Connection between the circumference/area of circles, and between the volume/surface area of spheres?...
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This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
Consider the formula for the area of a circle and the formula for its circumference. If one differentiates the formula of the area with respect to $r$ (the radius), the formula for the circle's circumference pops out.
The same applies to a sphere with it's volume and surface area: differentiate the formula for volume with respect to $r$, and you obtain the formula for surface area.
I have two questions about this phenomenon:
- Is this an unique property of the circle and sphere?
- Is there mathematical reason for this?
calculus geometry
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marked as duplicate by grand_chat, Arthur, Ethan Bolker, zipirovich, RRL Dec 27 '18 at 5:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
Consider the formula for the area of a circle and the formula for its circumference. If one differentiates the formula of the area with respect to $r$ (the radius), the formula for the circle's circumference pops out.
The same applies to a sphere with it's volume and surface area: differentiate the formula for volume with respect to $r$, and you obtain the formula for surface area.
I have two questions about this phenomenon:
- Is this an unique property of the circle and sphere?
- Is there mathematical reason for this?
calculus geometry
$endgroup$
marked as duplicate by grand_chat, Arthur, Ethan Bolker, zipirovich, RRL Dec 27 '18 at 5:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Thank you very much. This answers question no. 2.
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– Bulldocarx
Dec 26 '18 at 23:54
add a comment |
$begingroup$
This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
Consider the formula for the area of a circle and the formula for its circumference. If one differentiates the formula of the area with respect to $r$ (the radius), the formula for the circle's circumference pops out.
The same applies to a sphere with it's volume and surface area: differentiate the formula for volume with respect to $r$, and you obtain the formula for surface area.
I have two questions about this phenomenon:
- Is this an unique property of the circle and sphere?
- Is there mathematical reason for this?
calculus geometry
$endgroup$
This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
Consider the formula for the area of a circle and the formula for its circumference. If one differentiates the formula of the area with respect to $r$ (the radius), the formula for the circle's circumference pops out.
The same applies to a sphere with it's volume and surface area: differentiate the formula for volume with respect to $r$, and you obtain the formula for surface area.
I have two questions about this phenomenon:
- Is this an unique property of the circle and sphere?
- Is there mathematical reason for this?
This question already has an answer here:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
8 answers
calculus geometry
calculus geometry
edited Dec 27 '18 at 1:16
Eevee Trainer
6,0071936
6,0071936
asked Dec 26 '18 at 23:52
BulldocarxBulldocarx
104
104
marked as duplicate by grand_chat, Arthur, Ethan Bolker, zipirovich, RRL Dec 27 '18 at 5:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by grand_chat, Arthur, Ethan Bolker, zipirovich, RRL Dec 27 '18 at 5:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Thank you very much. This answers question no. 2.
$endgroup$
– Bulldocarx
Dec 26 '18 at 23:54
add a comment |
$begingroup$
Thank you very much. This answers question no. 2.
$endgroup$
– Bulldocarx
Dec 26 '18 at 23:54
$begingroup$
Thank you very much. This answers question no. 2.
$endgroup$
– Bulldocarx
Dec 26 '18 at 23:54
$begingroup$
Thank you very much. This answers question no. 2.
$endgroup$
– Bulldocarx
Dec 26 '18 at 23:54
add a comment |
1 Answer
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I'll begin by answering your questions in somewhat of a reverse order (though depending on your philosophical bend, you could take this first half of my answer as an answer to both). I take the core content from Wolfram MathWorld and Wikipedia.
A foreword, I'm generalizing this to the $n$-dimensional case, to show that this holds for spheres of all dimension.
We will let $V_n$ denote the $n$-dimensional analogue of volume, which the Wolfram article calls "content." To this, there is the analogue of surface area, which we'll call hyper-surface area, and denote $S_n$.
It can be shown that $S_n$ is given by
$$S_n = frac{2pi^{n/2}}{Gamma left(frac n2 right)} R^{n-1}$$
and that $V_n$ is given by
$$V_n = frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n$$
Okay, But Where Did The Formulas Come From?: It's only fair to wonder about where these formulas come from, instead of just taking me at my word. I'll link to some resources for the derivations; the explanations are a bit long for this post and may be above your head OP, assuming you're in an introductory calculus class as I suspect.
A derivation of the formula for volume can be found here. Dr. Peyam on YouTube did a derivation of the surface area formula, which can be found here, and includes a similar level of content (but at least more explanation). (He also touches on the volume a bit as well if you want a different explanation.)
If you're unfamiliar with the notation in the formulas above, $Gamma(n)$ is the gamma function, and is just a generalization of the notion of factorial to non-integers. It can be given by an integral, which might be a bit beyond the scope of this discussion. The relation for integers $n$ between $Gamma(n)$ and the factorial is
$$Gamma(n) = (n-1)!$$
For example, $Gamma(2) = (2-1)! = 1! = 1$. (The gamma function also has the property I see not used as well as it could be in the various links that $Gamma(n+1) = nGamma(n)$. This is essentially the recursion of the factorial, i.e. $n! = ncdot (n-1)!$.)
To convince yourself of these formulas, try a few $n$ individually: let $n=2$ to find $V_n$ (area of a circle) and $S_n$ (its circumference), for example.
In any event, to see that the derivative of content yields hyper-surface area here, note:
$$frac{d}{dR} V_n = frac{d}{dR} left( frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n right) = n cdot frac{2pi^{n/2}}{nGamma left(frac n2 right)} cdot R^{n-1} = frac{2pi^{n/2}}{Gamma left(frac n2 right)} cdot R^{n-1} = S_n$$
What this hints at is that this is a property of the $n$-dimensional sphere, i.e. that a property of $n$-spheres is precisely that their "content", differentiated, yields their hyper-surface area.
This presumably answers your second question regarding "is there a mathematical reason for this fact," that being it is a property of the $n$-sphere.
As for your first question, this is noted in several different ways in the question linked as a duplicate. I favor the answer by Helmer.Aslaksen, which cites a paper which can be found here:.
The bit essential to your first question is that, no, this is not a property unique to the sphere. For any shape with area able to be written as $A(s) = cs^2$ for some constant $c$ and parameter $s$ (such as side length or radius), and $P(s) = ks$ denoting the perimeter about it, a parameterization of $x = 2cs/k$ will let us have
$$frac{d}{dx} A(x) = P(x)$$
Whether this might hold in higher dimensions, I'm uncertain. At least in $2D$ space though, it holds for shapes such as squares and ellipses.
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'll begin by answering your questions in somewhat of a reverse order (though depending on your philosophical bend, you could take this first half of my answer as an answer to both). I take the core content from Wolfram MathWorld and Wikipedia.
A foreword, I'm generalizing this to the $n$-dimensional case, to show that this holds for spheres of all dimension.
We will let $V_n$ denote the $n$-dimensional analogue of volume, which the Wolfram article calls "content." To this, there is the analogue of surface area, which we'll call hyper-surface area, and denote $S_n$.
It can be shown that $S_n$ is given by
$$S_n = frac{2pi^{n/2}}{Gamma left(frac n2 right)} R^{n-1}$$
and that $V_n$ is given by
$$V_n = frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n$$
Okay, But Where Did The Formulas Come From?: It's only fair to wonder about where these formulas come from, instead of just taking me at my word. I'll link to some resources for the derivations; the explanations are a bit long for this post and may be above your head OP, assuming you're in an introductory calculus class as I suspect.
A derivation of the formula for volume can be found here. Dr. Peyam on YouTube did a derivation of the surface area formula, which can be found here, and includes a similar level of content (but at least more explanation). (He also touches on the volume a bit as well if you want a different explanation.)
If you're unfamiliar with the notation in the formulas above, $Gamma(n)$ is the gamma function, and is just a generalization of the notion of factorial to non-integers. It can be given by an integral, which might be a bit beyond the scope of this discussion. The relation for integers $n$ between $Gamma(n)$ and the factorial is
$$Gamma(n) = (n-1)!$$
For example, $Gamma(2) = (2-1)! = 1! = 1$. (The gamma function also has the property I see not used as well as it could be in the various links that $Gamma(n+1) = nGamma(n)$. This is essentially the recursion of the factorial, i.e. $n! = ncdot (n-1)!$.)
To convince yourself of these formulas, try a few $n$ individually: let $n=2$ to find $V_n$ (area of a circle) and $S_n$ (its circumference), for example.
In any event, to see that the derivative of content yields hyper-surface area here, note:
$$frac{d}{dR} V_n = frac{d}{dR} left( frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n right) = n cdot frac{2pi^{n/2}}{nGamma left(frac n2 right)} cdot R^{n-1} = frac{2pi^{n/2}}{Gamma left(frac n2 right)} cdot R^{n-1} = S_n$$
What this hints at is that this is a property of the $n$-dimensional sphere, i.e. that a property of $n$-spheres is precisely that their "content", differentiated, yields their hyper-surface area.
This presumably answers your second question regarding "is there a mathematical reason for this fact," that being it is a property of the $n$-sphere.
As for your first question, this is noted in several different ways in the question linked as a duplicate. I favor the answer by Helmer.Aslaksen, which cites a paper which can be found here:.
The bit essential to your first question is that, no, this is not a property unique to the sphere. For any shape with area able to be written as $A(s) = cs^2$ for some constant $c$ and parameter $s$ (such as side length or radius), and $P(s) = ks$ denoting the perimeter about it, a parameterization of $x = 2cs/k$ will let us have
$$frac{d}{dx} A(x) = P(x)$$
Whether this might hold in higher dimensions, I'm uncertain. At least in $2D$ space though, it holds for shapes such as squares and ellipses.
$endgroup$
add a comment |
$begingroup$
I'll begin by answering your questions in somewhat of a reverse order (though depending on your philosophical bend, you could take this first half of my answer as an answer to both). I take the core content from Wolfram MathWorld and Wikipedia.
A foreword, I'm generalizing this to the $n$-dimensional case, to show that this holds for spheres of all dimension.
We will let $V_n$ denote the $n$-dimensional analogue of volume, which the Wolfram article calls "content." To this, there is the analogue of surface area, which we'll call hyper-surface area, and denote $S_n$.
It can be shown that $S_n$ is given by
$$S_n = frac{2pi^{n/2}}{Gamma left(frac n2 right)} R^{n-1}$$
and that $V_n$ is given by
$$V_n = frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n$$
Okay, But Where Did The Formulas Come From?: It's only fair to wonder about where these formulas come from, instead of just taking me at my word. I'll link to some resources for the derivations; the explanations are a bit long for this post and may be above your head OP, assuming you're in an introductory calculus class as I suspect.
A derivation of the formula for volume can be found here. Dr. Peyam on YouTube did a derivation of the surface area formula, which can be found here, and includes a similar level of content (but at least more explanation). (He also touches on the volume a bit as well if you want a different explanation.)
If you're unfamiliar with the notation in the formulas above, $Gamma(n)$ is the gamma function, and is just a generalization of the notion of factorial to non-integers. It can be given by an integral, which might be a bit beyond the scope of this discussion. The relation for integers $n$ between $Gamma(n)$ and the factorial is
$$Gamma(n) = (n-1)!$$
For example, $Gamma(2) = (2-1)! = 1! = 1$. (The gamma function also has the property I see not used as well as it could be in the various links that $Gamma(n+1) = nGamma(n)$. This is essentially the recursion of the factorial, i.e. $n! = ncdot (n-1)!$.)
To convince yourself of these formulas, try a few $n$ individually: let $n=2$ to find $V_n$ (area of a circle) and $S_n$ (its circumference), for example.
In any event, to see that the derivative of content yields hyper-surface area here, note:
$$frac{d}{dR} V_n = frac{d}{dR} left( frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n right) = n cdot frac{2pi^{n/2}}{nGamma left(frac n2 right)} cdot R^{n-1} = frac{2pi^{n/2}}{Gamma left(frac n2 right)} cdot R^{n-1} = S_n$$
What this hints at is that this is a property of the $n$-dimensional sphere, i.e. that a property of $n$-spheres is precisely that their "content", differentiated, yields their hyper-surface area.
This presumably answers your second question regarding "is there a mathematical reason for this fact," that being it is a property of the $n$-sphere.
As for your first question, this is noted in several different ways in the question linked as a duplicate. I favor the answer by Helmer.Aslaksen, which cites a paper which can be found here:.
The bit essential to your first question is that, no, this is not a property unique to the sphere. For any shape with area able to be written as $A(s) = cs^2$ for some constant $c$ and parameter $s$ (such as side length or radius), and $P(s) = ks$ denoting the perimeter about it, a parameterization of $x = 2cs/k$ will let us have
$$frac{d}{dx} A(x) = P(x)$$
Whether this might hold in higher dimensions, I'm uncertain. At least in $2D$ space though, it holds for shapes such as squares and ellipses.
$endgroup$
add a comment |
$begingroup$
I'll begin by answering your questions in somewhat of a reverse order (though depending on your philosophical bend, you could take this first half of my answer as an answer to both). I take the core content from Wolfram MathWorld and Wikipedia.
A foreword, I'm generalizing this to the $n$-dimensional case, to show that this holds for spheres of all dimension.
We will let $V_n$ denote the $n$-dimensional analogue of volume, which the Wolfram article calls "content." To this, there is the analogue of surface area, which we'll call hyper-surface area, and denote $S_n$.
It can be shown that $S_n$ is given by
$$S_n = frac{2pi^{n/2}}{Gamma left(frac n2 right)} R^{n-1}$$
and that $V_n$ is given by
$$V_n = frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n$$
Okay, But Where Did The Formulas Come From?: It's only fair to wonder about where these formulas come from, instead of just taking me at my word. I'll link to some resources for the derivations; the explanations are a bit long for this post and may be above your head OP, assuming you're in an introductory calculus class as I suspect.
A derivation of the formula for volume can be found here. Dr. Peyam on YouTube did a derivation of the surface area formula, which can be found here, and includes a similar level of content (but at least more explanation). (He also touches on the volume a bit as well if you want a different explanation.)
If you're unfamiliar with the notation in the formulas above, $Gamma(n)$ is the gamma function, and is just a generalization of the notion of factorial to non-integers. It can be given by an integral, which might be a bit beyond the scope of this discussion. The relation for integers $n$ between $Gamma(n)$ and the factorial is
$$Gamma(n) = (n-1)!$$
For example, $Gamma(2) = (2-1)! = 1! = 1$. (The gamma function also has the property I see not used as well as it could be in the various links that $Gamma(n+1) = nGamma(n)$. This is essentially the recursion of the factorial, i.e. $n! = ncdot (n-1)!$.)
To convince yourself of these formulas, try a few $n$ individually: let $n=2$ to find $V_n$ (area of a circle) and $S_n$ (its circumference), for example.
In any event, to see that the derivative of content yields hyper-surface area here, note:
$$frac{d}{dR} V_n = frac{d}{dR} left( frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n right) = n cdot frac{2pi^{n/2}}{nGamma left(frac n2 right)} cdot R^{n-1} = frac{2pi^{n/2}}{Gamma left(frac n2 right)} cdot R^{n-1} = S_n$$
What this hints at is that this is a property of the $n$-dimensional sphere, i.e. that a property of $n$-spheres is precisely that their "content", differentiated, yields their hyper-surface area.
This presumably answers your second question regarding "is there a mathematical reason for this fact," that being it is a property of the $n$-sphere.
As for your first question, this is noted in several different ways in the question linked as a duplicate. I favor the answer by Helmer.Aslaksen, which cites a paper which can be found here:.
The bit essential to your first question is that, no, this is not a property unique to the sphere. For any shape with area able to be written as $A(s) = cs^2$ for some constant $c$ and parameter $s$ (such as side length or radius), and $P(s) = ks$ denoting the perimeter about it, a parameterization of $x = 2cs/k$ will let us have
$$frac{d}{dx} A(x) = P(x)$$
Whether this might hold in higher dimensions, I'm uncertain. At least in $2D$ space though, it holds for shapes such as squares and ellipses.
$endgroup$
I'll begin by answering your questions in somewhat of a reverse order (though depending on your philosophical bend, you could take this first half of my answer as an answer to both). I take the core content from Wolfram MathWorld and Wikipedia.
A foreword, I'm generalizing this to the $n$-dimensional case, to show that this holds for spheres of all dimension.
We will let $V_n$ denote the $n$-dimensional analogue of volume, which the Wolfram article calls "content." To this, there is the analogue of surface area, which we'll call hyper-surface area, and denote $S_n$.
It can be shown that $S_n$ is given by
$$S_n = frac{2pi^{n/2}}{Gamma left(frac n2 right)} R^{n-1}$$
and that $V_n$ is given by
$$V_n = frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n$$
Okay, But Where Did The Formulas Come From?: It's only fair to wonder about where these formulas come from, instead of just taking me at my word. I'll link to some resources for the derivations; the explanations are a bit long for this post and may be above your head OP, assuming you're in an introductory calculus class as I suspect.
A derivation of the formula for volume can be found here. Dr. Peyam on YouTube did a derivation of the surface area formula, which can be found here, and includes a similar level of content (but at least more explanation). (He also touches on the volume a bit as well if you want a different explanation.)
If you're unfamiliar with the notation in the formulas above, $Gamma(n)$ is the gamma function, and is just a generalization of the notion of factorial to non-integers. It can be given by an integral, which might be a bit beyond the scope of this discussion. The relation for integers $n$ between $Gamma(n)$ and the factorial is
$$Gamma(n) = (n-1)!$$
For example, $Gamma(2) = (2-1)! = 1! = 1$. (The gamma function also has the property I see not used as well as it could be in the various links that $Gamma(n+1) = nGamma(n)$. This is essentially the recursion of the factorial, i.e. $n! = ncdot (n-1)!$.)
To convince yourself of these formulas, try a few $n$ individually: let $n=2$ to find $V_n$ (area of a circle) and $S_n$ (its circumference), for example.
In any event, to see that the derivative of content yields hyper-surface area here, note:
$$frac{d}{dR} V_n = frac{d}{dR} left( frac{2pi^{n/2}}{nGamma left(frac n2 right)} R^n right) = n cdot frac{2pi^{n/2}}{nGamma left(frac n2 right)} cdot R^{n-1} = frac{2pi^{n/2}}{Gamma left(frac n2 right)} cdot R^{n-1} = S_n$$
What this hints at is that this is a property of the $n$-dimensional sphere, i.e. that a property of $n$-spheres is precisely that their "content", differentiated, yields their hyper-surface area.
This presumably answers your second question regarding "is there a mathematical reason for this fact," that being it is a property of the $n$-sphere.
As for your first question, this is noted in several different ways in the question linked as a duplicate. I favor the answer by Helmer.Aslaksen, which cites a paper which can be found here:.
The bit essential to your first question is that, no, this is not a property unique to the sphere. For any shape with area able to be written as $A(s) = cs^2$ for some constant $c$ and parameter $s$ (such as side length or radius), and $P(s) = ks$ denoting the perimeter about it, a parameterization of $x = 2cs/k$ will let us have
$$frac{d}{dx} A(x) = P(x)$$
Whether this might hold in higher dimensions, I'm uncertain. At least in $2D$ space though, it holds for shapes such as squares and ellipses.
answered Dec 27 '18 at 1:10
Eevee TrainerEevee Trainer
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Thank you very much. This answers question no. 2.
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– Bulldocarx
Dec 26 '18 at 23:54