VC3.10

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Given,\phi (x,y,z)=(x^{2}+y^{2}+z^{2})e^{{-(x^{2}+y^{2}+z^{2})^{{{\frac  {1}{2}}}}}}\,

Let this be equation 1.

Let r=xi+yj+zk,r^{2}=|r|^{2}=x^{2}+y^{2}+z^{2}\, [Equation 2]

Using (2),(1)=\phi =r^{2}e^{{-r}}\, [Equation 3]

From (2),{\frac  {\partial r}{\partial x}}={\frac  {x}{r}}\, [Equation 4]

Now \nabla \phi ={\frac  {\partial \phi }{\partial x}}i+{\frac  {\partial \phi }{\partial y}}j+{\frac  {\partial \phi }{\partial z}}k\, [Equation 5]

Here,{\frac  {\partial \phi }{\partial x}}={\frac  {\partial }{\partial x}}(r^{2}e^{{-r}})\,

={\frac  {\partial (r^{2}e^{{-r}})}{\partial r}}\cdot {\frac  {\partial r}{\partial x}}\,

=(2re^{{-r}}-r^{2}e^{{-r}})({\frac  {x}{r}})\, by (4)

=(2-r)e^{{-r}}x\, [Equation 6]

Similarly,{\frac  {\partial \phi }{\partial y}}=(2-r)e^{{-r}}y,{\frac  {\partial \phi }{\partial z}}=(2-r)e^{{-r}}z\, [Equation 7]

Using (6)and (7),(5)reduces to

\nabla \phi =(2-r)e^{{-r}}(xi+yj+zk)=(2-r)e^{{-r}}r\, by (2)

Therefore,|\nabla \phi |=|(2-r)e^{{-r}}r|=(2-r)e^{{-r}}r\,

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