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College calculus question

  1. Jan 19, 2004 #1
    Hi all,

    Came across this problem, but it has stumped me:


    Let S denote the unit sphere x^2 + y^2 + z^2 = 1, and let

    u = x + 2y + 3z

    be temperature at points everywhere in 3-space.

    Find the hottest and coldest points on the unit ball x^2 + y^2 + z^2 <= 1


    I figured out that we need to clculate the partial derivatives in each 3 co-ordinate directions and need them all to be zero... while ensuring that we are in the unit ball. Is this just down to trial and error or is there a trick involved? Any hints?

    Many Thanks,

    Sam
     
  2. jcsd
  3. Jan 19, 2004 #2
    I can think of a couple ways of solving this. One way is to solve the equation of the sphere for x, y and z respectively, and substitute those three values into the equation for u. This gets you three equations for u (one without x, one without y, one without z).

    Then find all the critical points of those equations, and test all of those points to find the global max and min.


    The other way would be to use Lagrange multipliers. That's probably the more elegant way of finding the critical points, but I can't think of a good explaination for using them off of the top of my head. You could try searching the net.
     
  4. Jan 19, 2004 #3

    Integral

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    It is clear that the extremes cannot exist on the interior of the sphere. To see this consider the temperature at any point inside the sphere, if you move in a direction toward the exterior the temp increases.

    Compute the gradient of the temp find the point the gradient vector intersects the sphere, that should be you max.


    As for the min, if you want an absolute min clearly the origin is 0 and is there for a min. If you want a min value then it will be at the point diametrically opposite the max.
     
  5. Jan 19, 2004 #4
    Hi, thanks for the replies... Integral I like your intuition...

    regards
     
  6. Jan 20, 2004 #5

    HallsofIvy

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    The gradient of the temperature function is the constant vector
    i+ 2j+ 3k. Since that points away from the origin, the maximum temperature "inside or on the surface" of the unit ball will be on the surface. The gradient vector will intersect the unit ball when
    y=2x, z= 3x and so x2+ 4x2+ 9x2= 15x2= 1 or x= 1/&radic;(15), y= 2/&radic;(15), and z= 3/&radic;(15) .

    The minimum temperature will be at (-1/&radic;(15),-2/&radic;(15),-3/&radic;(15)).
     
  7. Jan 20, 2004 #6

    Integral

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    Ah.. Come Halls' we need to leave them some fun!

    But you did! Last I checked 1+4+9=14 :smile:
     
  8. Jan 20, 2004 #7
    Only for small values of 1.
     
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