## Calculus 8th Edition

(a) $-\dfrac{40}{3\sqrt 3}$ (b) A vector pointing in the direction of the origin from the point $(x,y,z)$
(a) Our aim is to determine the maximum rate of change of $f(x,y,z)$.In order to find this, we have : $D_uf=|\nabla f(x,y,z)|$ or, $D_uT=|\nabla T(x,y,z)| \cdot u$ Thus, $\nabla T(x,y,z)=\lt -\dfrac{360x}{(x^2+y^2+z^2))^{3/2}},-\dfrac{360y}{(x^2+y^2+z^2))^{3/2}},-\dfrac{360z}{(x^2+y^2+z^2))^{3/2}} \gt$ From the given data :$f(x,y,z)=f(1,2,2)$ $\nabla T(1,2,2)=\lt -\dfrac{40}{3},-\dfrac{80}{3},-\dfrac{80}{3} \gt$ $D_uT(1,2,2) \cdot u=(\lt -\dfrac{40}{3},-\dfrac{80}{3},-\dfrac{80}{3} \gt) \lt \dfrac{1}{\sqrt 3},-\dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3})\gt=-\dfrac{40}{3\sqrt 3}+\dfrac{80}{3\sqrt 3}-\dfrac{80}{3\sqrt 3}=-\dfrac{40}{3\sqrt 3}$ b) Let us consider part (a) This implies $\nabla T(x,y,z)=\lt -\dfrac{360x}{(x^2+y^2+z^2)^{3/2}},-\dfrac{360y}{(x^2+y^2+z^2)^{3/2}},-\dfrac{360z}{(x^2+y^2+z^2)^{3/2}} \gt$ Also, $\nabla T(x,y,z)=\lt \dfrac{360}{(x^2+y^2+z^2))^{3/2}}(-x,-y,-z)\gt$ Hence, a vector pointing in the direction of the origin from the point $(x,y,z)$