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animalsavior94
 3 years ago
Best ResponseYou've already chosen the best response.0whatd your question?

animalsavior94
 3 years ago
Best ResponseYou've already chosen the best response.0i just did and ummmm im not very sure about it sorry:(

dabears900
 3 years ago
Best ResponseYou've already chosen the best response.0no probs! I hope someone might

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

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

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

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

dabears900
 3 years ago
Best ResponseYou've already chosen the best response.0yes, sorry I'm doing work at the same time

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0those look good for the gradient stuff

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

dabears900
 3 years ago
Best ResponseYou've already chosen the best response.0so eval at (1,1) fx=1 fy =1

dabears900
 3 years ago
Best ResponseYou've already chosen the best response.0don't I need a z point?

dabears900
 3 years ago
Best ResponseYou've already chosen the best response.0magnitude is \sqrt((a_1)^2 + (a_2)^2) so I use the 1,1 ?

dabears900
 3 years ago
Best ResponseYou've already chosen the best response.0so the magnitdue is sqrt(2) ?

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