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Relation between area and perimeter of an ellipse in terms of semi-major and semi-minor axes.
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We know the perimeter of a circunference (in terms of its radius $r$), $2 pi r$, can be obtained by differentiating its area (with respect to its radius), $pi r^2$. This fact generalizes to higher dimensions.
Using partial derivatives, is there something similar if we consider:
i) an ellipse ($fracx^2a^2 + fracy^2b^2 =1$) instead of a circunference and
ii) its semi-major and semi-minor axes $a$ and $b$ instead of the radius $r$?
Does it generalize to higher dimensions?
geometry multivariable-calculus differential-geometry
asked Aug 2 at 18:57
Lucas Amorim
576
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up vote
2
down vote
favorite
We know the perimeter of a circunference (in terms of its radius $r$), $2 pi r$, can be obtained by differentiating its area (with respect to its radius), $pi r^2$. This fact generalizes to higher dimensions.
Using partial derivatives, is there something similar if we consider:
i) an ellipse ($fracx^2a^2 + fracy^2b^2 =1$) instead of a circunference and
ii) its semi-major and semi-minor axes $a$ and $b$ instead of the radius $r$?
Does it generalize to higher dimensions?
geometry multivariable-calculus differential-geometry
asked Aug 2 at 18:57
Lucas Amorim
576
Maybe not because the area of an ellipse has closed form $pi a b$ while the perimeter doesn't? I'm not used to elliptic integrals enough to be definitive on this guess. But it this all there is to it? This end is somehow disappointing.
– Lucas Amorim
Aug 2 at 19:12
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
We know the perimeter of a circunference (in terms of its radius $r$), $2 pi r$, can be obtained by differentiating its area (with respect to its radius), $pi r^2$. This fact generalizes to higher dimensions.
Using partial derivatives, is there something similar if we consider:
i) an ellipse ($fracx^2a^2 + fracy^2b^2 =1$) instead of a circunference and
ii) its semi-major and semi-minor axes $a$ and $b$ instead of the radius $r$?
Does it generalize to higher dimensions?
geometry multivariable-calculus differential-geometry
asked Aug 2 at 18:57
Lucas Amorim
576
We know the perimeter of a circunference (in terms of its radius $r$), $2 pi r$, can be obtained by differentiating its area (with respect to its radius), $pi r^2$. This fact generalizes to higher dimensions.
Using partial derivatives, is there something similar if we consider:
i) an ellipse ($fracx^2a^2 + fracy^2b^2 =1$) instead of a circunference and
ii) its semi-major and semi-minor axes $a$ and $b$ instead of the radius $r$?
Does it generalize to higher dimensions?
geometry multivariable-calculus differential-geometry
asked Aug 2 at 18:57
Lucas Amorim
576
asked Aug 2 at 18:57
Lucas Amorim
576
asked Aug 2 at 18:57
Lucas Amorim
576
asked Aug 2 at 18:57
Lucas Amorim
576
576
Maybe not because the area of an ellipse has closed form $pi a b$ while the perimeter doesn't? I'm not used to elliptic integrals enough to be definitive on this guess. But it this all there is to it? This end is somehow disappointing.
– Lucas Amorim
Aug 2 at 19:12
add a comment |Â
Maybe not because the area of an ellipse has closed form $pi a b$ while the perimeter doesn't? I'm not used to elliptic integrals enough to be definitive on this guess. But it this all there is to it? This end is somehow disappointing.
– Lucas Amorim
Aug 2 at 19:12
Maybe not because the area of an ellipse has closed form $pi a b$ while the perimeter doesn't? I'm not used to elliptic integrals enough to be definitive on this guess. But it this all there is to it? This end is somehow disappointing.
– Lucas Amorim
Aug 2 at 19:12
Maybe not because the area of an ellipse has closed form $pi a b$ while the perimeter doesn't? I'm not used to elliptic integrals enough to be definitive on this guess. But it this all there is to it? This end is somehow disappointing.
– Lucas Amorim
Aug 2 at 19:12
add a comment |Â
1 Answer
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Part 1
As the OP has already said, the fact that the derivative of the volume of a circle is equal to its area, also generalises to spheres.
One way of
understanding why this is true, is to consider that when the volume of a sphere increases, it is like adding infinitely many thin (flat) shells to the outside of the sphere, each with an area of $4Àr^2$.
This line of reasoning can actually be used to to show why this is true for higher dimensional $n$-spheres!
What is less known, is that it also works for all regular polygons (squares, regular pentagons, etc). In such cases, you need to consider that the equivalent of the radius is the distance from the centre of the polygon to the midpoint of each side.
Combining both these properties, it is also true for cubes and hypercubes.
Part 2
Unfortunately, as the OP has stated, it does not hold for ellipses. The area of an ellipse is $pi a b$, but the circumference is given in terms of (surprise, surprise!) elliptic functions.
An excellent and very readable source for exact and approximate expressions for the circumference of the ellipse can be found at, 'Perimeter of ellipse".
One of my favourites for both elegance and amazing accuracy is :
$$ P simeq pi left( 3(a+b) - sqrt3(a+b)(a+3b) right) $$
answered Aug 2 at 23:35
Martin Roberts
1,194318
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Part 1
As the OP has already said, the fact that the derivative of the volume of a circle is equal to its area, also generalises to spheres.
One way of
understanding why this is true, is to consider that when the volume of a sphere increases, it is like adding infinitely many thin (flat) shells to the outside of the sphere, each with an area of $4Àr^2$.
This line of reasoning can actually be used to to show why this is true for higher dimensional $n$-spheres!
What is less known, is that it also works for all regular polygons (squares, regular pentagons, etc). In such cases, you need to consider that the equivalent of the radius is the distance from the centre of the polygon to the midpoint of each side.
Combining both these properties, it is also true for cubes and hypercubes.
Part 2
Unfortunately, as the OP has stated, it does not hold for ellipses. The area of an ellipse is $pi a b$, but the circumference is given in terms of (surprise, surprise!) elliptic functions.
An excellent and very readable source for exact and approximate expressions for the circumference of the ellipse can be found at, 'Perimeter of ellipse".
One of my favourites for both elegance and amazing accuracy is :
$$ P simeq pi left( 3(a+b) - sqrt3(a+b)(a+3b) right) $$
answered Aug 2 at 23:35
Martin Roberts
1,194318
add a comment |Â
up vote
0
down vote
Part 1
As the OP has already said, the fact that the derivative of the volume of a circle is equal to its area, also generalises to spheres.
One way of
understanding why this is true, is to consider that when the volume of a sphere increases, it is like adding infinitely many thin (flat) shells to the outside of the sphere, each with an area of $4Àr^2$.
This line of reasoning can actually be used to to show why this is true for higher dimensional $n$-spheres!
What is less known, is that it also works for all regular polygons (squares, regular pentagons, etc). In such cases, you need to consider that the equivalent of the radius is the distance from the centre of the polygon to the midpoint of each side.
Combining both these properties, it is also true for cubes and hypercubes.
Part 2
Unfortunately, as the OP has stated, it does not hold for ellipses. The area of an ellipse is $pi a b$, but the circumference is given in terms of (surprise, surprise!) elliptic functions.
An excellent and very readable source for exact and approximate expressions for the circumference of the ellipse can be found at, 'Perimeter of ellipse".
One of my favourites for both elegance and amazing accuracy is :
$$ P simeq pi left( 3(a+b) - sqrt3(a+b)(a+3b) right) $$
answered Aug 2 at 23:35
Martin Roberts
1,194318
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Part 1
As the OP has already said, the fact that the derivative of the volume of a circle is equal to its area, also generalises to spheres.
One way of
understanding why this is true, is to consider that when the volume of a sphere increases, it is like adding infinitely many thin (flat) shells to the outside of the sphere, each with an area of $4Àr^2$.
This line of reasoning can actually be used to to show why this is true for higher dimensional $n$-spheres!
What is less known, is that it also works for all regular polygons (squares, regular pentagons, etc). In such cases, you need to consider that the equivalent of the radius is the distance from the centre of the polygon to the midpoint of each side.
Combining both these properties, it is also true for cubes and hypercubes.
Part 2
Unfortunately, as the OP has stated, it does not hold for ellipses. The area of an ellipse is $pi a b$, but the circumference is given in terms of (surprise, surprise!) elliptic functions.
An excellent and very readable source for exact and approximate expressions for the circumference of the ellipse can be found at, 'Perimeter of ellipse".
One of my favourites for both elegance and amazing accuracy is :
$$ P simeq pi left( 3(a+b) - sqrt3(a+b)(a+3b) right) $$
answered Aug 2 at 23:35
Martin Roberts
1,194318
Part 1
As the OP has already said, the fact that the derivative of the volume of a circle is equal to its area, also generalises to spheres.
One way of
understanding why this is true, is to consider that when the volume of a sphere increases, it is like adding infinitely many thin (flat) shells to the outside of the sphere, each with an area of $4Àr^2$.
This line of reasoning can actually be used to to show why this is true for higher dimensional $n$-spheres!
What is less known, is that it also works for all regular polygons (squares, regular pentagons, etc). In such cases, you need to consider that the equivalent of the radius is the distance from the centre of the polygon to the midpoint of each side.
Combining both these properties, it is also true for cubes and hypercubes.
Part 2
Unfortunately, as the OP has stated, it does not hold for ellipses. The area of an ellipse is $pi a b$, but the circumference is given in terms of (surprise, surprise!) elliptic functions.
An excellent and very readable source for exact and approximate expressions for the circumference of the ellipse can be found at, 'Perimeter of ellipse".
One of my favourites for both elegance and amazing accuracy is :
$$ P simeq pi left( 3(a+b) - sqrt3(a+b)(a+3b) right) $$
answered Aug 2 at 23:35
Martin Roberts
1,194318
edited Aug 2 at 23:43
answered Aug 2 at 23:35
Martin Roberts
1,194318
answered Aug 2 at 23:35
Martin Roberts
1,194318
answered Aug 2 at 23:35
Martin Roberts
1,194318
1,194318
add a comment |Â
add a comment |Â
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Maybe not because the area of an ellipse has closed form $pi a b$ while the perimeter doesn't? I'm not used to elliptic integrals enough to be definitive on this guess. But it this all there is to it? This end is somehow disappointing.
– Lucas Amorim
Aug 2 at 19:12