Mathematiques d'Escalier
Nov. 20th, 2012 10:45 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
jackMany people have observed that, given the volume of a sphere radius r is V(r)=(4/3).pi.r3 and its surface area is S(r)=4.pi.r2, that, very conveniently:
dV(r)/dr = 3.(4/3).pi.r2 = 4.pi.r2 = S(r)
In fact, I use it to remember the surface area, since I only usually remember the volume. It seems to make sense sort of, but many people are not quite sure why.
Today someone pointed out, that the same thing works for a cube, if you take the side length as the diameter and half the side length as the radius.
V = (2r)3
S = 6.(2r)2
dV/dr = 2.3.(2r)2 = 6.(2r)2 = S
In fact, if you do the trigonometry, it turns out the same thing works for the tetrahedron and other platonic solids, although if you take the side length as d, r is no longer half d.
In fact, if you take d = ar for some unknown constant a, so long as the volume is something times r cubed, and the surface area is something times r squared, there's always some value of a that makes the derivative dV/dr = S exact.
It may or may not be immediately obvious what that value is, but in fact, it's the distance from the centre to the middle of one of the faces aka the radius of the largest sphere which fits inside.
At this point we were puzzled why, and it wasn't until I was at home that I saw the obvious way of thinking about it.
What does dV/dr mean? It means [V(r+δr)-V(r)]/δr (as δr->0).
That is, "imagine a solid with a slightly larger r, and subtract the original r" and ask what's left. What's left is thin slab covering each face, plus some neglible rods at each edge which have an extra factor of δr in so effectively vanish to zero. What's the volume of all those thin slabs? The areas of the faces, times the width of the slab. What's the width? The distance from one side to the other perpendicular to the face, ie. parallel with a line through the centre only at the centre of the face, ie. it is δr, so the volume of the slab is S.δr, and [V(r+δr)-V(r)]/δr is approximately S.
The same diagram apparently works for a sphere, if you imagine δV to be a thin shell covering the original sphere. What's the volume of the shell? Well, it's approximately "surface area times width", but the inner or outer surface area? Well, one's too small and one's too big, so the goldilocks answer is somewhere between S(r).δr <= δV <= S(r+δr).δr. But all the extra terms in the second one all have δr^2 in so they all vanish, and δV ~ S(r).δr.
I think I've probably seen that before but forgotten.
dV(r)/dr = 3.(4/3).pi.r2 = 4.pi.r2 = S(r)
In fact, I use it to remember the surface area, since I only usually remember the volume. It seems to make sense sort of, but many people are not quite sure why.
Today someone pointed out, that the same thing works for a cube, if you take the side length as the diameter and half the side length as the radius.
V = (2r)3
S = 6.(2r)2
dV/dr = 2.3.(2r)2 = 6.(2r)2 = S
In fact, if you do the trigonometry, it turns out the same thing works for the tetrahedron and other platonic solids, although if you take the side length as d, r is no longer half d.
In fact, if you take d = ar for some unknown constant a, so long as the volume is something times r cubed, and the surface area is something times r squared, there's always some value of a that makes the derivative dV/dr = S exact.
It may or may not be immediately obvious what that value is, but in fact, it's the distance from the centre to the middle of one of the faces aka the radius of the largest sphere which fits inside.
At this point we were puzzled why, and it wasn't until I was at home that I saw the obvious way of thinking about it.
What does dV/dr mean? It means [V(r+δr)-V(r)]/δr (as δr->0).
That is, "imagine a solid with a slightly larger r, and subtract the original r" and ask what's left. What's left is thin slab covering each face, plus some neglible rods at each edge which have an extra factor of δr in so effectively vanish to zero. What's the volume of all those thin slabs? The areas of the faces, times the width of the slab. What's the width? The distance from one side to the other perpendicular to the face, ie. parallel with a line through the centre only at the centre of the face, ie. it is δr, so the volume of the slab is S.δr, and [V(r+δr)-V(r)]/δr is approximately S.
The same diagram apparently works for a sphere, if you imagine δV to be a thin shell covering the original sphere. What's the volume of the shell? Well, it's approximately "surface area times width", but the inner or outer surface area? Well, one's too small and one's too big, so the goldilocks answer is somewhere between S(r).δr <= δV <= S(r+δr).δr. But all the extra terms in the second one all have δr^2 in so they all vanish, and δV ~ S(r).δr.
I think I've probably seen that before but forgotten.
no subject
Date: 2012-11-21 09:23 am (UTC)It confirms it, but not completely obviously. Suppose the cap is centred on the north pole of a unit sphere (discarding my previous measurements), and 0 < y < 1 is the distance from the north pole to the plane cutting off the cap. Let x be the radius of the circle which is the cross-section of the sphere at the point where the plane cuts it. I'm claiming that the surface area A of the cap is a fraction y/2 of the whole surface area 4π, that is, A = 2πy. Whereas Archimedes is saying that A = π(x2+y2).
But the two are equivalent, it turns out, because x and y are related by applying Pythagoras to the radius of the sphere: x2 + (1-y)2 = 1, which expands out to give x2+y2 = 2y.
In other words, I agree with Archimedes even though we've chosen to visualise and phrase our claims very differently; but precisely because of that very different visualisation, Archimedes' statement doesn't constitute an intuitive justification of my one :-)