## Thomas' Calculus 13th Edition

a) $\dfrac{736}{\sqrt {89}}^{\circ}C/m$ b) $736^{\circ}C/sec$
a. At the point $(8,6,-4)$ we have $D_uT=(\nabla T \cdot u) =[56 i+32 j-48 k \cdot \dfrac{8}{\sqrt 89}i+\dfrac{3}{\sqrt 89}j-\dfrac{4}{\sqrt 89}k]$ and $\dfrac{8}{\sqrt {89}} \cdot (56)+(32) \cdot (3)-(48) \cdot (-4)=\dfrac{736}{\sqrt {89}}736^{\circ}C/m$ b. Now, $\dfrac{dT}{dt}=|v| (D_uT) =(\dfrac{8}{\sqrt 89}i+\dfrac{3}{\sqrt 89}j-\dfrac{4}{\sqrt 89}k) \cdot (\dfrac{736}{\sqrt {89}})$ and $(\sqrt {89}) (\dfrac{736}{\sqrt {89}})=736^{\circ}C/sec$