College calculus question

In summary: The minimum temperature at (-1/√(15),-2/√(15),-3/√(15)) will not be at the origin, but at one of the other points on the sphere (depending on x,y, and z).
  • #1
sam2
22
0
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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  • #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 explanation for using them off of the top of my head. You could try searching the net.
 
  • #3
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.
 
  • #4
Hi, thanks for the replies... Integral I like your intuition...

regards
 
  • #5
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)).
 
  • #6
Ah.. Come Halls' we need to leave them some fun!

But you did! Last I checked 1+4+9=14 :smile:
 
  • #7
Originally posted by Integral
But you did! Last I checked 1+4+9=14 :smile:

Only for small values of 1.
 

What is college calculus?

College calculus is a branch of mathematics that deals with the study of change in continuous quantities. It involves the use of functions, limits, derivatives, and integrals to analyze and solve real-world problems.

Why is calculus important in college?

Calculus is important in college because it serves as the foundation for many other fields of study such as physics, engineering, and economics. It also helps develop critical thinking skills and problem-solving abilities.

How is college calculus different from high school calculus?

College calculus is more rigorous and theoretical than high school calculus. It covers more advanced topics such as multivariable calculus, vector calculus, and differential equations. It also requires a deeper understanding of mathematical concepts and the ability to apply them in various contexts.

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The key concepts in college calculus include limits, derivatives, integrals, and applications of these concepts such as optimization, related rates, and curve sketching. It also involves understanding the concept of infinity and its role in calculus.

Is college calculus difficult?

College calculus can be challenging for some students, as it requires a strong foundation in algebra and trigonometry. However, with dedication and practice, it is possible to master the concepts and excel in the subject. It is important to seek help from professors or tutors if needed.

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