Thomas' Calculus 13th Edition

$$\dfrac{\pi}{\sqrt 2}$$
Since, $D_u f = \nabla f \cdot u$ We have: $$r=\cos 3t i+\sin 3t j +3t k$$ $v(t)=\dfrac{dr}{dt}=-3 \sin 3t i+3 \cos 3t j +3t k$ or, $v(\dfrac{\pi}{3})=-3 \sin 3t i+3 \cos 3t j +3t k=-3j+3k$ Now, $f(-1,0, \pi) =yz i+xz j +xy k$ or, $f(-1,0, \pi) =-\pi \ j$ $\nabla f \cdot u = (- \pi j) \cdot (\dfrac{-1}{\sqrt 2} j+\dfrac{1}{\sqrt 2} k) = \dfrac{1}{\sqrt 2} \pi$