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Thread: Curvature problem

  1. #1
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    Curvature problem

    Let k(t), $\displaystyle t \geq 0 $ construct the curve a(t), for which the curvature at $\displaystyle t \geq 0 $ is equal to k(t) and which satisfies the initial conditions a(0) = (0,0) and a'(0) = (1,0).

    Prove that $\displaystyle a(t) = ( \int ^{t}_{0} cos ( \int ^{s}_{0}k(u)du)ds , \int ^{t}_{0} sin ( \int ^{s} _{0} k(u)du) ds ) $

    Questions:

    Now, if I substitute 0 in for t, I get (0,0). But taking the prime, should I take it relative to t or s?
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  2. #2
    Super Member Rebesques's Avatar
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    Just to get you going.

    Let the curve be $\displaystyle c(t)=(x(t),y(t)), t\geq 0$. Suppose (without loss of generality) c is already parametrized by arc length. Then the formula $\displaystyle |c'(t)|=1$ gives $\displaystyle x'(t)^2+y'(t)^2=1$, so there exists a function $\displaystyle \phi(t)$ such that $\displaystyle x'(t)=\cos(\phi(t)), \ y'(t)=\sin(\phi(t))$. To explicitly calculate this function, recall that $\displaystyle k(t)=x'(t)y''(t)-x''(t)y'(t)$.
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