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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 time? Thanks in advance.

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Ok, is the maximum value the length of grad f and the two directions are plus/minus the unit vector in the direction of graf f? –  Joe S Jun 19 '13 at 3:27
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1 Answer 1

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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As for the maximum rate of change - Find the gradient, then plug in the point, then find the magnitude of that. To stay in level, find the direction perpendicular to the gradient of f at the point. –  user51800 Jun 19 '13 at 3:42
    
Sorry for the bad formatting - I'll pick it up as soon as I can! –  user51800 Jun 19 '13 at 3:48
    
Please make sure I got it all correct. Regards –  Amzoti Jun 19 '13 at 3:52
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