Mixing
Finding a plane curve with curvature $kappa(s)$
$begingroup$
Given a differentiable function $kappa(s)$, $s in I$, show that the parametrized plane curve having $k(s)=k$ as curvature is given by:
$alpha(s)=(int cos(theta(s))ds+a,int sin(theta(s))ds+b)$
where
$theta(s) = int k(s)ds + gamma$
And that the curve is determined up to translation of the vector (a,b) and a rotation of the angel $gamma$
Solution:
$T'(s)=kappa(-sin(theta(s)),cos(theta(s))=kappa N$
Which is easy to see as $alpha(s)$ is arc length parameterized.
How do I show that this curve is uniquely determined up to translation of the vector (a,b) and rotation of the angel $gamma$?
differential-geometry
asked Jan 27 at 4:00
Mathematical MushroomMathematical Mushroom
848
$endgroup$
add a comment |
$begingroup$
Given a differentiable function $kappa(s)$, $s in I$, show that the parametrized plane curve having $k(s)=k$ as curvature is given by:
$alpha(s)=(int cos(theta(s))ds+a,int sin(theta(s))ds+b)$
where
$theta(s) = int k(s)ds + gamma$
And that the curve is determined up to translation of the vector (a,b) and a rotation of the angel $gamma$
Solution:
$T'(s)=kappa(-sin(theta(s)),cos(theta(s))=kappa N$
Which is easy to see as $alpha(s)$ is arc length parameterized.
How do I show that this curve is uniquely determined up to translation of the vector (a,b) and rotation of the angel $gamma$?
differential-geometry
asked Jan 27 at 4:00
Mathematical MushroomMathematical Mushroom
848
$endgroup$
add a comment |
$begingroup$
Given a differentiable function $kappa(s)$, $s in I$, show that the parametrized plane curve having $k(s)=k$ as curvature is given by:
$alpha(s)=(int cos(theta(s))ds+a,int sin(theta(s))ds+b)$
where
$theta(s) = int k(s)ds + gamma$
And that the curve is determined up to translation of the vector (a,b) and a rotation of the angel $gamma$
Solution:
$T'(s)=kappa(-sin(theta(s)),cos(theta(s))=kappa N$
Which is easy to see as $alpha(s)$ is arc length parameterized.
How do I show that this curve is uniquely determined up to translation of the vector (a,b) and rotation of the angel $gamma$?
differential-geometry
asked Jan 27 at 4:00
Mathematical MushroomMathematical Mushroom
848
$endgroup$
Given a differentiable function $kappa(s)$, $s in I$, show that the parametrized plane curve having $k(s)=k$ as curvature is given by:
$alpha(s)=(int cos(theta(s))ds+a,int sin(theta(s))ds+b)$
where
$theta(s) = int k(s)ds + gamma$
And that the curve is determined up to translation of the vector (a,b) and a rotation of the angel $gamma$
Solution:
$T'(s)=kappa(-sin(theta(s)),cos(theta(s))=kappa N$
Which is easy to see as $alpha(s)$ is arc length parameterized.
How do I show that this curve is uniquely determined up to translation of the vector (a,b) and rotation of the angel $gamma$?
differential-geometry
differential-geometry
asked Jan 27 at 4:00
Mathematical MushroomMathematical Mushroom
848
asked Jan 27 at 4:00
Mathematical MushroomMathematical Mushroom
848
asked Jan 27 at 4:00
Mathematical MushroomMathematical Mushroom
848
asked Jan 27 at 4:00
Mathematical MushroomMathematical Mushroom
848
asked Jan 27 at 4:00
Mathematical MushroomMathematical Mushroom
848
848
add a comment |
add a comment |
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