Quick ways of finding principle curvature in the Cartesian plane












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Say I have some polar curve $r=f(theta)$ in the Cartesian plane (smooth and twice-differential) and I've found a formula for the curvature, $k$. Does a clever trick exist (that isn't merely finding the maxima and minima of $k$) that allows us to find the principle curvature of our curve?



What about the mean curvature?










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  • 1




    $begingroup$
    What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
    $endgroup$
    – Vasily Mitch
    Dec 17 '18 at 21:04










  • $begingroup$
    I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
    $endgroup$
    – Jonathan Low
    Dec 17 '18 at 21:46










  • $begingroup$
    However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
    $endgroup$
    – Jonathan Low
    Dec 17 '18 at 21:47
















1












$begingroup$


Say I have some polar curve $r=f(theta)$ in the Cartesian plane (smooth and twice-differential) and I've found a formula for the curvature, $k$. Does a clever trick exist (that isn't merely finding the maxima and minima of $k$) that allows us to find the principle curvature of our curve?



What about the mean curvature?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
    $endgroup$
    – Vasily Mitch
    Dec 17 '18 at 21:04










  • $begingroup$
    I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
    $endgroup$
    – Jonathan Low
    Dec 17 '18 at 21:46










  • $begingroup$
    However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
    $endgroup$
    – Jonathan Low
    Dec 17 '18 at 21:47














1












1








1





$begingroup$


Say I have some polar curve $r=f(theta)$ in the Cartesian plane (smooth and twice-differential) and I've found a formula for the curvature, $k$. Does a clever trick exist (that isn't merely finding the maxima and minima of $k$) that allows us to find the principle curvature of our curve?



What about the mean curvature?










share|cite|improve this question









$endgroup$




Say I have some polar curve $r=f(theta)$ in the Cartesian plane (smooth and twice-differential) and I've found a formula for the curvature, $k$. Does a clever trick exist (that isn't merely finding the maxima and minima of $k$) that allows us to find the principle curvature of our curve?



What about the mean curvature?







polar-coordinates curvature






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 21:01









Jonathan LowJonathan Low

635




635








  • 1




    $begingroup$
    What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
    $endgroup$
    – Vasily Mitch
    Dec 17 '18 at 21:04










  • $begingroup$
    I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
    $endgroup$
    – Jonathan Low
    Dec 17 '18 at 21:46










  • $begingroup$
    However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
    $endgroup$
    – Jonathan Low
    Dec 17 '18 at 21:47














  • 1




    $begingroup$
    What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
    $endgroup$
    – Vasily Mitch
    Dec 17 '18 at 21:04










  • $begingroup$
    I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
    $endgroup$
    – Jonathan Low
    Dec 17 '18 at 21:46










  • $begingroup$
    However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
    $endgroup$
    – Jonathan Low
    Dec 17 '18 at 21:47








1




1




$begingroup$
What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
$endgroup$
– Vasily Mitch
Dec 17 '18 at 21:04




$begingroup$
What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
$endgroup$
– Vasily Mitch
Dec 17 '18 at 21:04












$begingroup$
I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:46




$begingroup$
I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:46












$begingroup$
However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:47




$begingroup$
However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:47










1 Answer
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$begingroup$

Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.



Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
So mean curvature is $k_a=2(pi/2)/lapprox0.41$






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    1 Answer
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    $begingroup$

    Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.



    Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
    So mean curvature is $k_a=2(pi/2)/lapprox0.41$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.



      Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
      So mean curvature is $k_a=2(pi/2)/lapprox0.41$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.



        Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
        So mean curvature is $k_a=2(pi/2)/lapprox0.41$






        share|cite|improve this answer









        $endgroup$



        Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.



        Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
        So mean curvature is $k_a=2(pi/2)/lapprox0.41$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 22:19









        Vasily MitchVasily Mitch

        2,6791312




        2,6791312






























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