Answer
a) $\frac{736}{\sqrt {89}}^{o}C/m$ b) $736^{o}C/sec$
Work Step by Step
a) Here, $u=\dfrac{8}{\sqrt 89}i+\dfrac{3}{\sqrt 89}j-\dfrac{4}{\sqrt 89}k$ and $\nabla T= 56 i+32 j-48 k$
Now, at $(8,6,-4)$ we have $D_uT=(\nabla T \cdot u) \implies (1/\sqrt {89})((56))8)+(32)(3)-(48)(-4))$
or, $=\frac{736}{\sqrt {89}}^{o}C/m$
b) Since, we have $u=\dfrac{8}{\sqrt 89}i+\dfrac{3}{\sqrt 89}j-\dfrac{4}{\sqrt 89}k$
From part (a):
Now, $\dfrac{dT}{dt}=|v| (D_uT) \implies (\sqrt {89}) (\dfrac{736}{\sqrt {89}})$
$=736^{o}C/sec$
