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anonymous
 5 years ago
Help!! I have no idea!
anonymous
 5 years ago
Help!! I have no idea!

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the direction of the gradient is the direction of greatest increase

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so take (dF/dx,dF/dy) at the point (1,1) and that vector points in the direction of greatest increase

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the magnitude of that vector is the greatest rate of change

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know how to find the partial derivatives?

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

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep they are right, now evaluate the derivatives at the specified point and you have the gradient vector. Then you can find magnitude.

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

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

anonymous
 5 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 ?

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

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