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anonymous

  • 5 years ago

find xy+(z^3)x-2yz=0 in terms of z

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  1. anonymous
    • 5 years ago
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    whts in terms of z mean?

  2. anonymous
    • 5 years ago
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    the answer should be z= f(x,y). so i am asking what f(x,y) is

  3. amistre64
    • 5 years ago
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    x.z^3 -2yz = -xy

  4. amistre64
    • 5 years ago
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    z(-2y +x.z^2) = -xy -2y +x.z^2 - -xy/z

  5. anonymous
    • 5 years ago
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    It's complicated to solve for z.

  6. anonymous
    • 5 years ago
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    yeah its a cubic eqn in z

  7. anonymous
    • 5 years ago
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    Could you write the original question, if it is not this?

  8. amistre64
    • 5 years ago
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    -2y +x.z^2 = -xy/z x.z^2 = (-xy+2yz)/z z^3 = (-xy+2yz)/x

  9. anonymous
    • 5 years ago
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    i am supposed to find dz/dx at the point (1,1,1) of that equation

  10. anonymous
    • 5 years ago
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    then say that

  11. anonymous
    • 5 years ago
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    spo i assumed i needed z as a function of x and y

  12. amistre64
    • 5 years ago
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    you dont need to solve for z then...

  13. anonymous
    • 5 years ago
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    u dont need to..i think not

  14. amistre64
    • 5 years ago
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    dz/dx = g(F).d(xbar)/dx

  15. anonymous
    • 5 years ago
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    ok, sorry about tha

  16. anonymous
    • 5 years ago
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    You don't have to solve for z.

  17. anonymous
    • 5 years ago
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    bt then how to do it???

  18. amistre64
    • 5 years ago
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    get the gradient vector but partialling the equation of the plane there

  19. amistre64
    • 5 years ago
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    get the gradient vector but partialling the equation of the plane there

  20. amistre64
    • 5 years ago
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    F=<a,b,c> a = dz/dx...but not that dz/dx...the funny looking one lol

  21. anonymous
    • 5 years ago
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    then do df/dx / df/dz?

  22. amistre64
    • 5 years ago
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    <y+z^3, x-2z, 3z^2-2y>

  23. amistre64
    • 5 years ago
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    plug in the values from your point know to get the gradianet vector

  24. amistre64
    • 5 years ago
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    <2,-1,1> is your normal vector/gradient..

  25. anonymous
    • 5 years ago
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    oh ok, i got the gradient vector, then plug in the point (1,1,1) to df/dx to get the answer?

  26. amistre64
    • 5 years ago
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    dot product that with the vecotr that is the point...<1,1,1>

  27. anonymous
    • 5 years ago
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    i got <2, -1, 1> yes

  28. amistre64
    • 5 years ago
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    2(x-1)-1(y-1)+1(z-1) = dz/dx i think :)

  29. anonymous
    • 5 years ago
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    is it right??

  30. anonymous
    • 5 years ago
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    is that the plane equation?

  31. anonymous
    • 5 years ago
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    i dont have the answer

  32. amistre64
    • 5 years ago
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    that is the plane equation

  33. anonymous
    • 5 years ago
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    doesnt the gradient vector give <df/dx, df/dy,df/dz>... in that case we can simply do df/dx / df/dz???

  34. anonymous
    • 5 years ago
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    the answer isnt an equation...its a value

  35. amistre64
    • 5 years ago
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    him; dunno havent read that far yet :)

  36. anonymous
    • 5 years ago
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    thats wht m sayin

  37. anonymous
    • 5 years ago
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    is the answer 2???

  38. amistre64
    • 5 years ago
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    the answer would be 2 then right?

  39. amistre64
    • 5 years ago
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    gradient . point

  40. anonymous
    • 5 years ago
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    i think so! thanks!

  41. anonymous
    • 5 years ago
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    is it 2??

  42. anonymous
    • 5 years ago
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    i dont know the value, but it seems right

  43. anonymous
    • 5 years ago
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    u understood how it ws done?

  44. amistre64
    • 5 years ago
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    i dont think I evenknow how it was done lol

  45. anonymous
    • 5 years ago
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    I got -2.

  46. anonymous
    • 5 years ago
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    yes, get the gradient, and plug in the x,y and z valuse of the point into the gradient and divide df/dx by df/dz to get dz/dx right?

  47. anonymous
    • 5 years ago
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    Not entirely sure though.

  48. anonymous
    • 5 years ago
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    i thnk it shld be right..there might b calculation errors bt b surev f d method

  49. amistre64
    • 5 years ago
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    <2,-1,1> <1, 1,1> --------- 2 -1 +1 = 2

  50. anonymous
    • 5 years ago
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    so i said

  51. anonymous
    • 5 years ago
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    david wht grade u in?

  52. anonymous
    • 5 years ago
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    im a sophomore in college. im studying for my calc 3 final

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