Answer
a) $0.935^{o}C/ft$ b) $1.87^{o}C/sec$
Work Step by Step
a) Here, $u=\dfrac{\sqrt 3}{2}i-\dfrac{1}{2} j$ and $\nabla T= \sin \sqrt 3 i+ \cos \sqrt 3 j$
Now, $D_uT=(\nabla T \cdot u) \implies (\dfrac{\sqrt 3}{2} \sin \sqrt 3-\dfrac{1}{2} \cos \sqrt 3)$
or, $\approx 0.935^{o}C/ft$
b) since, we have $u=\dfrac{\sqrt 3}{2}i-\dfrac{1}{2} j$
Now, $\dfrac{dT}{dt}=|v| (D_uT) \implies (2) (\dfrac{\sqrt 3}{2} \sin \sqrt 3-\dfrac{1}{2} \cos \sqrt 3)$
$\approx 1.87^{o}C/sec$
