Mixing
Best approximation of arc length of curve with only two line segments
$begingroup$
I know that arc length can be approximated with a certain number of line segments, as shown in this picture. http://en.wikipedia.org/wiki/File:Arclength.svg
The sum of the lengths of the line segments will approach the arc length of the curve as we approximate it with very large numbers of them.
But how do we minimize the error of the approximation -- that is, where should the stop and start points of the line segments be? I'm only interested in the situation where we're only limited to two line segments; three or more sounds complicated. Can we do the same thing for a curve in three dimensions?
...Intuitively, (and I really can't explain why I feel this is correct), I seem to think that the ideal approximation occurs when the total area enclosed by the curve and the approximative line segments is minimized. If this really is correct, why is it the case?
(This question seems related to mine, though it also seems more complicated: how to choose point spacing to approximate a parametric curve using line segments?)
calculus geometry multivariable-calculus
edited Apr 13 '17 at 12:19
Community♦
1
asked Nov 1 '14 at 3:57
Drew N Drew N
1459
$endgroup$
add a comment |
$begingroup$
I know that arc length can be approximated with a certain number of line segments, as shown in this picture. http://en.wikipedia.org/wiki/File:Arclength.svg
The sum of the lengths of the line segments will approach the arc length of the curve as we approximate it with very large numbers of them.
But how do we minimize the error of the approximation -- that is, where should the stop and start points of the line segments be? I'm only interested in the situation where we're only limited to two line segments; three or more sounds complicated. Can we do the same thing for a curve in three dimensions?
...Intuitively, (and I really can't explain why I feel this is correct), I seem to think that the ideal approximation occurs when the total area enclosed by the curve and the approximative line segments is minimized. If this really is correct, why is it the case?
(This question seems related to mine, though it also seems more complicated: how to choose point spacing to approximate a parametric curve using line segments?)
calculus geometry multivariable-calculus
edited Apr 13 '17 at 12:19
Community♦
1
asked Nov 1 '14 at 3:57
Drew N Drew N
1459
$endgroup$
$begingroup$
If you are expecting general solutions, think about function $f(x)=xsin(1/x)$.
$endgroup$
– Przemysław Scherwentke
Nov 1 '14 at 4:08
$begingroup$
Well...should I be thinking that it has infinite arc length, over any interval that includes 0? I can't quite convince myself of that because of the x in front of the sin(1/x), and I'm not sure how I would. Maybe an alternating series of some sort. Why didn't you just use plain old f(x) = sin(1/x)?
$endgroup$
– Drew N
Nov 2 '14 at 4:39
$begingroup$
The arc lenghth is infinite, beacause harmonic series is divergent, and the lenght is greater than the sum of heights of extrema. This function is better than $sin(1/x)$ because it is een continuous everywhere. But in both cases choosing an aproximative line segments seems to be hopeless.
$endgroup$
– Przemysław Scherwentke
Nov 2 '14 at 4:50
add a comment |
$begingroup$
I know that arc length can be approximated with a certain number of line segments, as shown in this picture. http://en.wikipedia.org/wiki/File:Arclength.svg
The sum of the lengths of the line segments will approach the arc length of the curve as we approximate it with very large numbers of them.
But how do we minimize the error of the approximation -- that is, where should the stop and start points of the line segments be? I'm only interested in the situation where we're only limited to two line segments; three or more sounds complicated. Can we do the same thing for a curve in three dimensions?
...Intuitively, (and I really can't explain why I feel this is correct), I seem to think that the ideal approximation occurs when the total area enclosed by the curve and the approximative line segments is minimized. If this really is correct, why is it the case?
(This question seems related to mine, though it also seems more complicated: how to choose point spacing to approximate a parametric curve using line segments?)
calculus geometry multivariable-calculus
edited Apr 13 '17 at 12:19
Community♦
1
asked Nov 1 '14 at 3:57
Drew N Drew N
1459
$endgroup$
I know that arc length can be approximated with a certain number of line segments, as shown in this picture. http://en.wikipedia.org/wiki/File:Arclength.svg
The sum of the lengths of the line segments will approach the arc length of the curve as we approximate it with very large numbers of them.
But how do we minimize the error of the approximation -- that is, where should the stop and start points of the line segments be? I'm only interested in the situation where we're only limited to two line segments; three or more sounds complicated. Can we do the same thing for a curve in three dimensions?
...Intuitively, (and I really can't explain why I feel this is correct), I seem to think that the ideal approximation occurs when the total area enclosed by the curve and the approximative line segments is minimized. If this really is correct, why is it the case?
(This question seems related to mine, though it also seems more complicated: how to choose point spacing to approximate a parametric curve using line segments?)
calculus geometry multivariable-calculus
calculus geometry multivariable-calculus
edited Apr 13 '17 at 12:19
Community♦
1
asked Nov 1 '14 at 3:57
Drew N Drew N
1459
edited Apr 13 '17 at 12:19
Community♦
1
asked Nov 1 '14 at 3:57
Drew N Drew N
1459
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Nov 1 '14 at 3:57
Drew N Drew N
1459
asked Nov 1 '14 at 3:57
Drew N Drew N
1459
asked Nov 1 '14 at 3:57
Drew N Drew N
1459
1459
$begingroup$
If you are expecting general solutions, think about function $f(x)=xsin(1/x)$.
$endgroup$
– Przemysław Scherwentke
Nov 1 '14 at 4:08
$begingroup$
Well...should I be thinking that it has infinite arc length, over any interval that includes 0? I can't quite convince myself of that because of the x in front of the sin(1/x), and I'm not sure how I would. Maybe an alternating series of some sort. Why didn't you just use plain old f(x) = sin(1/x)?
$endgroup$
– Drew N
Nov 2 '14 at 4:39
$begingroup$
The arc lenghth is infinite, beacause harmonic series is divergent, and the lenght is greater than the sum of heights of extrema. This function is better than $sin(1/x)$ because it is een continuous everywhere. But in both cases choosing an aproximative line segments seems to be hopeless.
$endgroup$
– Przemysław Scherwentke
Nov 2 '14 at 4:50
add a comment |
$begingroup$
If you are expecting general solutions, think about function $f(x)=xsin(1/x)$.
$endgroup$
– Przemysław Scherwentke
Nov 1 '14 at 4:08
$begingroup$
Well...should I be thinking that it has infinite arc length, over any interval that includes 0? I can't quite convince myself of that because of the x in front of the sin(1/x), and I'm not sure how I would. Maybe an alternating series of some sort. Why didn't you just use plain old f(x) = sin(1/x)?
$endgroup$
– Drew N
Nov 2 '14 at 4:39
$begingroup$
The arc lenghth is infinite, beacause harmonic series is divergent, and the lenght is greater than the sum of heights of extrema. This function is better than $sin(1/x)$ because it is een continuous everywhere. But in both cases choosing an aproximative line segments seems to be hopeless.
$endgroup$
– Przemysław Scherwentke
Nov 2 '14 at 4:50
$begingroup$
If you are expecting general solutions, think about function $f(x)=xsin(1/x)$.
$endgroup$
– Przemysław Scherwentke
Nov 1 '14 at 4:08
$begingroup$
If you are expecting general solutions, think about function $f(x)=xsin(1/x)$.
$endgroup$
– Przemysław Scherwentke
Nov 1 '14 at 4:08
$begingroup$
Well...should I be thinking that it has infinite arc length, over any interval that includes 0? I can't quite convince myself of that because of the x in front of the sin(1/x), and I'm not sure how I would. Maybe an alternating series of some sort. Why didn't you just use plain old f(x) = sin(1/x)?
$endgroup$
– Drew N
Nov 2 '14 at 4:39
$begingroup$
Well...should I be thinking that it has infinite arc length, over any interval that includes 0? I can't quite convince myself of that because of the x in front of the sin(1/x), and I'm not sure how I would. Maybe an alternating series of some sort. Why didn't you just use plain old f(x) = sin(1/x)?
$endgroup$
– Drew N
Nov 2 '14 at 4:39
$begingroup$
The arc lenghth is infinite, beacause harmonic series is divergent, and the lenght is greater than the sum of heights of extrema. This function is better than $sin(1/x)$ because it is een continuous everywhere. But in both cases choosing an aproximative line segments seems to be hopeless.
$endgroup$
– Przemysław Scherwentke
Nov 2 '14 at 4:50
$begingroup$
The arc lenghth is infinite, beacause harmonic series is divergent, and the lenght is greater than the sum of heights of extrema. This function is better than $sin(1/x)$ because it is een continuous everywhere. But in both cases choosing an aproximative line segments seems to be hopeless.
$endgroup$
– Przemysław Scherwentke
Nov 2 '14 at 4:50
add a comment |
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$begingroup$
If you are expecting general solutions, think about function $f(x)=xsin(1/x)$.
$endgroup$
– Przemysław Scherwentke
Nov 1 '14 at 4:08
$begingroup$
Well...should I be thinking that it has infinite arc length, over any interval that includes 0? I can't quite convince myself of that because of the x in front of the sin(1/x), and I'm not sure how I would. Maybe an alternating series of some sort. Why didn't you just use plain old f(x) = sin(1/x)?
$endgroup$
– Drew N
Nov 2 '14 at 4:39
$begingroup$
The arc lenghth is infinite, beacause harmonic series is divergent, and the lenght is greater than the sum of heights of extrema. This function is better than $sin(1/x)$ because it is een continuous everywhere. But in both cases choosing an aproximative line segments seems to be hopeless.
$endgroup$
– Przemysław Scherwentke
Nov 2 '14 at 4:50