Mixing
Understanding geometric centroids
Assume I have an arbitrary closed parametric curve $r(t) = <X(t), Y(t)>$ and density function 1 for the area inside of that curve.
Wolfram defines the geometric centroid to be:
$$(bar x, bar y) = bigg(frac{intint xdA}{M}, frac{intint ydA}{M}bigg)$$
For an object with constant density function. Or put into simple english, the $bar x$ centroid coordinate is the average of the x's and the $bar y$ coordinate is the average of the y's
So that would mean that the centroid of the curve $r$ is $$c = bigg<frac{int X(t)dt}{M}, frac{int Y(t)dt}{M}bigg>$$
And $M = text{arclength}(r)$
Is my reasoning correct?
calculus geometry multivariable-calculus differential-geometry
asked Nov 21 '18 at 15:29
Makogan
751217
add a comment |
Assume I have an arbitrary closed parametric curve $r(t) = <X(t), Y(t)>$ and density function 1 for the area inside of that curve.
Wolfram defines the geometric centroid to be:
$$(bar x, bar y) = bigg(frac{intint xdA}{M}, frac{intint ydA}{M}bigg)$$
For an object with constant density function. Or put into simple english, the $bar x$ centroid coordinate is the average of the x's and the $bar y$ coordinate is the average of the y's
So that would mean that the centroid of the curve $r$ is $$c = bigg<frac{int X(t)dt}{M}, frac{int Y(t)dt}{M}bigg>$$
And $M = text{arclength}(r)$
Is my reasoning correct?
calculus geometry multivariable-calculus differential-geometry
asked Nov 21 '18 at 15:29
Makogan
751217
Sounds right to me....
– Mostafa Ayaz
Nov 21 '18 at 17:20
add a comment |
Assume I have an arbitrary closed parametric curve $r(t) = <X(t), Y(t)>$ and density function 1 for the area inside of that curve.
Wolfram defines the geometric centroid to be:
$$(bar x, bar y) = bigg(frac{intint xdA}{M}, frac{intint ydA}{M}bigg)$$
For an object with constant density function. Or put into simple english, the $bar x$ centroid coordinate is the average of the x's and the $bar y$ coordinate is the average of the y's
So that would mean that the centroid of the curve $r$ is $$c = bigg<frac{int X(t)dt}{M}, frac{int Y(t)dt}{M}bigg>$$
And $M = text{arclength}(r)$
Is my reasoning correct?
calculus geometry multivariable-calculus differential-geometry
asked Nov 21 '18 at 15:29
Makogan
751217
Assume I have an arbitrary closed parametric curve $r(t) = <X(t), Y(t)>$ and density function 1 for the area inside of that curve.
Wolfram defines the geometric centroid to be:
$$(bar x, bar y) = bigg(frac{intint xdA}{M}, frac{intint ydA}{M}bigg)$$
For an object with constant density function. Or put into simple english, the $bar x$ centroid coordinate is the average of the x's and the $bar y$ coordinate is the average of the y's
So that would mean that the centroid of the curve $r$ is $$c = bigg<frac{int X(t)dt}{M}, frac{int Y(t)dt}{M}bigg>$$
And $M = text{arclength}(r)$
Is my reasoning correct?
calculus geometry multivariable-calculus differential-geometry
calculus geometry multivariable-calculus differential-geometry
asked Nov 21 '18 at 15:29
Makogan
751217
asked Nov 21 '18 at 15:29
Makogan
751217
asked Nov 21 '18 at 15:29
Makogan
751217
asked Nov 21 '18 at 15:29
Makogan
751217
asked Nov 21 '18 at 15:29
Makogan
751217
751217
Sounds right to me....
– Mostafa Ayaz
Nov 21 '18 at 17:20
add a comment |
Sounds right to me....
– Mostafa Ayaz
Nov 21 '18 at 17:20
Sounds right to me....
– Mostafa Ayaz
Nov 21 '18 at 17:20
Sounds right to me....
– Mostafa Ayaz
Nov 21 '18 at 17:20
add a comment |
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Sounds right to me....
– Mostafa Ayaz
Nov 21 '18 at 17:20