Question

Please solve question no 2

Answer 1

Hey friend, Harish here.

Here is your answer:

Given:

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

To Find:

$\mathrm{ The \ gradient\ of\ r^n}$

Solution:

$r^2 = x^2 + y^2 + z^2 \ \ \ - \ \ (Given) \\ \\ r = \bigl( x^2 +y^2+z^2 \bigr)^{\frac{1}{2}} \\ \\ r^n = \bigl( x^2 +y^2+z^2 \bigr)^{\frac{n}{2}} \\ \\ \mathrm{Then,} \\ \\ \frac{\partial}{\partial x}r^n = \frac{\partial}{\partial x} \bigl( x^2 +y^2+z^2 \bigr)^{\frac{n}{2}} \\ \\ \to 2x .\frac{n}{2}.\bigl( x^2 +y^2+z^2 \bigr)^{\frac{n}{2}-1} \\ \\ \to nxr^{n-2} \\ \\ \mathrm{Now\ using\ the\ symmetry\ of\ r\ with\ respect\ to\ x,y,z\ ;} \\ \\ \to \frac{\partial}{\partial y}r^n = nyr^{n-2}$
$\to \frac{\partial}{\partial z}r^n = nzr^{n-2} \\ \\ \mathrm{Thus,} \\ \\ \nabla(r^n) = \Bigl[ \bigl( \frac{\partial}{\partial x}r^n \bigr ),\bigl( \frac{\partial}{\partial y}r^n \bigr ),\bigl( \frac{\partial}{\partial z}r^n \bigr )\Bigr ]\\ \\ \to \Bigl[ (nxr^{n-2}) , (nyr^{n-2}),(nzr^{n-2}) \Bigr] \\ \\ \to \boxed{\bold{n.r^{n-2}.r}}$
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Hope my answer is helpful to you.