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anonymous

  • 5 years ago

tangent plane of (x^2/a^2)+ (y^2/b^2)+(z^2/c^2)=1 ?

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  1. anonymous
    • 5 years ago
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    \[xx ^{*} /a ^{2} + yy^*/b^2 + zz^*/c^2 =1 ????\]

  2. anonymous
    • 5 years ago
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    tangent plane at what point?

  3. anonymous
    • 5 years ago
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    at a general point x^* y^* z^*

  4. anonymous
    • 5 years ago
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    okay change of answer it is x^*/a^2 +y^*/b^2 +z^*/c^2 =1 i think thats the correc tangent plane eq

  5. anonymous
    • 5 years ago
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    you need to take derivative of the parabola to get the tangent

  6. anonymous
    • 5 years ago
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    a parabola? btw its cal 2 and vector algebra

  7. anonymous
    • 5 years ago
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    x y and z are independent of each other.

  8. anonymous
    • 5 years ago
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    scrap that last post.

  9. anonymous
    • 5 years ago
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    rsaad, you there?

  10. anonymous
    • 5 years ago
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    yea

  11. anonymous
    • 5 years ago
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    You can find the equation for the tangent plane at the point (x_0,y_0,z_0) by taking the gradient of your function, evaluating at that point (the grad. will give you a vector that is perpendicular to the plane) and sub. in those values into the equation for a plane.

  12. anonymous
    • 5 years ago
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    So...

  13. anonymous
    • 5 years ago
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    i did calcu;at the equation of tangent plane.

  14. anonymous
    • 5 years ago
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    \[f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1=0\]

  15. anonymous
    • 5 years ago
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    Oh...what's wrong then?

  16. anonymous
    • 5 years ago
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    the problem i s i am getting the volume in negative.. =(

  17. anonymous
    • 5 years ago
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    \[\nabla f=(\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2})\]

  18. anonymous
    • 5 years ago
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    Volume? You don't need volume here.

  19. anonymous
    • 5 years ago
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    i took the triple product. should i attach my cal. ?

  20. anonymous
    • 5 years ago
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    Wait and see what I get, then you can compare.

  21. anonymous
    • 5 years ago
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    i need a volume of a tetrahedron which is formed when x=0,y=0,z=0 and with thi stangent planbe.

  22. anonymous
    • 5 years ago
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    At the point (x_0,y_0,z_0), the normal to the tangent plane will be\[\nabla f(x_0,y_0,z_0)=(\frac{2x_0}{a^2},\frac{2y_0}{b^2},\frac{2z_0}{c^2})\]and so, at the same point, the equation of the tangent plane will be,\[\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)+\frac{2z_0}{c^2}(z-z_0)=0\]that is,\[\frac{x_0}{a^2}(x-x_0)+\frac{y_0}{b^2}(y-y_0)+\frac{z_0}{c^2}(z-z_0)=0\]

  23. anonymous
    • 5 years ago
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    i got these coordinates \[(-a ^{2}/x , b ^{2}/y, 0), (-a ^{2}/x,0,c ^{2}/z),(-a ^{2}/x,0,0)\]

  24. anonymous
    • 5 years ago
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    yes and on simplifying that equation you get:\[xx _{o}^{}/a^2 +yy _{o}^{}/b^2 +zz _{o}^{}/c^2 =1\]

  25. anonymous
    • 5 years ago
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    then i used the tripple product to get vol. but its negative

  26. anonymous
    • 5 years ago
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    Yeah, see, I'm thinking the plane has to form one side of the tetrahedron. Given that, it will cut the x-, y- and z-axes at the points,\[(\frac{a^2}{x_0},0,0)\]\[(0,\frac{b^2}{y_0},0)\]\[(0,0,\frac{c^2}{z_0})\]

  27. anonymous
    • 5 years ago
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  28. anonymous
    • 5 years ago
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    hey why did you take -x,-y,-z? how did you know its negative?

  29. anonymous
    • 5 years ago
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    ?

  30. anonymous
    • 5 years ago
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    as in i took x y z . that is the positive axis. why did u take the negative ones?

  31. anonymous
    • 5 years ago
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    I haven't taken negative axes.

  32. anonymous
    • 5 years ago
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    sorry my bad i was jumping to the triple product... so with the coordinates u ve mentioned, i should be taking triple product of these? or as i have done in my calculaatios in the attachment?

  33. anonymous
    • 5 years ago
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    Well, it's hard for me to say because I don't have the entire question. I'm only piecing together the fact you have a tangent plane that's to form one surface of a tetrahedron, and with the other three sides lying in the x-y, x-z and x-y planes.

  34. anonymous
    • 5 years ago
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    The triple product is only going to find the volume of a parallelopiped.

  35. anonymous
    • 5 years ago
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    ok its find the min vol bounded by the planes, xy, yz,xz, and a plane tangent to the folowing: x^2/a^2+(y/b)^2 + (z/c)^2 =1

  36. anonymous
    • 5 years ago
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    Okay, I guessed right.

  37. anonymous
    • 5 years ago
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  38. anonymous
    • 5 years ago
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    whats wrong with my calculations =(

  39. anonymous
    • 5 years ago
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    I'd have to do the problem to see that. When do you need this? I have some things I need to take care of.

  40. anonymous
    • 5 years ago
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    the attachment is not opening here.

  41. anonymous
    • 5 years ago
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    it finally opened.

  42. anonymous
    • 5 years ago
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  43. anonymous
    • 5 years ago
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    its hw question. anyway i'll further brood on it.

  44. anonymous
    • 5 years ago
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    thank u!

  45. anonymous
    • 5 years ago
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    ok. if i get a chance, i'll come back to it :)

  46. anonymous
    • 5 years ago
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    rsaad, if you're there, I have the answer for you.

  47. anonymous
    • 5 years ago
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    It's a long problem...

  48. anonymous
    • 5 years ago
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    There are a couple of ways of doing this (e.g. Lagrange multipliers), but substitution works fine too.

  49. anonymous
    • 5 years ago
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    I'll upload the first sheet where I've found the appropriate determinant using the coordinates I found before:

  50. anonymous
    • 5 years ago
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  51. anonymous
    • 5 years ago
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    Now, the x, y and z are points that satisfy the equation of your surface, so we may eliminate z as\[z=c \sqrt{1-\frac{x^2}{a^2}+\frac{y^2}{b^2}}\]

  52. anonymous
    • 5 years ago
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    and sub. this in for z in the volume equation. Now we have a function in x and y only. We can find the extrema by finding\[\frac{\partial V}{\partial x}=0\]and\[\frac{\partial V}{\partial y}=0\]

  53. anonymous
    • 5 years ago
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    lol, loki, you've got a neat writing

  54. anonymous
    • 5 years ago
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    You may solve for x^2 in the first to find\[x^2=\frac{(b^2-y^2)a^2}{2b^2}\]

  55. anonymous
    • 5 years ago
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    hehe, thanks :)

  56. anonymous
    • 5 years ago
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    ^_^ np

  57. anonymous
    • 5 years ago
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    and sub. this into the partial derivative found for y to find:\[y^2=\frac{b^2}{3}\]Substituting this back into the first expression for x^2 we have,\[x^2=\frac{a^2}{3}\]

  58. anonymous
    • 5 years ago
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    To find z, we only need to realise that z is constrained by the surface, so substituting these in, we find,\[z^2=c^2(1-\frac{x^2}{a^2}-\frac{y^2}{b^2})=c^2(1-\frac{a^2/3}{a^2}-\frac{b^2/3}{b^2})=\frac{c^2}{3}\]

  59. anonymous
    • 5 years ago
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    Substituting these into the expression for volume, we have\[V=\frac{1}{6}\frac{a^2b^2c^2}{xyz}=\frac{1}{6}\frac{a^2b^2c^2}{\frac{a}{\sqrt{3}}\frac{b}{\sqrt{3}}\frac{c}{\sqrt{3}}}=\frac{\sqrt{3}}{2}abc\]

  60. anonymous
    • 5 years ago
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    You have to check that these values for x and y do yield a maximum in V by using the appropriate test\[f_{xx}<0\]and\[f_{xx}f_{yy}-f_{xy}^2>0\]

  61. anonymous
    • 5 years ago
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    And, um, give us a fan point if you haven't already ;p

  62. anonymous
    • 5 years ago
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    thanks lokisan =) but i've a question.why did you use the surface equation when actually it is the tangent plane which is bounding the tetrahedron?

  63. anonymous
    • 5 years ago
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    surface eq to eliminate z. why not the tangent plane equation? and also, why in calculating the detrminent, u used a-b, b-c,c-d where d was 0,0,0. on one website, that i consukted the formula was d-a,d-b,d-c... i am confused.

  64. anonymous
    • 5 years ago
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    You have to think of the surface equation as being the set of values x, y and z are allowed to take. Remember, the x, y, and z used in the volume formula were obtained from the plane equation, and that particular point was the point on the surface where the tangent plane was generated.

  65. anonymous
    • 5 years ago
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    I dropped the notation,\[V=\frac{1}{6}\frac{a^2b^2c^2}{x_0y_0z_0}\]for\[V=\frac{1}{6}\frac{a^2b^2c^2}{xyz}\]

  66. anonymous
    • 5 years ago
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    okay.

  67. anonymous
    • 5 years ago
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    You're not convinced...

  68. anonymous
    • 5 years ago
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    what about my 2nd q:and also, why in calculating the detrminent, u used a-b, b-c,c-d where d was 0,0,0. on one website, that i consukted the formula was d-a,d-b,d-c... i am confused.

  69. anonymous
    • 5 years ago
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    hang on...brb

  70. anonymous
    • 5 years ago
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    well yes. cuz when i plotted the tetrahedron, the sid was flanked by tangent PLANE. so why not set of points satisfied by that tangent plane? i mean if we look at the surface as a whole, wont that be misleading?

  71. anonymous
    • 5 years ago
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    For the determinant, I used the formula (1/6)·det(a−b, b−c, c−d), where a, b, c and d are the vertices. This one makes no specific mention of assuming a vertex (e.g. d) to be the origin.

  72. anonymous
    • 5 years ago
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    but like i said, i checked internet earlier and there it was mentioned d=000

  73. anonymous
    • 5 years ago
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    Hang on, this is easy to deal with. I want to talk about your other question.

  74. anonymous
    • 5 years ago
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    ok just tell me if we have found the 3 vertices, we simply subtract each of them to get 3 eqs?

  75. anonymous
    • 5 years ago
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    ok.

  76. anonymous
    • 5 years ago
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    You asked, "so why not set of points satisfied by that tangent plane? i mean if we look at the surface as a whole, wont that be misleading?"

  77. anonymous
    • 5 years ago
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    Okay...the volume is determined by the tangent plane, and the tangent plane is determined by the surface. You're being asked to find the specific coordinates (i.e. point on surface which generates the tangent plane) that leads to a tangent plane that gives you the maximal volume. That's why the x, y and z must satisfy the surface equation, because we're generating tangent planes for various points (x,y,z) on the surface. We can't choose arbitrary x, y AND z, since these won't necessarily be on the surface, and therefore, won't give us a tangent plane to the surface. We can choose x and y, but the surface equation will dictate the allowable z we may take.

  78. anonymous
    • 5 years ago
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    O......!!! now i see! =) thanks!!!

  79. anonymous
    • 5 years ago
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    good :)

  80. anonymous
    • 5 years ago
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    Now, as for the equation for the volume...I used a formula from one of my books, and I've also seen it on wikipedia: http://en.wikipedia.org/wiki/Tetrahedron Look under 'volume'.

  81. anonymous
    • 5 years ago
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    You can derive this for yourself if you need convincing by understanding that the volume of any pyramid is 1/3 x base x perpendicular height.

  82. anonymous
    • 5 years ago
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    ahan. let me see

  83. anonymous
    • 5 years ago
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    right. so then it follows what you have done / thanks a lot! btw which uni r u from? i m curious 'cuz u r able to answer all my questions!

  84. anonymous
    • 5 years ago
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    A very old university... ;)

  85. anonymous
    • 5 years ago
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    IV league?

  86. anonymous
    • 5 years ago
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    Were all your questions answered?

  87. anonymous
    • 5 years ago
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    well yes. i'd posted some problems earlier and u solved them and so i'd fan-ed you already =)

  88. anonymous
    • 5 years ago
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    Ah...good...you caught me just in time, too...I was about to log out -.- Happy mathing :D

  89. anonymous
    • 5 years ago
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    =D lucky me! hey come on which uni r u from? i m curious! caltech?

  90. anonymous
    • 5 years ago
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    no... :)

  91. anonymous
    • 5 years ago
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    then? iv league?

  92. anonymous
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    I'm a citizen of the UK

  93. anonymous
    • 5 years ago
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    oxford? cambridge?

  94. anonymous
    • 5 years ago
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    Well, it's possibly one of the two ;)

  95. anonymous
    • 5 years ago
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    aha! no doubt you are really good at math! =P ok thanks a lot!

  96. anonymous
    • 5 years ago
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    no probs :D

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