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dabears900

  • 3 years ago

Help!! I have no idea!

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  1. dabears900
    • 3 years ago
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  2. animalsavior94
    • 3 years ago
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    whatd your question?

  3. dabears900
    • 3 years ago
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    see attached ^^^

  4. animalsavior94
    • 3 years ago
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    i just did and ummmm im not very sure about it sorry:(

  5. dabears900
    • 3 years ago
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    no probs! I hope someone might

  6. rsvitale
    • 3 years ago
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    the direction of the gradient is the direction of greatest increase

  7. rsvitale
    • 3 years ago
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    so take (dF/dx,dF/dy) at the point (-1,1) and that vector points in the direction of greatest increase

  8. rsvitale
    • 3 years ago
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    the magnitude of that vector is the greatest rate of change

  9. rsvitale
    • 3 years ago
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    do you know how to find the partial derivatives?

  10. dabears900
    • 3 years ago
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    yes, sorry I'm doing work at the same time

  11. dabears900
    • 3 years ago
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    fx= 2x +y fy = x

  12. dabears900
    • 3 years ago
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    ?

  13. amistre64
    • 3 years ago
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    those look good for the gradient stuff

  14. rsvitale
    • 3 years ago
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    yep they are right, now evaluate the derivatives at the specified point and you have the gradient vector. Then you can find magnitude.

  15. dabears900
    • 3 years ago
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    so eval at (-1,1) fx=-1 fy =-1

  16. dabears900
    • 3 years ago
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    don't I need a z point?

  17. dabears900
    • 3 years ago
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    magnitude is \sqrt((a_1)^2 + (a_2)^2) so I use the -1,-1 ?

  18. dabears900
    • 3 years ago
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    so the magnitdue is sqrt(2) ?

  19. chaguanas
    • 3 years ago
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    \[<-1/\sqrt{2}, 1/\sqrt{2}>\]

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