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How to find the most far and most close ( points on curve ) to another point ?

  1. Jan 14, 2009 #1
    how to find the "most far and most close" ( points on curve ) to another point ?

    i'm studying a chapter on how to find maxima and minima values of a function using partial derivatives.

    one of the problems is the following:
    "if plane [itex]z=x+y+1[/itex] intersects cone [itex]z^2=x^2+y^2[/itex]
    let the curve that result from intersection, is C
    What is the most far and most close points on C, with respect to (0,0,0) ? and also with respect to [itex](x_1,y_1,z_1)[/itex]"

    i think that that curve would be something like a circle, and that there would be some function that depends on the length between "(0,0,0) or [itex](x_1,y_1,z_1)[/itex]", and "any point on C".

    But what is that function ?
    And how to work out that problem ?

    * i've exam in that chapter after about 10 hours, so please try to answer me with detailed answer as i've no time for discussions for now, may be i do that later.
     
  2. jcsd
  3. Jan 14, 2009 #2

    tiny-tim

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    Hi AbuYusufEg! :smile:

    Hint: the distance2 from (0,0,0) to (x,y,z) is x2 + y2 + z2

    and to help find the intersection, I suggest you rearrange z = x + y + 1 and then square it. :wink:
     
  4. Jan 14, 2009 #3
    Re: how to find the "most far and most close" ( points on curve ) to another point ?

    yes i got that, but i want the points that ONLY on the curve C, i think that (x,y,z) is any point.
    So, How can i get the points that only on the curve C ?
     
  5. Jan 15, 2009 #4

    tiny-tim

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    Rearrange z = x + y + 1 and then square it, and compare with z2 = x2 + y2 :smile:
     
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