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Prove that the planar curve obtained by projecting $alpha$ into its osculating plane at $P$ has the same...
$begingroup$
Let $alpha(s)$ be a regular curve with $kappaneq0$ at P, where $kappa$ is the curvature. Prove that the planar curve obtained by projecting $alpha(s)$ into its osculating plane at $P$ has the same curvature at $P$ as $alpha(s)$.
(This is problem 1.2.14 in Ted Shifrin's differential geometry notes, (http://alpha.math.uga.edu/~shifrin/ShifrinDiffGeo.pdf).)
The local projection onto the osculating plane is given by $(s,frac{s^2kappa}{2})$. However, i'm not sure if this equation holds only for when curves are parametrized by arc length.
Anyway, lets say that this projection is the correct projection for $alpha(s)$ even though it's not stated to be parameterized by arc length.
The curvature is given by $frac{|alpha'(s) times alpha''(s)|}{alpha'(s)}$.
Let $gamma=(s,frac{s^2kappa}{2})$.
$rightarrow$ $gamma'=(1,skappa+frac{s^2kappa'}{2})$
$rightarrow$ $gamma''=(0,kappa+skappa'+skappa'+frac{s^2kappa''}{2})$
$rightarrow$ $|gamma'|=(1^2+(sk)^2+(frac{s^2kappa'}{2})^2+s^3kappakappa')^{1/2}$
So, if I apply the equation for curvature to $gamma$, in theory I should get $kappa$. Is this reasoning correct? Thanks!
differential-geometry
edited Jan 22 at 3:42
Sarah T.
1933
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add a comment |
$begingroup$
Let $alpha(s)$ be a regular curve with $kappaneq0$ at P, where $kappa$ is the curvature. Prove that the planar curve obtained by projecting $alpha(s)$ into its osculating plane at $P$ has the same curvature at $P$ as $alpha(s)$.
(This is problem 1.2.14 in Ted Shifrin's differential geometry notes, (http://alpha.math.uga.edu/~shifrin/ShifrinDiffGeo.pdf).)
The local projection onto the osculating plane is given by $(s,frac{s^2kappa}{2})$. However, i'm not sure if this equation holds only for when curves are parametrized by arc length.
Anyway, lets say that this projection is the correct projection for $alpha(s)$ even though it's not stated to be parameterized by arc length.
The curvature is given by $frac{|alpha'(s) times alpha''(s)|}{alpha'(s)}$.
Let $gamma=(s,frac{s^2kappa}{2})$.
$rightarrow$ $gamma'=(1,skappa+frac{s^2kappa'}{2})$
$rightarrow$ $gamma''=(0,kappa+skappa'+skappa'+frac{s^2kappa''}{2})$
$rightarrow$ $|gamma'|=(1^2+(sk)^2+(frac{s^2kappa'}{2})^2+s^3kappakappa')^{1/2}$
So, if I apply the equation for curvature to $gamma$, in theory I should get $kappa$. Is this reasoning correct? Thanks!
differential-geometry
edited Jan 22 at 3:42
Sarah T.
1933
$endgroup$
$begingroup$
Where you say that the local projection into the osculating plane is given by $left(s, frac{kappa}{2}s^2 right)$, you should actually have the parameterization $(s + O(s^3), frac{kappa_0}{2}s^2 + O(s^3)$, where $kappa_0$ is the curvature of $s$ at $P$. In particular, $kappa_0$ is a constant and should not be differentiated. This will make the rest of the computations a lot simpler.
$endgroup$
– Sarah T.
Jan 22 at 1:36
$begingroup$
Incidentally, this is recognizably a problem from Ted Shifrin's differential geometry notes. I've edited to add a reference.
$endgroup$
– Sarah T.
Jan 22 at 1:41
$begingroup$
Okay cool. But since the parameterization is local, shouldn't we be able to ignore the higher order terms?
$endgroup$
– user624065
Jan 22 at 1:41
$begingroup$
Yes, higher order terms can be ignored.
$endgroup$
– Sarah T.
Jan 22 at 1:44
$begingroup$
Okay, if $gamma(s)=(s,frac{ks^2}{2})$ then I got that $gamma'(s) times gamma''(s)=kappa$. However, the denominator $|gamma'(s)|$^3 is not behaving well... Is there a reason that $gamma$ is arc length parameterized because $alpha(s)$ is? Otherwise, the donominator will be $1+3ks+3(ks)^2+(ks)^3$
$endgroup$
– user624065
Jan 22 at 21:28
add a comment |
$begingroup$
Let $alpha(s)$ be a regular curve with $kappaneq0$ at P, where $kappa$ is the curvature. Prove that the planar curve obtained by projecting $alpha(s)$ into its osculating plane at $P$ has the same curvature at $P$ as $alpha(s)$.
(This is problem 1.2.14 in Ted Shifrin's differential geometry notes, (http://alpha.math.uga.edu/~shifrin/ShifrinDiffGeo.pdf).)
The local projection onto the osculating plane is given by $(s,frac{s^2kappa}{2})$. However, i'm not sure if this equation holds only for when curves are parametrized by arc length.
Anyway, lets say that this projection is the correct projection for $alpha(s)$ even though it's not stated to be parameterized by arc length.
The curvature is given by $frac{|alpha'(s) times alpha''(s)|}{alpha'(s)}$.
Let $gamma=(s,frac{s^2kappa}{2})$.
$rightarrow$ $gamma'=(1,skappa+frac{s^2kappa'}{2})$
$rightarrow$ $gamma''=(0,kappa+skappa'+skappa'+frac{s^2kappa''}{2})$
$rightarrow$ $|gamma'|=(1^2+(sk)^2+(frac{s^2kappa'}{2})^2+s^3kappakappa')^{1/2}$
So, if I apply the equation for curvature to $gamma$, in theory I should get $kappa$. Is this reasoning correct? Thanks!
differential-geometry
edited Jan 22 at 3:42
Sarah T.
1933
$endgroup$
Let $alpha(s)$ be a regular curve with $kappaneq0$ at P, where $kappa$ is the curvature. Prove that the planar curve obtained by projecting $alpha(s)$ into its osculating plane at $P$ has the same curvature at $P$ as $alpha(s)$.
(This is problem 1.2.14 in Ted Shifrin's differential geometry notes, (http://alpha.math.uga.edu/~shifrin/ShifrinDiffGeo.pdf).)
The local projection onto the osculating plane is given by $(s,frac{s^2kappa}{2})$. However, i'm not sure if this equation holds only for when curves are parametrized by arc length.
Anyway, lets say that this projection is the correct projection for $alpha(s)$ even though it's not stated to be parameterized by arc length.
The curvature is given by $frac{|alpha'(s) times alpha''(s)|}{alpha'(s)}$.
Let $gamma=(s,frac{s^2kappa}{2})$.
$rightarrow$ $gamma'=(1,skappa+frac{s^2kappa'}{2})$
$rightarrow$ $gamma''=(0,kappa+skappa'+skappa'+frac{s^2kappa''}{2})$
$rightarrow$ $|gamma'|=(1^2+(sk)^2+(frac{s^2kappa'}{2})^2+s^3kappakappa')^{1/2}$
So, if I apply the equation for curvature to $gamma$, in theory I should get $kappa$. Is this reasoning correct? Thanks!
differential-geometry
differential-geometry
edited Jan 22 at 3:42
Sarah T.
1933
edited Jan 22 at 3:42
Sarah T.
1933
edited Jan 22 at 3:42
Sarah T.
1933
edited Jan 22 at 3:42
Sarah T.
1933
edited Jan 22 at 3:42
Sarah T.
1933
1933
asked Jan 22 at 1:22
user624065
$begingroup$
Where you say that the local projection into the osculating plane is given by $left(s, frac{kappa}{2}s^2 right)$, you should actually have the parameterization $(s + O(s^3), frac{kappa_0}{2}s^2 + O(s^3)$, where $kappa_0$ is the curvature of $s$ at $P$. In particular, $kappa_0$ is a constant and should not be differentiated. This will make the rest of the computations a lot simpler.
$endgroup$
– Sarah T.
Jan 22 at 1:36
$begingroup$
Incidentally, this is recognizably a problem from Ted Shifrin's differential geometry notes. I've edited to add a reference.
$endgroup$
– Sarah T.
Jan 22 at 1:41
$begingroup$
Okay cool. But since the parameterization is local, shouldn't we be able to ignore the higher order terms?
$endgroup$
– user624065
Jan 22 at 1:41
$begingroup$
Yes, higher order terms can be ignored.
$endgroup$
– Sarah T.
Jan 22 at 1:44
$begingroup$
Okay, if $gamma(s)=(s,frac{ks^2}{2})$ then I got that $gamma'(s) times gamma''(s)=kappa$. However, the denominator $|gamma'(s)|$^3 is not behaving well... Is there a reason that $gamma$ is arc length parameterized because $alpha(s)$ is? Otherwise, the donominator will be $1+3ks+3(ks)^2+(ks)^3$
$endgroup$
– user624065
Jan 22 at 21:28
add a comment |
$begingroup$
Where you say that the local projection into the osculating plane is given by $left(s, frac{kappa}{2}s^2 right)$, you should actually have the parameterization $(s + O(s^3), frac{kappa_0}{2}s^2 + O(s^3)$, where $kappa_0$ is the curvature of $s$ at $P$. In particular, $kappa_0$ is a constant and should not be differentiated. This will make the rest of the computations a lot simpler.
$endgroup$
– Sarah T.
Jan 22 at 1:36
$begingroup$
Incidentally, this is recognizably a problem from Ted Shifrin's differential geometry notes. I've edited to add a reference.
$endgroup$
– Sarah T.
Jan 22 at 1:41
$begingroup$
Okay cool. But since the parameterization is local, shouldn't we be able to ignore the higher order terms?
$endgroup$
– user624065
Jan 22 at 1:41
$begingroup$
Yes, higher order terms can be ignored.
$endgroup$
– Sarah T.
Jan 22 at 1:44
$begingroup$
Okay, if $gamma(s)=(s,frac{ks^2}{2})$ then I got that $gamma'(s) times gamma''(s)=kappa$. However, the denominator $|gamma'(s)|$^3 is not behaving well... Is there a reason that $gamma$ is arc length parameterized because $alpha(s)$ is? Otherwise, the donominator will be $1+3ks+3(ks)^2+(ks)^3$
$endgroup$
– user624065
Jan 22 at 21:28
$begingroup$
Where you say that the local projection into the osculating plane is given by $left(s, frac{kappa}{2}s^2 right)$, you should actually have the parameterization $(s + O(s^3), frac{kappa_0}{2}s^2 + O(s^3)$, where $kappa_0$ is the curvature of $s$ at $P$. In particular, $kappa_0$ is a constant and should not be differentiated. This will make the rest of the computations a lot simpler.
$endgroup$
– Sarah T.
Jan 22 at 1:36
$begingroup$
Where you say that the local projection into the osculating plane is given by $left(s, frac{kappa}{2}s^2 right)$, you should actually have the parameterization $(s + O(s^3), frac{kappa_0}{2}s^2 + O(s^3)$, where $kappa_0$ is the curvature of $s$ at $P$. In particular, $kappa_0$ is a constant and should not be differentiated. This will make the rest of the computations a lot simpler.
$endgroup$
– Sarah T.
Jan 22 at 1:36
$begingroup$
Incidentally, this is recognizably a problem from Ted Shifrin's differential geometry notes. I've edited to add a reference.
$endgroup$
– Sarah T.
Jan 22 at 1:41
$begingroup$
Incidentally, this is recognizably a problem from Ted Shifrin's differential geometry notes. I've edited to add a reference.
$endgroup$
– Sarah T.
Jan 22 at 1:41
$begingroup$
Okay cool. But since the parameterization is local, shouldn't we be able to ignore the higher order terms?
$endgroup$
– user624065
Jan 22 at 1:41
$begingroup$
Okay cool. But since the parameterization is local, shouldn't we be able to ignore the higher order terms?
$endgroup$
– user624065
Jan 22 at 1:41
$begingroup$
Yes, higher order terms can be ignored.
$endgroup$
– Sarah T.
Jan 22 at 1:44
$begingroup$
Yes, higher order terms can be ignored.
$endgroup$
– Sarah T.
Jan 22 at 1:44
$begingroup$
Okay, if $gamma(s)=(s,frac{ks^2}{2})$ then I got that $gamma'(s) times gamma''(s)=kappa$. However, the denominator $|gamma'(s)|$^3 is not behaving well... Is there a reason that $gamma$ is arc length parameterized because $alpha(s)$ is? Otherwise, the donominator will be $1+3ks+3(ks)^2+(ks)^3$
$endgroup$
– user624065
Jan 22 at 21:28
$begingroup$
Okay, if $gamma(s)=(s,frac{ks^2}{2})$ then I got that $gamma'(s) times gamma''(s)=kappa$. However, the denominator $|gamma'(s)|$^3 is not behaving well... Is there a reason that $gamma$ is arc length parameterized because $alpha(s)$ is? Otherwise, the donominator will be $1+3ks+3(ks)^2+(ks)^3$
$endgroup$
– user624065
Jan 22 at 21:28
add a comment |
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$begingroup$
Where you say that the local projection into the osculating plane is given by $left(s, frac{kappa}{2}s^2 right)$, you should actually have the parameterization $(s + O(s^3), frac{kappa_0}{2}s^2 + O(s^3)$, where $kappa_0$ is the curvature of $s$ at $P$. In particular, $kappa_0$ is a constant and should not be differentiated. This will make the rest of the computations a lot simpler.
$endgroup$
– Sarah T.
Jan 22 at 1:36
$begingroup$
Incidentally, this is recognizably a problem from Ted Shifrin's differential geometry notes. I've edited to add a reference.
$endgroup$
– Sarah T.
Jan 22 at 1:41
$begingroup$
Okay cool. But since the parameterization is local, shouldn't we be able to ignore the higher order terms?
$endgroup$
– user624065
Jan 22 at 1:41
$begingroup$
Yes, higher order terms can be ignored.
$endgroup$
– Sarah T.
Jan 22 at 1:44
$begingroup$
Okay, if $gamma(s)=(s,frac{ks^2}{2})$ then I got that $gamma'(s) times gamma''(s)=kappa$. However, the denominator $|gamma'(s)|$^3 is not behaving well... Is there a reason that $gamma$ is arc length parameterized because $alpha(s)$ is? Otherwise, the donominator will be $1+3ks+3(ks)^2+(ks)^3$
$endgroup$
– user624065
Jan 22 at 21:28