Answer
(a) $-\dfrac{40}{3\sqrt 3}$
(b) A vector pointing in the direction of the origin from the point $(x,y,z)$
Work Step by Step
(a) Formula to calculate the maximum rate of change of $f$: $D_uT=|\nabla T(x,y,z)| \cdot u$
$\nabla T(x,y,z)=\lt -\dfrac{360x}{(x^2+y^2+z^2))^{3/2}},-\dfrac{360y}{(x^2+y^2+z^2))^{3/2}},-\dfrac{360z}{(x^2+y^2+z^2))^{3/2}} \gt$
$\nabla T(1,2,2)=\lt -\dfrac{40}{3},-\dfrac{80}{3},-\dfrac{80}{3} \gt$
$D_uT(1,2,2) \cdot u=(\lt -\dfrac{40}{3},-\dfrac{80}{3},-\dfrac{80}{3} \gt) \lt \dfrac{1}{\sqrt 3},-\dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3})\gt=-\dfrac{40}{3\sqrt 3}+\dfrac{80}{3\sqrt 3}-\dfrac{80}{3\sqrt 3}=-\dfrac{40}{3\sqrt 3}$
(b) From part (a), we have
$\nabla T(x,y,z)=\lt -\dfrac{360x}{(x^2+y^2+z^2)^{3/2}},-\dfrac{360y}{(x^2+y^2+z^2)^{3/2}},-\dfrac{360z}{(x^2+y^2+z^2)^{3/2}} \gt$
or,
$\nabla T(x,y,z)=\lt \dfrac{360}{(x^2+y^2+z^2))^{3/2}}(-x,-y,-z)\gt$
Hence, a vector pointing in the direction of the origin from the point $(x,y,z)$
