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.