What Are the Hottest and Coldest Points on the Unit Ball?

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    Calculus College
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Discussion Overview

The discussion revolves around finding the hottest and coldest points on the unit ball defined by the equation x^2 + y^2 + z^2 ≤ 1, given a temperature function u = x + 2y + 3z. Participants explore various methods to approach this problem, including calculus techniques and geometric reasoning.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests calculating partial derivatives in each coordinate direction and setting them to zero to find critical points within the unit ball.
  • Another proposes solving the sphere's equation for x, y, and z, substituting these into the temperature function to derive equations for u, then finding critical points.
  • It is noted that extremes likely do not exist in the interior of the sphere, as moving outward generally increases temperature.
  • A participant mentions computing the gradient of the temperature function, indicating that the maximum temperature will occur on the surface of the unit ball.
  • Specific coordinates for maximum and minimum temperatures are calculated, with one participant asserting that the minimum temperature occurs at the origin and the maximum at a point on the surface derived from the gradient.
  • Humor is introduced in the discussion, with participants playfully correcting each other's calculations.

Areas of Agreement / Disagreement

Participants generally agree that the maximum temperature occurs on the surface of the unit ball, but there is some contention regarding the exact calculations and methods used to derive the maximum and minimum points. The discussion remains unresolved regarding the best approach to find these points.

Contextual Notes

Some calculations presented may depend on assumptions about the behavior of the temperature function and the geometry of the unit ball. There are unresolved mathematical steps and potential inaccuracies in the calculations discussed.

sam2
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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
 
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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 explanation for using them off of the top of my head. You could try searching the net.
 
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.
 
Hi, thanks for the replies... Integral I like your intuition...

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

But you did! Last I checked 1+4+9=14 :smile:
 
Originally posted by Integral
But you did! Last I checked 1+4+9=14 :smile:

Only for small values of 1.
 

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