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animalsavior94Best ResponseYou've already chosen the best response.0
whatd your question?
 2 years ago

animalsavior94Best ResponseYou've already chosen the best response.0
i just did and ummmm im not very sure about it sorry:(
 2 years ago

dabears900Best ResponseYou've already chosen the best response.0
no probs! I hope someone might
 2 years ago

rsvitaleBest ResponseYou've already chosen the best response.2
the direction of the gradient is the direction of greatest increase
 2 years ago

rsvitaleBest ResponseYou've already chosen the best response.2
so take (dF/dx,dF/dy) at the point (1,1) and that vector points in the direction of greatest increase
 2 years ago

rsvitaleBest ResponseYou've already chosen the best response.2
the magnitude of that vector is the greatest rate of change
 2 years ago

rsvitaleBest ResponseYou've already chosen the best response.2
do you know how to find the partial derivatives?
 2 years ago

dabears900Best ResponseYou've already chosen the best response.0
yes, sorry I'm doing work at the same time
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
those look good for the gradient stuff
 2 years ago

rsvitaleBest ResponseYou've already chosen the best response.2
yep they are right, now evaluate the derivatives at the specified point and you have the gradient vector. Then you can find magnitude.
 2 years ago

dabears900Best ResponseYou've already chosen the best response.0
so eval at (1,1) fx=1 fy =1
 2 years ago

dabears900Best ResponseYou've already chosen the best response.0
don't I need a z point?
 2 years ago

dabears900Best ResponseYou've already chosen the best response.0
magnitude is \sqrt((a_1)^2 + (a_2)^2) so I use the 1,1 ?
 2 years ago

dabears900Best ResponseYou've already chosen the best response.0
so the magnitdue is sqrt(2) ?
 2 years ago

chaguanasBest ResponseYou've already chosen the best response.0
\[<1/\sqrt{2}, 1/\sqrt{2}>\]
 2 years ago
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