At the point $(0, 1, 2)$ in which direction does the function $f(x,y,z) =xy^2z$ increase most rapidly? What is the rate of change of $f$ in this direction? At the point $(1, 1, 0)$, what is the derivative of $f$ in the direction of the vector $2\hat i +3 \hat j+ 6 \hat k$?
I assumed that the rate of change is the same as the gradient of the function, namely $\bigtriangledown f$. Calculating this gave me:
$\bigtriangledown f = \frac{\partial (xy^2z)}{\partial x} \hat i + \frac{\partial (xy^2z)}{\partial y} \hat j +\frac{\partial (xy^2 z)}{\partial z} \hat k$.
$ \space\space\space\space\space\space= y^2z\space \hat i+2xz \space\hat j+ xy^2 \space\hat k $
Evaluating at point:
$\bigtriangledown f(0,1,2)= 2 \space\hat i$
Hence, the function increases most rapidly in the $x$ direction.
I am uncertain of how to approach solving the third part of the question, should I evaluate the rate of change at $(1, 1, 0)$ and then find the difference between that and the vector $2\hat i +3 \hat j+ 6 \hat k$?
Hint: The directional derivative of $f$, in the direction of vector $\vec u$, is just: $$\nabla f\cdot \vec u,\quad\text{or}\quad \nabla f\cdot \frac {\vec u}{\lVert\vec u\rVert}$$ (there are different conventions, according to context).