Mixing
Showing the co-planar evolute of an arc-length parameterized planar curve is unique.
$begingroup$
If $alpha$ is an arc-length parameterized plane curve, prove that the curve $b(t)$ given by:
$b(s)=alpha(s) + frac{N(s)}{kappa(s)}$
is the unique evolute of $alpha$ lying in the same plane as $alpha$.
Furthermore, prove that this curve is regular if $kappa'(s) neq 0$
So i'm a bit puzzled at this one. Because $alpha$ is an involute of $b(s)$ we have:
$alpha(s) = frac{b'(s)}{|b'(s)|}(c-s)+b(s)$ and $alpha'(s) cdot b'(s) = 0$
I can show that the $b(s)$ given in the problem statement is indeed an involute by taking the dot product of it with $alpha'(s)$ and showing that it is zero. Furthermore, $b(s)$ must be in the same plane as $alpha(s)$, because of how it is defined it is simple to see that there is no torsion and so will not leave the plane. I do not know how to show uniqueness.
To show regularity, note that:
$b'(s) = alpha'(s) - T - frac{N(s) kappa'(s)}{kappa(s)^2}$
= $-frac{N(s) kappa'(s)}{kappa(s)^2}$
The fact that $b(t)$ is regular means its derivative will never be zero. $kappa'(s)$ is given to never be zero, but why must $N(s)$ never be zero?
Thanks in advance!
differential-geometry
asked Jan 26 at 1:50
Mathematical MushroomMathematical Mushroom
847
$endgroup$
add a comment |
$begingroup$
If $alpha$ is an arc-length parameterized plane curve, prove that the curve $b(t)$ given by:
$b(s)=alpha(s) + frac{N(s)}{kappa(s)}$
is the unique evolute of $alpha$ lying in the same plane as $alpha$.
Furthermore, prove that this curve is regular if $kappa'(s) neq 0$
So i'm a bit puzzled at this one. Because $alpha$ is an involute of $b(s)$ we have:
$alpha(s) = frac{b'(s)}{|b'(s)|}(c-s)+b(s)$ and $alpha'(s) cdot b'(s) = 0$
I can show that the $b(s)$ given in the problem statement is indeed an involute by taking the dot product of it with $alpha'(s)$ and showing that it is zero. Furthermore, $b(s)$ must be in the same plane as $alpha(s)$, because of how it is defined it is simple to see that there is no torsion and so will not leave the plane. I do not know how to show uniqueness.
To show regularity, note that:
$b'(s) = alpha'(s) - T - frac{N(s) kappa'(s)}{kappa(s)^2}$
= $-frac{N(s) kappa'(s)}{kappa(s)^2}$
The fact that $b(t)$ is regular means its derivative will never be zero. $kappa'(s)$ is given to never be zero, but why must $N(s)$ never be zero?
Thanks in advance!
differential-geometry
asked Jan 26 at 1:50
Mathematical MushroomMathematical Mushroom
847
$endgroup$
$begingroup$
Very cool question, +1!!! A question of my own: how $c$ defined? Thanks!
$endgroup$
– Robert Lewis
Jan 26 at 4:01
add a comment |
$begingroup$
If $alpha$ is an arc-length parameterized plane curve, prove that the curve $b(t)$ given by:
$b(s)=alpha(s) + frac{N(s)}{kappa(s)}$
is the unique evolute of $alpha$ lying in the same plane as $alpha$.
Furthermore, prove that this curve is regular if $kappa'(s) neq 0$
So i'm a bit puzzled at this one. Because $alpha$ is an involute of $b(s)$ we have:
$alpha(s) = frac{b'(s)}{|b'(s)|}(c-s)+b(s)$ and $alpha'(s) cdot b'(s) = 0$
I can show that the $b(s)$ given in the problem statement is indeed an involute by taking the dot product of it with $alpha'(s)$ and showing that it is zero. Furthermore, $b(s)$ must be in the same plane as $alpha(s)$, because of how it is defined it is simple to see that there is no torsion and so will not leave the plane. I do not know how to show uniqueness.
To show regularity, note that:
$b'(s) = alpha'(s) - T - frac{N(s) kappa'(s)}{kappa(s)^2}$
= $-frac{N(s) kappa'(s)}{kappa(s)^2}$
The fact that $b(t)$ is regular means its derivative will never be zero. $kappa'(s)$ is given to never be zero, but why must $N(s)$ never be zero?
Thanks in advance!
differential-geometry
asked Jan 26 at 1:50
Mathematical MushroomMathematical Mushroom
847
$endgroup$
If $alpha$ is an arc-length parameterized plane curve, prove that the curve $b(t)$ given by:
$b(s)=alpha(s) + frac{N(s)}{kappa(s)}$
is the unique evolute of $alpha$ lying in the same plane as $alpha$.
Furthermore, prove that this curve is regular if $kappa'(s) neq 0$
So i'm a bit puzzled at this one. Because $alpha$ is an involute of $b(s)$ we have:
$alpha(s) = frac{b'(s)}{|b'(s)|}(c-s)+b(s)$ and $alpha'(s) cdot b'(s) = 0$
I can show that the $b(s)$ given in the problem statement is indeed an involute by taking the dot product of it with $alpha'(s)$ and showing that it is zero. Furthermore, $b(s)$ must be in the same plane as $alpha(s)$, because of how it is defined it is simple to see that there is no torsion and so will not leave the plane. I do not know how to show uniqueness.
To show regularity, note that:
$b'(s) = alpha'(s) - T - frac{N(s) kappa'(s)}{kappa(s)^2}$
= $-frac{N(s) kappa'(s)}{kappa(s)^2}$
The fact that $b(t)$ is regular means its derivative will never be zero. $kappa'(s)$ is given to never be zero, but why must $N(s)$ never be zero?
Thanks in advance!
differential-geometry
differential-geometry
asked Jan 26 at 1:50
Mathematical MushroomMathematical Mushroom
847
asked Jan 26 at 1:50
Mathematical MushroomMathematical Mushroom
847
asked Jan 26 at 1:50
Mathematical MushroomMathematical Mushroom
847
asked Jan 26 at 1:50
Mathematical MushroomMathematical Mushroom
847
asked Jan 26 at 1:50
Mathematical MushroomMathematical Mushroom
847
847
$begingroup$
Very cool question, +1!!! A question of my own: how $c$ defined? Thanks!
$endgroup$
– Robert Lewis
Jan 26 at 4:01
add a comment |
$begingroup$
Very cool question, +1!!! A question of my own: how $c$ defined? Thanks!
$endgroup$
– Robert Lewis
Jan 26 at 4:01
$begingroup$
Very cool question, +1!!! A question of my own: how $c$ defined? Thanks!
$endgroup$
– Robert Lewis
Jan 26 at 4:01
$begingroup$
Very cool question, +1!!! A question of my own: how $c$ defined? Thanks!
$endgroup$
– Robert Lewis
Jan 26 at 4:01
add a comment |
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$begingroup$
Very cool question, +1!!! A question of my own: how $c$ defined? Thanks!
$endgroup$
– Robert Lewis
Jan 26 at 4:01