Mixing
Limit of a continuous angle
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Let $gamma :[0,1] to mathbb{R}^2$ be an injective continuous map. Consider point $P$ in $mathbb{R}^2$ outside of the curve $gamma$ and $A,B$ be the two ending points $gamma(0)$, $gamma(1)$ of the curve respectly. We fix point $M$ on the curve somewhere between $A$ and $B$. Now point $C$ on the curve moves continuously from $A$ to $B$ crossing $M$.
What can we say about both left and right limits of $lim_{C to M} angle PCM$ ? (the angle is signed with respect to clockwork orienration).
geometry limits
edited Jan 27 at 0:11
Bungo
13.7k22148
asked Jan 27 at 0:09
MasMMasM
11611
$endgroup$
add a comment |
$begingroup$
Let $gamma :[0,1] to mathbb{R}^2$ be an injective continuous map. Consider point $P$ in $mathbb{R}^2$ outside of the curve $gamma$ and $A,B$ be the two ending points $gamma(0)$, $gamma(1)$ of the curve respectly. We fix point $M$ on the curve somewhere between $A$ and $B$. Now point $C$ on the curve moves continuously from $A$ to $B$ crossing $M$.
What can we say about both left and right limits of $lim_{C to M} angle PCM$ ? (the angle is signed with respect to clockwork orienration).
geometry limits
edited Jan 27 at 0:11
Bungo
13.7k22148
asked Jan 27 at 0:09
MasMMasM
11611
$endgroup$
$begingroup$
I would guess they represent the angle at which the line segment PM crosses the image of $gamma$.
$endgroup$
– Stan Tendijck
Jan 27 at 0:28
add a comment |
$begingroup$
Let $gamma :[0,1] to mathbb{R}^2$ be an injective continuous map. Consider point $P$ in $mathbb{R}^2$ outside of the curve $gamma$ and $A,B$ be the two ending points $gamma(0)$, $gamma(1)$ of the curve respectly. We fix point $M$ on the curve somewhere between $A$ and $B$. Now point $C$ on the curve moves continuously from $A$ to $B$ crossing $M$.
What can we say about both left and right limits of $lim_{C to M} angle PCM$ ? (the angle is signed with respect to clockwork orienration).
geometry limits
edited Jan 27 at 0:11
Bungo
13.7k22148
asked Jan 27 at 0:09
MasMMasM
11611
$endgroup$
Let $gamma :[0,1] to mathbb{R}^2$ be an injective continuous map. Consider point $P$ in $mathbb{R}^2$ outside of the curve $gamma$ and $A,B$ be the two ending points $gamma(0)$, $gamma(1)$ of the curve respectly. We fix point $M$ on the curve somewhere between $A$ and $B$. Now point $C$ on the curve moves continuously from $A$ to $B$ crossing $M$.
What can we say about both left and right limits of $lim_{C to M} angle PCM$ ? (the angle is signed with respect to clockwork orienration).
geometry limits
geometry limits
edited Jan 27 at 0:11
Bungo
13.7k22148
asked Jan 27 at 0:09
MasMMasM
11611
edited Jan 27 at 0:11
Bungo
13.7k22148
asked Jan 27 at 0:09
MasMMasM
11611
edited Jan 27 at 0:11
Bungo
13.7k22148
edited Jan 27 at 0:11
Bungo
13.7k22148
edited Jan 27 at 0:11
Bungo
13.7k22148
13.7k22148
asked Jan 27 at 0:09
MasMMasM
11611
asked Jan 27 at 0:09
MasMMasM
11611
asked Jan 27 at 0:09
MasMMasM
11611
11611
$begingroup$
I would guess they represent the angle at which the line segment PM crosses the image of $gamma$.
$endgroup$
– Stan Tendijck
Jan 27 at 0:28
add a comment |
$begingroup$
I would guess they represent the angle at which the line segment PM crosses the image of $gamma$.
$endgroup$
– Stan Tendijck
Jan 27 at 0:28
$begingroup$
I would guess they represent the angle at which the line segment PM crosses the image of $gamma$.
$endgroup$
– Stan Tendijck
Jan 27 at 0:28
$begingroup$
I would guess they represent the angle at which the line segment PM crosses the image of $gamma$.
$endgroup$
– Stan Tendijck
Jan 27 at 0:28
add a comment |
1 Answer
1
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oldest
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$begingroup$
Line $CM$, as $Cto M$, tends to the line tangent at $M$ to the curve (provided that tangent exists, i.e. if $gamma$ is differentiable). Hence $angle PCM$ tends to the angle between line $PM$ and the tangent at $M$. Of course left and right limit give the two (different) angles formed by such lines.
answered Jan 27 at 9:24
AretinoAretino
25.5k21445
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add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Line $CM$, as $Cto M$, tends to the line tangent at $M$ to the curve (provided that tangent exists, i.e. if $gamma$ is differentiable). Hence $angle PCM$ tends to the angle between line $PM$ and the tangent at $M$. Of course left and right limit give the two (different) angles formed by such lines.
answered Jan 27 at 9:24
AretinoAretino
25.5k21445
$endgroup$
add a comment |
$begingroup$
Line $CM$, as $Cto M$, tends to the line tangent at $M$ to the curve (provided that tangent exists, i.e. if $gamma$ is differentiable). Hence $angle PCM$ tends to the angle between line $PM$ and the tangent at $M$. Of course left and right limit give the two (different) angles formed by such lines.
answered Jan 27 at 9:24
AretinoAretino
25.5k21445
$endgroup$
add a comment |
$begingroup$
Line $CM$, as $Cto M$, tends to the line tangent at $M$ to the curve (provided that tangent exists, i.e. if $gamma$ is differentiable). Hence $angle PCM$ tends to the angle between line $PM$ and the tangent at $M$. Of course left and right limit give the two (different) angles formed by such lines.
answered Jan 27 at 9:24
AretinoAretino
25.5k21445
$endgroup$
Line $CM$, as $Cto M$, tends to the line tangent at $M$ to the curve (provided that tangent exists, i.e. if $gamma$ is differentiable). Hence $angle PCM$ tends to the angle between line $PM$ and the tangent at $M$. Of course left and right limit give the two (different) angles formed by such lines.
answered Jan 27 at 9:24
AretinoAretino
25.5k21445
edited Jan 28 at 11:38
answered Jan 27 at 9:24
AretinoAretino
25.5k21445
answered Jan 27 at 9:24
AretinoAretino
25.5k21445
answered Jan 27 at 9:24
AretinoAretino
25.5k21445
25.5k21445
add a comment |
add a comment |
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$begingroup$
I would guess they represent the angle at which the line segment PM crosses the image of $gamma$.
$endgroup$
– Stan Tendijck
Jan 27 at 0:28