Did you try maximizing the distance among all points constrained to z = Sqrt[ 1 - x^2 - y^2 ], using calculus?
I've been given the metric spaces R^3 and S^2=\{(x,y,z) | x^2+y^2+z^2=1\} And i need to find the diameter of A which is A\subseteq S^2\subseteq R^3 Where A^2=\{(x,y,z) \in S^2 | z>0\}
I'm thinking that since S^2 is the unit-sphere and z has to be greater than 0, its down to half the unitsphere, thus making the diameter 2?. But im not sure on how to show it.
Nope not yet, care to elaborate a tad more?
I know the distance between points is given by d((x1,y1,z1),(x2,y2,z2) = Sqrt[(x1-x2)^2+(y1-y2)^2+(z1-z2)^2 ], And then i was thinking plotting in the maximum points hence getting sqrt[(1-(-1)^2+(1-(-1)^2+(1-0)^2] = 3. But the z-coordinate 0 is wrong right? as z > 0. Or am i totally off here?
You can't just put in arbitrary points, necessarily. It doesn't prove that there isn't some other optimal configuration.
Instead, you can try using Lagrange Multipliers on the function
f(x1, y1, z1, x2, y2, z2) = Sqrt[(x1-x2)^2+(y1-y2)^2+(z1-z2)^2 ] subject to constraints
z1 = Sqrt[ 1 - x1^2 - y1^2 ]
z2 = Sqrt[ 1 - x2^2 - y2^2 ]
You are guaranteed to get some maxima and minima.
Lagrange Multipliers is the first method that comes to my mind.
If you don't like optimization, you might be able to appeal to the definition of a sphere and argue geometrically. It might not be as straightforward as doing an optimization procedure, though.
Even though it's not in my syllabus, you intrique me to go look that stuff up I will look into whatever i can, and get back to this thread within a couple of days.
I'm thankful for your reply, allthough our professor is pretty straightforward when it comes to stuff 'outside our syllabus'. Mostly it isnt accepted. But ill look into it anyhow
Ok, i read up on this, and it was actually pretty simple, also i was on the right track (thats always good), so here is my solution:
We're obviously looking at half the unit-sphere, when z>0. thus we're looking for the maximum distance between 2 points on the unitsphere when z -> 0, and thus we can set either x or y =0.
Pick either x=0 og y=0 and use the distance formula: Choose y for example and get sqrtx1-x2)^2+(0-0)^2+(z1-z2)^2 --> sqrt(1-(-1)^2 = 2 for z-> 0. So the diameter is 2. QED