Answer to Question #275801 in Calculus for Hafsa Tahsin

Question

[SADT10] Let r=x hat i +y hat j +z hat k and r = ||r||

Show that:

nabla(lnr)= r r^ 2 .

and

nabla*(r^ n r)=0 .

Answer 1

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

$\nabla(lnr)=\frac{2xi+2yj+2zk}{2(x^2+y^2+z^2)}=r/||r||^2$

$r^ n r=(x^2+y^2+z^2)^{n/2}(xi+yj+zk)$

$\nabla\times(r^ n r)=(\frac{\partial (r^ n r)_z}{\partial y}-\frac{\partial (r^ n r)_y}{\partial z})i+(\frac{\partial (r^ n r)_x}{\partial z}-\frac{\partial (r^ n r)_z}{\partial x})j+(\frac{\partial (r^ n r)_y}{\partial x}-\frac{\partial (r^ n r)_x}{\partial y})k$

$\frac{\partial (r^ n r)_z}{\partial y}=\frac{\partial (r^ n r)_y}{\partial z}=yzn (x^2+y^2+z^2)^{n/2-1}$

$\frac{\partial (r^ n r)_x}{\partial z}=\frac{\partial (r^ n r)_z}{\partial x}=xzn (x^2+y^2+z^2)^{n/2-1}$

$\frac{\partial (r^ n r)_y}{\partial x}=\frac{\partial (r^ n r)_x}{\partial y}=xyn (x^2+y^2+z^2)^{n/2-1}$

$\nabla\times(r^ n r)=0$