H(\frac{{\partial }^{2}H}{\partial X^2}+\frac{{\partial }^{2}H}{\partial Y^2}+\frac{{\partial }^{2}H}{\partial Z^2})=2$$$$\Rightarrow \frac{{\partial }^{2}H}{\partial X^2}+\frac{{\partial }^{2}H}{\partial Y^2}+\frac{{\partial }^{2}H}{\partial Z^2}=\frac{2}{H}$$

Commented by aliesam last updated on 13/May/19

$$thankyousir$$

Answered by tanmay last updated on 13/May/19

$$plsrecheckquestion...$$

Commented by aliesam last updated on 13/May/19

$$H^2$$

Commented by aliesam last updated on 13/May/19

$$=X^2+Y^2+Z^2$$

Answered by tanmay last updated on 13/May/19

$$x^2+y^2+z^2=H^2$$$$H=\sqrt{x^2+y^2+z^2}$$$$\frac{\partial H}{\partial x}=\frac{1}{2\sqrt{x^2+y^2+z^2}}\times 2x$$$$\frac{{\partial }^{2}H}{\partial x^2}=\frac{\sqrt{x^2+y^2+z^2}\times \frac{\partial x}{\partial x}-x\times \frac{\partial }{\partial x}(\sqrt{x^2+y^2+z^2}}{(x^2+y^2+z^2)}$$$$\frac{{\partial }^{2}H}{\partial x^2}=\frac{\sqrt{x^2+y^2+z^2}-\frac{x}{2\sqrt{x^2+y^2+z^2}}\times 2x}{(x^2+y^2+z^2)}$$$$\frac{{\partial }^{2}H}{\partial x^2}=\frac{y^2+z^2}{{(x^2+y^2+z^2)}^{\frac{3}{2}}}$$$$so\frac{{\partial }^{2}H}{\partial x^2}+\frac{{\partial }^{2}H}{\partial y^2}+\frac{{\partial }^{2}H}{\partial z^2}$$$$=\frac{(y^2+z^2)+(x^2+z^2)+(x^2+z^2)}{{(x^2+y^2+z^2)}^{\frac{3}{2}}}$$$$=\frac{2(x^2+y^2+z^2)}{{(x^2+y^2+z^2)}^{\frac{3}{2}}}$$$$.\frac{2H^2}{{\left(H^2\right)}^{\frac{3}{2}}}$$$$=\frac{2}{H}$$

Commented by aliesam last updated on 13/May/19

$$genius$$

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