Conditioned Extremes (cos(t),sen(t),sen(t/2))

[xoninhas]

Ok, I have really big problems with these kinds of exercises cause I don't even know how to begin.

Determine the points in the line, defined by (cos(t),sen(t),sen(t/2)); farther away from the origin. t belongs to R

ok I know the function to use for the lagrange method is the distance of the point: sqrt(x^2+y^2+z^2) but for the sake of deriving it, it is possible to use also the square of the distance: (x^2 + y^2 + z^2)

f(x,y,z) = x^2 + y^2 + z^2
my condition is F(t) = (cos(t), sen(t), sen(t/2))

so how do I even start?? :S

 

[tkhunny]

Are you SURE you're supposed to talk to LaGrange on this one?

Why not just substitute into F and see what happens?

(x(t),y(t),z(t)) = (cos(t),sen(t),sen(t/2))

 

[xoninhas]

Ok, it is supposed to be talking about conditioned extremes. And since there is a constrain of belonging to the line while being the most farther away from the origin I supposed it would be LaGrange. But in this exercise there is no mention that LaGrange maethod HAS TO be used. So reply to you, no I'm not sure.

As for replacing into F I don't know what do you mean... how do I do that? Do I:

F(x^2, y^2, z^2) ?
or

well honestly I don't have any idea... can you help me out see the light better?

Thanks!

 

[tkhunny]

Oops, I did say "F", didn't I? My mistake. use 'f'.

\(\displaystyle f(x,y,z) = f(t) = cos^{2}(t) + sin^{2}(t) + sin^{2}(t/2)\)

Simplify and proceed.

"Line"?

 

[xoninhas]

Ok, simplified I got:

f(t) = 1 + (sin(t/2))^2

Now I don't really know how to proceed... but I would guess that there is no use for Lagrange and you just do it old-school style...

df/dt = 0 <=> sin(t)/2 = 0 <=> t = 0 + kPi, k belongs to Z

therefore trying on f(t) you get for t = (0 + 2kPi) => f(0) = 1 (maximum) and t = (Pi + 2kPi) => f(Pi) = -1 (minimum)

Is that it??

P.S.- Line, curve, path... Here those words are thrown around in common situations (except a few, not this one though). Maybe correct english math you should say curve?