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2022-11-13

Calculating Rate of Change
At the point (0,1,2) in which direction does the function $f\left(x,y,z\right)=x{y}^{2}z$ 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 $2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}$?
I assumed that the rate of change is the same as the gradient of the function, namely $▽f$. Calculating this gave me:
$▽f=\frac{\mathrm{\partial }\left(x{y}^{2}z\right)}{\mathrm{\partial }x}\stackrel{^}{i}+\frac{\mathrm{\partial }\left(x{y}^{2}z\right)}{\mathrm{\partial }y}\stackrel{^}{j}+\frac{\mathrm{\partial }\left(x{y}^{2}z\right)}{\mathrm{\partial }z}\stackrel{^}{k}$

Evaluating at point:

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 $2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}$?

Miah Carlson

Expert

Hint: The directional derivative of f, in the direction of vector $\stackrel{\to }{u}$, is just:
$\mathrm{\nabla }f\cdot \stackrel{\to }{u},\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mathrm{\nabla }f\cdot \frac{\stackrel{\to }{u}}{‖\stackrel{\to }{u}‖}$
(there are different conventions, according to context).

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