Answer
a) $0.935^{\circ}C/ft$ b) $1.87^{\circ}C/sec$
Work Step by Step
a. Here, we have $u=(\dfrac{\sqrt 3}{2})i-(\dfrac{1}{2})j$; $\nabla T= \sin (\sqrt 3) i+ \cos (\sqrt 3) j$
Then, we get $D_uT=(\nabla T \cdot u)= [(\dfrac{\sqrt 3}{2}) \sin \sqrt 3-(\dfrac{1}{2}) \cos \sqrt 3] \approx 0.935^{\circ}C/ft$
b. Here, we have $u=(\dfrac{\sqrt 3}{2})i-(\dfrac{1}{2})j$
Then, we get $\dfrac{dT}{dt}=|v| (D_uT)$\
and $(2) (\dfrac{\sqrt 3}{2} \sin \sqrt 3-\dfrac{1}{2} \cos \sqrt 3)\approx 1.87^{\circ}C/sec$
