Consider the surface $h(x,y) = 1+2x\sqrt{y}$
(i) Find the maximum rate of change of $h$ at $(3,4)$.
(ii) Find the (two) directions one could begin to move to stay level if one is standing on the surface at $(3,4,13)$.
Could someone please give me some pointers on how to approach a question of this kind? Thanks in advance.
$\endgroup$ 11 Answer
$\begingroup$For part i)...
i) For the maximum rate of change, try taking the gradient.
The gradient vector is $<2y^{1/2}, xy^{-1/2}>$.
The maximum rate of change will occur in the direction of $<2*(4)^{1/2}, 3*(4)^{-1/2}> = <4, 3/2>$.
The maximum rate of change is then $\sqrt {4^2 + (3/2)^{2}} = \sqrt {73}/2$
As for part ii)...
To find the (two) directions one could begin to move to stay level if one is standing on the surface at $(3,4,13)$, we must find the direction perpendicular to the gradient of $f$ at $(3, 4)$, and this would be $\pm <3/2, -4>$.
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