So I'm going to clarify the general idea behind your questions. What I see you asking is that if you have a family of shapes that grow in size, how are:
area vs perimeter (in 2D)
volume vs surface area (in 3D)
related through differentiation. More specifically, can you always find a way to measure the "size" of a shape such that
rate of change of area = (constant) * perimeter
rate of change of volume = (same constant) * surface area
Anyways, let me now answer your questions. If I'm misinterpreting what you are asking, let me know.
I'm going to straighten one thing out first. The hidden issue is that you need to ask what variable you are differentiating with respect to.
For the circle and sphere, you use the radius r.
Circle: A = πr^2 -> A'(r) = 2πr = perimeter
Sphere: V = 4/3πr^3 -> V'(r) = 4πr^2 = surface area.
For the square and cube, you are using the edge length a:
Square: A = a^2 -> A'(a) = 2a = 1/2 * perimeter
Cube: V = a^3 -> V'(a) = 3a^2 = 1/2 * surface area.
So the mismatch (what x is) is coming purely from the choice of parameter. You can define the square and cube using a "radius" from the center (the inradius), as such:
Square: If the square is centered at the origin, let r be the distance from the center to a side. So the square goes from x = -r to x = r, and the same for y. Thus the side length is 2r. Then the area = (2r)^2 = 4r^2. And the perimeter is 4(2r) = 8r. You'll notice that if you differentiate area, you get dA / dr = 8r = perimeter. This matches the circle perfectly (x = 1).
Cube: This uses the same idea. Let r be the distance from the center to a face. So the side length is 2r. Then the volume is (2r)^3 = 8r^3. The surface area is 6(2r)^2 = 24r^2. If you differentiate volume, you get: dV / dr = 24r^2 = surface area. Again, this is a perfect match (x = 1).
For some reason, people think defining these shapes using the side length is easier (and I guess it is), but this should showcase my point about this mismatch coming from parameterization.
The general principle, which should answer most of your question, is that if you scale any shape uniformly by a factor t, the
To see why intuitively, let's look at 2D. Suppose you scale a shape by t. Then every length becomes t * bigger. Area is made of length x length, so
For example, if you have a square with side 1, then the area is 1. If you scale by 3, then the side is 3, and the area becomes 9 = 3^2.
In 2D, volume is length * length * length, so we get volume -> t^3.
Again, we can do an example. If you have a cube with side 1, then the volume is 1. If you scale by 3, then the side is 3, and the volume = 27 = 3^3.
This always works. To see why intuitively, note that any shape can be approximated by tiny boxes. Each box scales in all directions. So the total area / volume picks up these powers.
More formally, take a given shape, and scale it up or down uniformly. What I am essentially saying is that if you have a parameter t, where t = 1 is the original shape, t = 2 is twice as big, t = 0.5 is half as big, and so on, then
where C_2 is the original area (for a 2D shape), and C_3 is the original volume (for a 3D shape).
Note that everything scales proportionally. This works regardless of the shape. If you start with a square of side length 1, then the area = 1. If you scale by t, then the side length is t and the area is t^2, so here, C_2 = 1. If you start with a circle of radius 1, then the area is π. If you scale by t, then the radius is t, and the area is πt^2. So C_2 = π.
If you then differentiate:
A'(t) = 2 * C_2 * t = 2 * A(t) / t
V'(t) = 3 * C_3 * t^2 = 3 * V(t) / t
So A'(t) is proportional to A(t) by a factor of 2 / t, and V'(t) is proportional to V(t) by a factor of 3 / t.
There's an important fact we can use here:
If t is the inradius (distance from center to side length), then
V = 1/3 * surface area * t
To see why this is true (for 2D) think of a polygon. Break it into triangles from the center. Each triangle has a height of t (the inradius), and the base is the side of the polygon. So:
Area of one triangle = 1/2 * side * t
Add them all up: A = 1/2 * perimeter * t
So plugging in A'(t) = perimeter and V'(t) = surface area.
For 3D, it's the same idea. Break the solid into pyramids from the center. Each has a height of t, and a base is the face. So
Volume of a pyramid = 1/3 * face * t
Add them up: V = 1/3 * surface area * t
This works for shapes that admit a common inradius decomposition (e.g., regular polygons, spheres, and certain symmetric convex solids).
Now plug this into the derivative formula. We had
Now substitute A = 1/2 * perimeter * t
It's the same idea in 3D. We had:
Substitute V = 1/3 * surface area * t
So using the inradius makes your x = 1.
I just showed that if t = inradius, then x = 1. If we instead used something like side length, then we would get something different.
We'll use a square as an example.
Let t = a (side length). Then area = a^2, derivative of area with respect to side length is 2a, and the perimeter is 4a. So x = 1/2.
What happened was that you changed how you measure size. That rescales the derivative, which changes x.
If t = side length (square case) -> x = 1/2
If t = circumradius -> another constant depending on n
For a fixed geometric parameter, x is determined by the shape. But by reparameterizing the scaling variable, you can make x any positive constant.
So to answer your main question about always creating a 3D surface where dV / dt = x * (surface area), yes you can.
For any family of similar shapes, dV / dt = 3 / t V = (constant depending on parameter) * surface area. And if you choose inradius, dV / dt = surface area.
You did ask explicitly about what parameter you should use for an n-gon. There's three natural choices that I see. If you use the side length, it's simple, but it'll give weird constants for x.
You could also use the circumradius (like the inradius, but goes from the center to the corner), which matches circle nicely as n -> ∞, but I would recommend the inradius since that makes everything unify cleanly, namely:
A = 1/2 * (perimeter) * (inradius)
directly and again makes x = 1.
Building a 3D Analogue of Regular N-gons
Now your more interesting structural question, namely "What does the series of 3D surfaces corresponding to the series of n-gons look like?"
In 2D, a regular n-gon approaches a circle as n -> ∞.
However, in 3D, there is no infinite family of "regular polyhedra" (there are only 5 platonic solids). In other words, there isn't a clear analogue. However, you can use prisms, pyramids, or surfaces of revolutions to make a series of 3D shapes where they contain at least 1 n-gon (just not every face being an n-gon).
So one sequence could be a triangular prism, then rectangular prism, then pentagonal prism, and so on.
Another sequence could be a tetrahedron, a square-based pyramid, a pentagon-based pyramid, a hexagon-based pyramid, and so on.
But you want a sequence that includes a cube and converges to a sphere. This is possible, but you need to be careful with how you construct it.
One clean method is to start with a cube, then repeatedly subdivide each square face into smaller squares. Then "push" the new vertices outward so they lie on a sphere. This will give a cube, then a refined cube, then a more refined cube, and so on, and then it will converge to a sphere.
Now for your question about everything having straight edges. You can (I just did with the refined square example above), but with an important caveat.
You can build polyhedra with only flat faces and approximate a sphere arbitrarily well. However, exact equality V'(t) = surface area still depends on your parameter choice. Smoothness is not required for the scaling law to work.
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