Mixing
Unique evolute for arc-parameterized plane curve.
$begingroup$
Let $a(s)$ be an arc paramterized plane curve. The goal of this problem is to show that that the unique evolute for $a(s)$ that is in the same plane as $a(s)$ is given by:
$b(s)= a(s) + frac{N}{kappa}$
where $N$ is the unit normal vector for $a(s)$ and $kappa$ is the curvature.
Since $b(s)$ is the evolute of $a(s)$, $a(s)$ is the involute of $b(s)$. I showed earlier that this means that $a(s)$ is of the form:
$a(s)= frac{b'(s)}{|b'(s)|}(c-s)+b(s)$
Also:
$a'(s) cdot b'(s)=0$.
where $b'(s)$ is given by:
$b'(s)=a'(s) + frac{N'}{kappa}-frac{Nkappa'}{kappa^2}$,
Anyway, I would appreciate some insight into this. How do i show uniqueness?
So:
$a'(s) cdot b'(s)=a'(s)cdot a'(s)+a'(s) cdot frac{N'}{kappa}- a'(s) cdot frac{Nkappa'}{kappa^2}$
= $1+a'(s) cdot frac{(-kappa a'(s))}{kappa}+0$
= $1-1=0$
so $b(s)$ appears be an evolute of $a(s)$... but how do we show uniqueness? and what gaurentees that $b(s)$ is in the same plane as $a(s)$?
differential-geometry
asked Jan 22 at 21:10
Math is hardMath is hard
822211
$endgroup$
add a comment |
$begingroup$
Let $a(s)$ be an arc paramterized plane curve. The goal of this problem is to show that that the unique evolute for $a(s)$ that is in the same plane as $a(s)$ is given by:
$b(s)= a(s) + frac{N}{kappa}$
where $N$ is the unit normal vector for $a(s)$ and $kappa$ is the curvature.
Since $b(s)$ is the evolute of $a(s)$, $a(s)$ is the involute of $b(s)$. I showed earlier that this means that $a(s)$ is of the form:
$a(s)= frac{b'(s)}{|b'(s)|}(c-s)+b(s)$
Also:
$a'(s) cdot b'(s)=0$.
where $b'(s)$ is given by:
$b'(s)=a'(s) + frac{N'}{kappa}-frac{Nkappa'}{kappa^2}$,
Anyway, I would appreciate some insight into this. How do i show uniqueness?
So:
$a'(s) cdot b'(s)=a'(s)cdot a'(s)+a'(s) cdot frac{N'}{kappa}- a'(s) cdot frac{Nkappa'}{kappa^2}$
= $1+a'(s) cdot frac{(-kappa a'(s))}{kappa}+0$
= $1-1=0$
so $b(s)$ appears be an evolute of $a(s)$... but how do we show uniqueness? and what gaurentees that $b(s)$ is in the same plane as $a(s)$?
differential-geometry
asked Jan 22 at 21:10
Math is hardMath is hard
822211
$endgroup$
add a comment |
$begingroup$
Let $a(s)$ be an arc paramterized plane curve. The goal of this problem is to show that that the unique evolute for $a(s)$ that is in the same plane as $a(s)$ is given by:
$b(s)= a(s) + frac{N}{kappa}$
where $N$ is the unit normal vector for $a(s)$ and $kappa$ is the curvature.
Since $b(s)$ is the evolute of $a(s)$, $a(s)$ is the involute of $b(s)$. I showed earlier that this means that $a(s)$ is of the form:
$a(s)= frac{b'(s)}{|b'(s)|}(c-s)+b(s)$
Also:
$a'(s) cdot b'(s)=0$.
where $b'(s)$ is given by:
$b'(s)=a'(s) + frac{N'}{kappa}-frac{Nkappa'}{kappa^2}$,
Anyway, I would appreciate some insight into this. How do i show uniqueness?
So:
$a'(s) cdot b'(s)=a'(s)cdot a'(s)+a'(s) cdot frac{N'}{kappa}- a'(s) cdot frac{Nkappa'}{kappa^2}$
= $1+a'(s) cdot frac{(-kappa a'(s))}{kappa}+0$
= $1-1=0$
so $b(s)$ appears be an evolute of $a(s)$... but how do we show uniqueness? and what gaurentees that $b(s)$ is in the same plane as $a(s)$?
differential-geometry
asked Jan 22 at 21:10
Math is hardMath is hard
822211
$endgroup$
Let $a(s)$ be an arc paramterized plane curve. The goal of this problem is to show that that the unique evolute for $a(s)$ that is in the same plane as $a(s)$ is given by:
$b(s)= a(s) + frac{N}{kappa}$
where $N$ is the unit normal vector for $a(s)$ and $kappa$ is the curvature.
Since $b(s)$ is the evolute of $a(s)$, $a(s)$ is the involute of $b(s)$. I showed earlier that this means that $a(s)$ is of the form:
$a(s)= frac{b'(s)}{|b'(s)|}(c-s)+b(s)$
Also:
$a'(s) cdot b'(s)=0$.
where $b'(s)$ is given by:
$b'(s)=a'(s) + frac{N'}{kappa}-frac{Nkappa'}{kappa^2}$,
Anyway, I would appreciate some insight into this. How do i show uniqueness?
So:
$a'(s) cdot b'(s)=a'(s)cdot a'(s)+a'(s) cdot frac{N'}{kappa}- a'(s) cdot frac{Nkappa'}{kappa^2}$
= $1+a'(s) cdot frac{(-kappa a'(s))}{kappa}+0$
= $1-1=0$
so $b(s)$ appears be an evolute of $a(s)$... but how do we show uniqueness? and what gaurentees that $b(s)$ is in the same plane as $a(s)$?
differential-geometry
differential-geometry
asked Jan 22 at 21:10
Math is hardMath is hard
822211
asked Jan 22 at 21:10
Math is hardMath is hard
822211
asked Jan 22 at 21:10
Math is hardMath is hard
822211
asked Jan 22 at 21:10
Math is hardMath is hard
822211
asked Jan 22 at 21:10
Math is hardMath is hard
822211
822211
add a comment |
add a comment |
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