Why are the derivatives of a circle's area its perimeter, why the deriv. of a spheres vol. is its surface area

Question

also why for the squares and cubes the ratio of the derivative of square's area to its perimeter is 1:2; and for cubes volume to surface area is 1:2

Answer 1

Let's say that some solid has a volume which is a function of some length somewhere in it, so that we have volume = V(x). Let's suppose further that an increase in x by ∆x, so that we have volume V(x + ∆x). The difference between the volumes divided by ∆x is (V(x + ∆x) - V(x)) / ∆x, or dV/dx. But if the surface area S is multiplied by ∆x to give a volume, and if it's the same as V(x + ∆x) - V(x)), then we know that S = dV/dx. For example, d(4/3 π x³) / dx = 4 π x². Or d(8 x³) / dx = 24 x². Did I just now pull a fast one? No, because this is true only if ∆x is the thickness of successive increases of volume V(x + ∆x). For a sphere, that means x is the radius, and for a cube, it means the distance from the center to a face. That's why you wondered why the "ratio is 2:1 for cubes", because you were using the "diameter" of the cube, and not the "radius". There aren't too many other solids that have this property where ∆x is the thickness of successive increases in volume, but when this condition is met, then the area is dV/dx.

All regular Platonic solids, for example, should have this property.

Addendum: For the radius r of the inscribed sphere, the volume V(r) and surface area S(r) of a dodecahedron are:

V(r) = ( 2000(15 + 7√5) / (250 + 110√5)^(3/2) ) r³
S(r) = ( (1200(√(25 + 10√5)) / (250 + 110√5) ) r²

and it works out that dV(r)/dr = S(r).

Answer 2

I suppose the circle you are thinking of is the lip of the cone. [Also supposed is the convention that said lip is the perimeter of the base and the 'top' is the viewable area when the cone is set base down.] If the cone were split and flattened to form a wedge of a circle, the lip perimeter is an arc of the circle. Its proportion to the full circle is the same as that of the area of the cone to that of the whole circle, the radius of that whole circle being the slant height of the cone. With h for r and Arc for perimeter, Arc / 2πh = Suface / πh². Surface = πh² * Arc / 2πh = (h/2) * Arc. If you have the slant height and the lip perimeter, it's an easy computation. When h = 2, the magnitude of perimeter and surface are the same. [The linear aspect of h crossing with the linear aspect of the perimeter take the measurement from linear to the squared units of area.]

Answer 3

The area of a circle is obtained by performing the following double integral:
A = ∫{0,2π} ∫{0,R} r dr dθ
= 2π∫{0,R} r dr
= 2π∫{0,R} d(r^2 /2)

The perimeter is obtained from
S = ∫{0,2π} ∫{0,R} dr dθ
= 2π∫{0,R} dr

So it works out that both are similar functions
A = f(R^2 /2)
S = f(R)
It's only coincidence that dA/dR = S. It works out that way because f is a linear function.