a) Para encontrarmos a taxa de variação, realizaremos os cálculos abaixo:

\(\begin{align} & {{T}_{x}}=\frac{-360x}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}} \\ & \\ & {{T}_{y}}=\frac{-360y}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}} \\ & \\ & {{T}_{z}}=\frac{-360z}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}} \\ & \\ & \nabla T=\left( {{T}_{x}},{{T}_{y}},{{T}_{z}} \right) \\ & \nabla T=\left( \frac{-360\cdot 1}{{{(6)}^{3/2}}},\frac{-360\cdot 2}{{{(6)}^{3/2}}},\frac{-360\cdot 2}{{{(6)}^{3/2}}} \right) \\ & \nabla T=\left( \frac{-40}{3},\frac{-80}{3},\frac{-80}{3} \right) \\ & \\ & u=\frac{v}{|v|} \\ & u=\left( \frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right) \\ & \\ & {{D}_{u}}T(1,2,2)=\left( \frac{-40}{3},\frac{-80}{3},\frac{-80}{3} \right)\left( \frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}} \right) \\ & {{D}_{u}}T(1,2,2)=\frac{-40}{3\sqrt{3}} \\ \end{align}\ \)

b) 

\(\begin{align} & {{T}_{x}}=\frac{-360x}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}} \\ & \\ & {{T}_{y}}=\frac{-360y}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}} \\ & \\ & {{T}_{z}}=\frac{-360z}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}} \\ & \\ & \nabla T=\left( {{T}_{x}},{{T}_{y}},{{T}_{z}} \right) \\ & \nabla T=\left( \frac{-360x}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}},\frac{-360y}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}},\frac{-360z}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}} \right) \\ & \nabla T=\frac{360}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}}\left( -x,-y,-z \right) \\ & \nabla T=\frac{360}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}}\left( \frac{0-x}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}},\frac{0-y}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}},\frac{0-z}{{{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{3/2}}} \right) \\ \end{align}\ \)