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anonymous

  • 5 years ago

Find the point on the graph of z=4x^2−2y^2 at which vector n=<8,8,−1> is normal to the tangent plane. P= ?

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  1. amistre64
    • 5 years ago
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    well, lets take the gradient vector

  2. amistre64
    • 5 years ago
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    maybe?

  3. amistre64
    • 5 years ago
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    fx(x,y) = 8x fy(x,y) = -4y

  4. amistre64
    • 5 years ago
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    <8x,-4y> is an equation for the normal to every point (x,y) if i recall correctly

  5. anonymous
    • 5 years ago
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    yes

  6. amistre64
    • 5 years ago
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    8x(x-Px) -4y(y-Py) +0(z-Pz) then... maybe?

  7. amistre64
    • 5 years ago
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    i should prolly include a -1 for the z

  8. amistre64
    • 5 years ago
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    df/dz = -1 so its part of the gradient 8x(x-Px) -4y(y-Py) -1(z-Pz)

  9. amistre64
    • 5 years ago
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    simply put; 8x=8 , and -4y=8

  10. anonymous
    • 5 years ago
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    cross multiply?

  11. amistre64
    • 5 years ago
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    x=2. y=-2

  12. anonymous
    • 5 years ago
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    haha, nm

  13. amistre64
    • 5 years ago
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    put those into your z equation to get the point for z

  14. amistre64
    • 5 years ago
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    z=4(2)^2−2(-2)^2 z = 16 -8 = 8; P(2,-2,8)

  15. amistre64
    • 5 years ago
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    am i right?

  16. anonymous
    • 5 years ago
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    i follow your thought process and agree with it but it says that the first and 3rd coordinates are incorrect

  17. amistre64
    • 5 years ago
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    its possible :)

  18. anonymous
    • 5 years ago
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    haha thats what I said! let me recheck i wrote the correct problem

  19. amistre64
    • 5 years ago
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    we could plug in our normal and try to work it backwards...

  20. anonymous
    • 5 years ago
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    my answer was 2,-2,2

  21. anonymous
    • 5 years ago
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    okay

  22. amistre64
    • 5 years ago
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    8(x-Px)+8(y-Py)-1(z-z0) = 0 look good?

  23. anonymous
    • 5 years ago
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    yes, that looks right

  24. amistre64
    • 5 years ago
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    thats the equation of our tangent plane right?

  25. amistre64
    • 5 years ago
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    z0 = Pz lol mixing memories lol

  26. anonymous
    • 5 years ago
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    yes my book has: z=f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b)

  27. amistre64
    • 5 years ago
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    2,-2,2 you say was the answer?

  28. anonymous
    • 5 years ago
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    no, thats what I got when i tried to work the problem

  29. anonymous
    • 5 years ago
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    which was also incorrect

  30. amistre64
    • 5 years ago
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    ohhh..... do we have an answer to go by :)

  31. anonymous
    • 5 years ago
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    unfortunetly, no. I do have an example problem similiar to this which I used as model that has an answer

  32. amistre64
    • 5 years ago
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    f(x,y,z) = 4x^2 -2y^2-z = 0 is our original right?

  33. anonymous
    • 5 years ago
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    its z=3x^2-4y^2 n=<3,2,2> and the answer was (-1/4,1/8,1/8)

  34. anonymous
    • 5 years ago
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    yes z=4x^2-2y^2 which is eqivalent to what you said

  35. amistre64
    • 5 years ago
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    and the gradient should give us a parameteric for any normal to any point right?

  36. amistre64
    • 5 years ago
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    <8x,-4y,-1> at any given P(x,y,z)

  37. amistre64
    • 5 years ago
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    z=3x^2-4y^2 n=<3,2,2> <6x,-8y,-1> ( px, py, pz) ------------ <3 , 2, 2> if my thougths are right

  38. amistre64
    • 5 years ago
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    that made no sense lol

  39. anonymous
    • 5 years ago
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    i'm not sure about the gradient, my book doesn't mentioned it. but it says: "vector i normal to the plane: v=<fx(a,b),fy(a,b),-1>

  40. amistre64
    • 5 years ago
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    fx(a,b) is df/dx which is the gradient part... the a,b seems odd to me tho; but they seem to be generic fillins for the inputs that are usually x,y

  41. amistre64
    • 5 years ago
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    your equation has no a or b variables so im assuming it means x,y

  42. anonymous
    • 5 years ago
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    in the example probelm it just converts it fx(x,y)=6x => fx(a,b)=6a

  43. amistre64
    • 5 years ago
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    yeah ....... thats what i was thinking :)

  44. amistre64
    • 5 years ago
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    <6x,-8y,-1> as our 'generic' normal

  45. amistre64
    • 5 years ago
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    <6x,-8y,-1> = <3,2,2> how?

  46. amistre64
    • 5 years ago
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    when x=1/2 and y=-1/4 it appears; but lets see if that makes sense to find z :)

  47. amistre64
    • 5 years ago
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    z = 1/2 when we input those; and we are looking to match: (-1/4,1/8,1/8) (1/2,-1/4,1/2) i wonder if we scale the gradient.... -2<6x,-8y,-1> = <-12x,16y,2> perhaps?

  48. amistre64
    • 5 years ago
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    got it lol

  49. amistre64
    • 5 years ago
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    -12x=3 ; x = -1/4 16y = 2 ; y = 1/8

  50. amistre64
    • 5 years ago
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    8x=8; x=1 .... not 2 lol -4y =8; y=-2 4(1)^2 -2(-2^2) = 4-8 = -4 P(1,2,-4)

  51. amistre64
    • 5 years ago
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    lets see if I can type this without a typo.... P(1,-2,-4) yay!!

  52. anonymous
    • 5 years ago
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    100%!

  53. anonymous
    • 5 years ago
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    you are awesome :)

  54. amistre64
    • 5 years ago
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    thnx :)

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