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anonymous

  • 5 years ago

Find the point(s) on the grapf of z=3(x^2) -4(y^2) at which the vector <3,2,2> is normal to the tangent plane.

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  1. amistre64
    • 5 years ago
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    so you want the gradiennt to be <3,2,2>

  2. amistre64
    • 5 years ago
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    F(x,y,z)=3x^2 -4y^2 -z = 0

  3. amistre64
    • 5 years ago
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    dF/dx = 6x dF/dy = -8y dF/dz = -1

  4. anonymous
    • 5 years ago
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    that is as far as i have gotten

  5. amistre64
    • 5 years ago
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    6x = 3 -8y = 2 -1 = 2 ....gonna have to scale it i spose. <3,2,2>(-1/2) = <-3/2, -1,-1> for this to work

  6. amistre64
    • 5 years ago
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    6x = -3/2 ; x = -1/4 -8y = -1 ; y = -1/8 -1 = -1 ; z=-1

  7. anonymous
    • 5 years ago
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    ok cool. thanks alot! i had no idea where to start

  8. amistre64
    • 5 years ago
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    lets work this back in again to see if it gets us something good :)

  9. amistre64
    • 5 years ago
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    -1=3((-1/4)^2) -4((-1/8)^2).. is it 3x^2 or 3(x^2)?

  10. anonymous
    • 5 years ago
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    3(x^2)

  11. amistre64
    • 5 years ago
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    been loooking at these to long...that the same thing lol

  12. anonymous
    • 5 years ago
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    three x squared

  13. amistre64
    • 5 years ago
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    3/16 -4/64 = -1?

  14. anonymous
    • 5 years ago
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    what is that?

  15. amistre64
    • 5 years ago
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    dunno, just a thought really

  16. amistre64
    • 5 years ago
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    i think the point is (2,1,8) 8=3(2^2) -4(1^2) 8 = 12-4 8=8 seems better

  17. anonymous
    • 5 years ago
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    how did you get that?

  18. amistre64
    • 5 years ago
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    i used the point i got to begin with and scaled it back to size again... i gotta see if this is right tho..

  19. amistre64
    • 5 years ago
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    cant tell......

  20. anonymous
    • 5 years ago
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    ok, so you are sure we want the gradient to be the normal vector?

  21. amistre64
    • 5 years ago
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    well, the gradient and the normal are the same thing...

  22. amistre64
    • 5 years ago
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    they would have to equal each other.....

  23. amistre64
    • 5 years ago
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    suppose we scale the gradient fo match the normal; does that sound like a good thing to do?

  24. anonymous
    • 5 years ago
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    sure, lets try it

  25. amistre64
    • 5 years ago
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    if we scale it by -2; that will make the -1 = 2

  26. anonymous
    • 5 years ago
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    yeah

  27. anonymous
    • 5 years ago
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    i got the same thing

  28. amistre64
    • 5 years ago
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    but that gets me the same results for x and y; <-1/4,1/8>

  29. amistre64
    • 5 years ago
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    lets plug those in then and see what we get for a z i guess...

  30. amistre64
    • 5 years ago
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    i betcha thats our answers then..... (-1/4,-1/8,z) (-1/4,-1/8,z) (-1/4,1/8,z) (1/4,-1/8,z) (1/4,1/8,z)...whatcha think?

  31. amistre64
    • 5 years ago
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    got a little hyper on that one lol

  32. anonymous
    • 5 years ago
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    haha, so we leave z as a variable?

  33. amistre64
    • 5 years ago
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    z is dependant on x and y right? and it will be the same ragardless if the combo we use for that

  34. anonymous
    • 5 years ago
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    ok that makes sense

  35. amistre64
    • 5 years ago
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    what does z look like...it seems hyperbolic to me....

  36. anonymous
    • 5 years ago
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    its a saddle shape graph

  37. amistre64
    • 5 years ago
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    z=1/8

  38. anonymous
    • 5 years ago
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    the graph looks like we should only have 1 answer

  39. amistre64
    • 5 years ago
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    ...... you sure?

  40. amistre64
    • 5 years ago
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    the vector can be positioned anywhere...its not anchored into anthing

  41. anonymous
    • 5 years ago
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    im not sure, i just think it looks like that

  42. amistre64
    • 5 years ago
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    checking it out on wolframs graphing thing

  43. anonymous
    • 5 years ago
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    i have a different graphing software im using

  44. amistre64
    • 5 years ago
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  45. amistre64
    • 5 years ago
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    well lets see yours :)

  46. anonymous
    • 5 years ago
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    ok, let me see if i can send it in a word document

  47. anonymous
    • 5 years ago
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  48. anonymous
    • 5 years ago
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    ok, well i need to go. Thanks alot for working with me on this problem!

  49. amistre64
    • 5 years ago
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    heres what i have deduced: The gradient is a replica of the normal; for example: y=2x+8; 0 = 2x-y+8 the gradient is <2,-1,0>; we can find the normal by picking any point on the line and using it against the gradient like this: g(Px,Py,Pz) = n<x,y,z> g=2,-1,0 P=4,16,0 --------- n=2,-1,0 simply becasue there is no place to use the point in 'g'. Hence the gradient and the normal are scalar vectors of one another

  50. amistre64
    • 5 years ago
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    z = 3x^2 -4y^2; 0=3x^2-4y^2-z g(x,y,z)=6x,-8y,-1 P=(x0,y0,z0) 6(x0),-8(y0),-1 is parallel to <3,2,2>; so lets scale this normal vector to suit our needs:

  51. amistre64
    • 5 years ago
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    -1/2<3,2,2> = <-3/2,-1,-1> 6x,-8y,-1 x0, y0, z0 ---------- -3/2,-1,-1 x0=-1/4 y0= 1/8 z0 = 3(-1/4)^2 -4(1/8)^2 z0 = 3/16 - 4/64 = 1/8 P(-1/4,1/8,1/8) is a point that is on the tangent plane; with the normal vector <3,2,2>

  52. amistre64
    • 5 years ago
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    the equation of the tangent plane is: 3(x+1/4) +2(y-1/8)+2(z-1/8)=0

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