Proof- Question about Rolle's Theorem

[fireb]

Homework Statement

Consider a differentiable curve r: [a,b]-> R(3) such that r(a)= r(b). show that there is a value t belongs [a,b] such that r(t) is orthogonal to r(prime)(t).

Homework Equations

The Attempt at a Solution

My answer: Since r(a)= r(b) the curve must reach a max/min point somewhere in [a,b] then there is a value r(prime) = 0. so r(t) dot r(prime)(t)=0 .

Official answer: Define f(t)= |r(t)|^2 then f is a differentiable function of one function with derivative =[2r(t)] r(prime)(t).
since f(a)= f(b), by rolle's theorem there is a point t belongs [a,b] such that f(prime) is 0. Therefore r(t) dot r(prime)(t) = 0.

Am I completely wrong? It seems like pretty much the same answer to me... can someone explain to me the difference?

 

[LCKurtz]

The most glaring problem with your answer is that there is no meaning to a "maximum" for a space curve twisting through 3d space. Your r(t) is a vector function. So, yes, you are "completely wrong".