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mathstina Group Title

Show that the sum of the x, y, z intercepts of any tangent plane to the surface √x+√y+√z= √c is a constant.

  • 2 years ago
  • 2 years ago

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  1. TuringTest Group Title
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    similar to the last one in approach; get the equation of the tangent plane for a point on the surface P=(p,q,r)

    • 2 years ago
  2. mathstina Group Title
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    i ve get the eqn for tangent plane?

    • 2 years ago
  3. TuringTest Group Title
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    no, you need to find it you need the gradient at that point (normal to the plane) and a point on the plane

    • 2 years ago
  4. mathstina Group Title
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    \[1/2[1/\sqrt{p} +1/\sqrt{q}+1/\sqrt{r}]=0\]

    • 2 years ago
  5. TuringTest Group Title
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    yes, now dot that with a general vector in the plane to get the equation of the tangent plane

    • 2 years ago
  6. TuringTest Group Title
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    actually you should have written that as a vector\[\nabla f(p,q,r)=\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\]

    • 2 years ago
  7. mathstina Group Title
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    \[1/2{p/\sqrt{p}(x-p) + q/\sqrt{q}(y-q) +r/\sqrt{r}(z-r)} =0\]

    • 2 years ago
  8. mathstina Group Title
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    how to simplify?

    • 2 years ago
  9. TuringTest Group Title
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    I think you got an extra p,q and r somehow\[\nabla f(p,q,r)=\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]

    • 2 years ago
  10. mathstina Group Title
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    ok.

    • 2 years ago
  11. TuringTest Group Title
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    \[\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]\[p^{-1/2}x+q^{-1/2}y+r^{-1/2}z=p^{1/2}+q^{1/2}+r^{1/2}\]

    • 2 years ago
  12. mathstina Group Title
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    hw to get the second eqn?

    • 2 years ago
  13. TuringTest Group Title
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    you mean the part on the right? I just simplified the dot product and moved all the constants to one side

    • 2 years ago
  14. mathstina Group Title
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    ok!

    • 2 years ago
  15. TuringTest Group Title
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    \[\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]\[p^{-1/2}x-p^{1/2}+q^{-1/2}y-q^{1/2}+r^{-1/2}z-r^{1/2}=0\]\[p^{-1/2}x+q^{-1/2}y+r^{-1/2}z=p^{1/2}+q^{1/2}+r^{1/2}\]

    • 2 years ago
  16. TuringTest Group Title
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    now plug in for each intercept x=0, y=0, z=0 add the three results, what do you get?

    • 2 years ago
  17. mathstina Group Title
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    what happened to the 1/2 for the dot product?

    • 2 years ago
  18. TuringTest Group Title
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    all that stuff was =0, so multiplying both sides by 2 it is still =0, but simpler

    • 2 years ago
  19. TuringTest Group Title
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    \[\frac12\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]\[\implies\langle p^{-1/2},q^{-1/2},r^{-1/2}\rangle\cdot\langle x-p,y-q,z-r\rangle=0\]

    • 2 years ago
  20. mathstina Group Title
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    k. hw do i sub for each intercept?

    • 2 years ago
  21. mathstina Group Title
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    is it (x, y,z) =1 respectively

    • 2 years ago
  22. TuringTest Group Title
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    x=0 will give you one equation, y=0 another, z=0 another add them all up

    • 2 years ago
  23. TuringTest Group Title
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    sorry, I guess I mean xy=0, yz=0, xz=0

    • 2 years ago
  24. mathstina Group Title
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    im stuck

    • 2 years ago
  25. TuringTest Group Title
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    to find the x-intercept, plug in y=0 and z=0 what do you get?

    • 2 years ago
  26. mathstina Group Title
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    r −1/2 z=p 1/2 +q 1/2 +r 1/2

    • 2 years ago
  27. TuringTest Group Title
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    you plugged in x=0 and y=0, so this is going to be the z-intercept solve for z

    • 2 years ago
  28. mathstina Group Title
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    z= 1

    • 2 years ago
  29. TuringTest Group Title
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    how do you get that?

    • 2 years ago
  30. mathstina Group Title
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    r −1/2 z=r 1/2

    • 2 years ago
  31. TuringTest Group Title
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    where did p and q go?

    • 2 years ago
  32. TuringTest Group Title
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    besides, what you wrote would imply that z=r

    • 2 years ago
  33. mathstina Group Title
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    as x and y is 0

    • 2 years ago
  34. mathstina Group Title
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    show me an eg

    • 2 years ago
  35. TuringTest Group Title
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    but x and y being zero does not change the value of p and q

    • 2 years ago
  36. TuringTest Group Title
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    p, q, and r are the coordinates of the point *on the shape* from which we got our tangent plane x, y, and z can be anywhere in the plane, but the equation of the plane still depends on p, q, and r

    • 2 years ago
  37. TuringTest Group Title
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    the x-intercept *of the plane* will be found by plugging in y=z=0 to our formula *for the plane*

    • 2 years ago
  38. TuringTest Group Title
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    x-intercept:\[p^{-1/2}x=p^{1/2}+q^{1/2}+r^{1/2}\implies x=p+p^{1/2}q^{1/2}+q^{1/2}r^{1/2}\]now find the y and z-intercepts

    • 2 years ago
  39. TuringTest Group Title
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    sorry, typo\[p^{-1/2}x=p^{1/2}+q^{1/2}+r^{1/2}\implies x=p+p^{1/2}q^{1/2}+p^{1/2}r^{1/2}\]

    • 2 years ago
  40. mathstina Group Title
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    q −1/2 y=p 1/2 +q 1/2 +r 1/2 ⟹y=p1/2q1q+p 1/2 r 1/2

    • 2 years ago
  41. mathstina Group Title
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    how to solve for the unknowns?

    • 2 years ago
  42. TuringTest Group Title
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    you don't, we are trying to show that this is true for *all* points that are on the surface so we have to leave it general. same reason we went ahead with the last one in terms of a,b, and c

    • 2 years ago
  43. TuringTest Group Title
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    so get the z-intercept, then add the x, y, and z-intercept together

    • 2 years ago
  44. mathstina Group Title
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    hw to get the sum?

    • 2 years ago
  45. TuringTest Group Title
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    just add them; it will look ugly at first, but a miracle will happen...

    • 2 years ago
  46. TuringTest Group Title
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    erm, just add the intercepts...

    • 2 years ago
  47. mathstina Group Title
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    eg?

    • 2 years ago
  48. TuringTest Group Title
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    \[x_0=p+p^{1/2}q+p^{1/2}r^{1/2}\]\[y_0=p^{1/2}q^{1/2}+q+q^{1/2}r^{1/2}\]\[z_0=p^{1/2}r^{1/2}+q^{1/2}r^{1/2}+r\]

    • 2 years ago
  49. TuringTest Group Title
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    the sum of the intercepts is then\[x_0+y_0+z_0=p+q+r+2p^{1/2}q^{1/2}+2p^{1/2}r^{1/2}+2q^{1/2}r^{1/2}\]

    • 2 years ago
  50. TuringTest Group Title
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    this looks ugly, but it can be factored look at the original function... what is the relationship?

    • 2 years ago
  51. mathstina Group Title
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    sry nt sure

    • 2 years ago
  52. TuringTest Group Title
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    \[p+q+r+2p^{1/2}q^{1/2}+2p^{1/2}r^{1/2}+2q^{1/2}r^{1/2}=(\sqrt p+\sqrt q+\sqrt r)^2\]now do you see how this is connected to the original function?

    • 2 years ago
  53. mathstina Group Title
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    wow hw do do this?

    • 2 years ago
  54. TuringTest Group Title
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    to do what?

    • 2 years ago
  55. mathstina Group Title
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    simplify hw?

    • 2 years ago
  56. TuringTest Group Title
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    how did I know the above you mean? practice and recognizing forms

    • 2 years ago
  57. TuringTest Group Title
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    do you see what to do from here at all?

    • 2 years ago
  58. mathstina Group Title
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    no.pls

    • 2 years ago
  59. TuringTest Group Title
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    remember that p, q, and r satisfy our original equation, so\[\sqrt p+\sqrt q+\sqrt r=\sqrt c\]

    • 2 years ago
  60. TuringTest Group Title
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    combine that with the fact that the sum of the intercepts we have shown to be\[(\sqrt p+\sqrt q+\sqrt r)^2=(\sqrt c)^2=c\]and you see that all intercepts for any tangent plane sum to exactly c QED

    • 2 years ago
  61. mathstina Group Title
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    how to get simplified to (p √ +q √ +r √ ) 2 ?

    • 2 years ago
  62. TuringTest Group Title
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    I did it by observation and recognizing the form, but... look at the intercepts\[x_0=p+p^{1/2}q+p^{1/2}r^{1/2}=p^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[y_0=p^{1/2}q^{1/2}+q+q^{1/2}r^{1/2}=q^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[z_0=p^{1/2}r^{1/2}+q^{1/2}r^{1/2}+r=r^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]

    • 2 years ago
  63. TuringTest Group Title
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    \[x_0+y_0+z_0\]\[=p^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})+q^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})+r^{1/2}(p^{1/2}+q^{1/2}+r^{1/2})\]\[=(p^{1/2}+q^{1/2}+r^{1/2})(p^{1/2}+q^{1/2}+r^{1/2})\]\[=(p^{1/2}+q^{1/2}+r^{1/2})^2\]

    • 2 years ago
  64. mathstina Group Title
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    ok .thanks a lot!!!!!

    • 2 years ago
  65. TuringTest Group Title
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    welcome, this was a good mental exercise to start my day :)

    • 2 years ago
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