Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   Calculus & Beyond Homework (http://www.physicsforums.com/forumdisplay.php?f=156)
-   -   Calculus 3 question (http://www.physicsforums.com/showthread.php?t=151561)

feelau Jan15-07 07:12 PM

Calculus 3 question
 
1. The problem statement, all variables and given/known data
So we're suppose to show that the intervals connecting vertices of a tetrahedron with centers of gravity of opposite sides interesect at one point, namely the center of gravity of a normal tetrahedron. The center of gravity is (P+Q+R+S)/4.


2. Relevant equations
All I know is we have to somehow work with vectors so we at the end, we end up with result.


I really have no idea how to start. Please help :(

Tom1992 Jan15-07 07:39 PM

you have to use triple integration. it's not really a proof, but a calculation.

feelau Jan15-07 09:07 PM

That can't be right, we just barely started this class and we're only learning about vectors and haven't done anything with integration :bugeye:

Tom1992 Jan15-07 09:18 PM

you said center of gravity of a normal tetrahedron. that must involve triple integration, unless the density is uniform throughout the tetrahedron, which you did not state.

if the density is uniform, simply intersect two lines, each of which go from one vertex to the center of the opposite triangle, which is at one third the height of the triangle.

feelau Jan15-07 09:23 PM

Well we show that the lines going from one vertice to the center of gravity of opposite sides will end up crossing each other right in the middle of the tetrahedron(the very center). The problem says that at that point, we know the answer is (P+Q+R+S)/4 where the letters correspond to vertices. We're suppose to use vectors to show that, at that middle point, we'll get (P+Q+R+S)/4. I think you're interpreting it another way, perhaps this made more sense?

Tom1992 Jan15-07 09:36 PM

you can think along the lines of use the center of gravity formula for 4 masses:

cg = [m1(P) + m2(Q) +m3(R) + m4(S)]/(m1+m2+m3+m4)

think of what m1,m2,m3,m4 would be in your special case.

but the symmetry argument using the interesection of two lines should also work too.

feelau Jan15-07 09:46 PM

well there would be four lines all crossing at the center of gravity and i'm suppose to show that with vectors though cuz this is a vector problem. I know the physics concept of it(learned it last semester) but I need to show it with vectors. Do you have any ideas?

Tom1992 Jan15-07 10:02 PM

using nothing but vectors?

how about place the center of gravity at the origin, by symmetry (which we can use due to the simple conditions of the object) the four vertices must be equidistant from the origin. you can figure out the other symmetry properties to argue that the sum of the four vectors must be zero.

i don't like this solution very much, but i haven't taken vector courses since i was 11.

feelau Jan15-07 10:08 PM

hm....ok...i'll look into it. thanks


All times are GMT -5. The time now is 02:50 PM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums