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Proof- Question about Rolle's Theorem

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    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).

    2. Relevant equations

    3. 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?
  2. jcsd
  3. Nov 19, 2009 #2


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    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".
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