Mixing
Why circle of curvature is defined to be in the inside part of the circle?
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We know that for a curve at a certain point,its circle of curvature is defined to be a circle whose curvature is same as the curve at that point and both the curve and the circle has same tangent at that point.But i couldn't understand why is the circle of curvature is defined to be always in the inside part of the curve? If the circle is on the outside part of the circle(convex part), it still holds the above two criteria? Can anyone clarify my doubt?
calculus multivariable-calculus curvature
asked Jan 26 at 6:51
Soumyajit DebSoumyajit Deb
1
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add a comment |
$begingroup$
We know that for a curve at a certain point,its circle of curvature is defined to be a circle whose curvature is same as the curve at that point and both the curve and the circle has same tangent at that point.But i couldn't understand why is the circle of curvature is defined to be always in the inside part of the curve? If the circle is on the outside part of the circle(convex part), it still holds the above two criteria? Can anyone clarify my doubt?
calculus multivariable-calculus curvature
asked Jan 26 at 6:51
Soumyajit DebSoumyajit Deb
1
$endgroup$
add a comment |
$begingroup$
We know that for a curve at a certain point,its circle of curvature is defined to be a circle whose curvature is same as the curve at that point and both the curve and the circle has same tangent at that point.But i couldn't understand why is the circle of curvature is defined to be always in the inside part of the curve? If the circle is on the outside part of the circle(convex part), it still holds the above two criteria? Can anyone clarify my doubt?
calculus multivariable-calculus curvature
asked Jan 26 at 6:51
Soumyajit DebSoumyajit Deb
1
$endgroup$
We know that for a curve at a certain point,its circle of curvature is defined to be a circle whose curvature is same as the curve at that point and both the curve and the circle has same tangent at that point.But i couldn't understand why is the circle of curvature is defined to be always in the inside part of the curve? If the circle is on the outside part of the circle(convex part), it still holds the above two criteria? Can anyone clarify my doubt?
calculus multivariable-calculus curvature
calculus multivariable-calculus curvature
asked Jan 26 at 6:51
Soumyajit DebSoumyajit Deb
1
asked Jan 26 at 6:51
Soumyajit DebSoumyajit Deb
1
asked Jan 26 at 6:51
Soumyajit DebSoumyajit Deb
1
asked Jan 26 at 6:51
Soumyajit DebSoumyajit Deb
1
asked Jan 26 at 6:51
Soumyajit DebSoumyajit Deb
1
1
add a comment |
add a comment |
1 Answer
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Your definition is incomplete. The osculating circle has its center at the intersection of infinitesimally close normals drawn at points on both sides of the given point, and the normals intersect on the inside. Yet another way to define the osculating circle is by taking two more points on the curve on either side of the given point and consider the limit position of the (unique) circle through the three of them. In this way you obviously get a circle on the inside of the curve.
answered Jan 26 at 7:15
GReyesGReyes
2,22315
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$begingroup$
what you said is true but can you clarify it more?
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– Soumyajit Deb
Jan 26 at 9:22
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your definition is incomplete. The osculating circle has its center at the intersection of infinitesimally close normals drawn at points on both sides of the given point, and the normals intersect on the inside. Yet another way to define the osculating circle is by taking two more points on the curve on either side of the given point and consider the limit position of the (unique) circle through the three of them. In this way you obviously get a circle on the inside of the curve.
answered Jan 26 at 7:15
GReyesGReyes
2,22315
$endgroup$
$begingroup$
what you said is true but can you clarify it more?
$endgroup$
– Soumyajit Deb
Jan 26 at 9:22
add a comment |
$begingroup$
Your definition is incomplete. The osculating circle has its center at the intersection of infinitesimally close normals drawn at points on both sides of the given point, and the normals intersect on the inside. Yet another way to define the osculating circle is by taking two more points on the curve on either side of the given point and consider the limit position of the (unique) circle through the three of them. In this way you obviously get a circle on the inside of the curve.
answered Jan 26 at 7:15
GReyesGReyes
2,22315
$endgroup$
$begingroup$
what you said is true but can you clarify it more?
$endgroup$
– Soumyajit Deb
Jan 26 at 9:22
add a comment |
$begingroup$
Your definition is incomplete. The osculating circle has its center at the intersection of infinitesimally close normals drawn at points on both sides of the given point, and the normals intersect on the inside. Yet another way to define the osculating circle is by taking two more points on the curve on either side of the given point and consider the limit position of the (unique) circle through the three of them. In this way you obviously get a circle on the inside of the curve.
answered Jan 26 at 7:15
GReyesGReyes
2,22315
$endgroup$
Your definition is incomplete. The osculating circle has its center at the intersection of infinitesimally close normals drawn at points on both sides of the given point, and the normals intersect on the inside. Yet another way to define the osculating circle is by taking two more points on the curve on either side of the given point and consider the limit position of the (unique) circle through the three of them. In this way you obviously get a circle on the inside of the curve.
answered Jan 26 at 7:15
GReyesGReyes
2,22315
answered Jan 26 at 7:15
GReyesGReyes
2,22315
answered Jan 26 at 7:15
GReyesGReyes
2,22315
answered Jan 26 at 7:15
GReyesGReyes
2,22315
2,22315
$begingroup$
what you said is true but can you clarify it more?
$endgroup$
– Soumyajit Deb
Jan 26 at 9:22
add a comment |
$begingroup$
what you said is true but can you clarify it more?
$endgroup$
– Soumyajit Deb
Jan 26 at 9:22
$begingroup$
what you said is true but can you clarify it more?
$endgroup$
– Soumyajit Deb
Jan 26 at 9:22
$begingroup$
what you said is true but can you clarify it more?
$endgroup$
– Soumyajit Deb
Jan 26 at 9:22
add a comment |
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